WEBVTT 00:00:00.000 --> 00:00:01.976 align:middle line:90% [SQUEAKING] 00:00:01.976 --> 00:00:03.458 align:middle line:90% [RUSTLING] 00:00:03.458 --> 00:00:05.434 align:middle line:90% [CLICKING] 00:00:05.434 --> 00:00:13.350 align:middle line:90% 00:00:13.350 --> 00:00:17.190 align:middle line:84% CASEY RODRIGUEZ: OK, so let's continue with our discussion 00:00:17.190 --> 00:00:22.020 align:middle line:84% about continuous functions, no pun intended. 00:00:22.020 --> 00:00:23.715 align:middle line:84% First, let me recall a few facts. 00:00:23.715 --> 00:00:29.740 align:middle line:90% 00:00:29.740 --> 00:00:31.240 align:middle line:90% Well, this is not a definition. 00:00:31.240 --> 00:00:33.290 align:middle line:84% But this was a theorem we proved last time. 00:00:33.290 --> 00:00:35.590 align:middle line:84% So I won't recall the definition of continuity. 00:00:35.590 --> 00:00:39.130 align:middle line:84% But we proved this theorem characterizing continuity. 00:00:39.130 --> 00:00:42.430 align:middle line:90% 00:00:42.430 --> 00:00:48.670 align:middle line:84% And if we have a subset of R, an element in S, 00:00:48.670 --> 00:01:06.440 align:middle line:84% and a function f going from S to R, then f is continuous at C 00:01:06.440 --> 00:01:13.702 align:middle line:84% if and only if for every sequence x sub n, 00:01:13.702 --> 00:01:22.190 align:middle line:84% we have elements of S such that xn converges to C. 00:01:22.190 --> 00:01:28.600 align:middle line:84% We have limit as n goes to infinity 00:01:28.600 --> 00:01:34.000 align:middle line:84% if x sub n equals f of C. So this was 00:01:34.000 --> 00:01:35.890 align:middle line:90% a theorem we proved last time. 00:01:35.890 --> 00:01:39.550 align:middle line:84% And let me recall an older theorem, which I said 00:01:39.550 --> 00:01:42.850 align:middle line:84% was very powerful, but never really gave you 00:01:42.850 --> 00:01:48.970 align:middle line:84% any powerful application, which we'll 00:01:48.970 --> 00:01:51.610 align:middle line:84% start giving some powerful applications now. 00:01:51.610 --> 00:02:00.570 align:middle line:84% But this was the Bolzano-Weierstrass theorem, 00:02:00.570 --> 00:02:11.550 align:middle line:84% which states that every bounded sequence of real numbers 00:02:11.550 --> 00:02:14.340 align:middle line:90% has a convergent sub-sequence. 00:02:14.340 --> 00:02:27.770 align:middle line:90% 00:02:27.770 --> 00:02:31.070 align:middle line:84% And we're going to use these two theorems today 00:02:31.070 --> 00:02:36.470 align:middle line:84% and in the next lecture to prove that a continuous function 00:02:36.470 --> 00:02:41.240 align:middle line:84% on a closed and bounded interval is very well-behaved. 00:02:41.240 --> 00:02:53.870 align:middle line:84% So what we're going to show, let's call it the theme, 00:02:53.870 --> 00:03:15.880 align:middle line:84% is that so if f from a closed and bounded interval 00:03:15.880 --> 00:03:24.940 align:middle line:84% is continuous, meaning it's continuous at every point, 00:03:24.940 --> 00:03:34.640 align:middle line:90% then it is well-behaved. 00:03:34.640 --> 00:03:38.525 align:middle line:84% And for example, the first thing that we're going to prove, 00:03:38.525 --> 00:03:40.700 align:middle line:84% this will be a combination of two theorems, 00:03:40.700 --> 00:03:49.100 align:middle line:84% is that the image of a closed and bounded interval 00:03:49.100 --> 00:03:55.900 align:middle line:84% by a continuous function is another closed and bounded 00:03:55.900 --> 00:03:58.270 align:middle line:90% interval. 00:03:58.270 --> 00:04:01.220 align:middle line:84% And E could equal D. So this could just be a single point. 00:04:01.220 --> 00:04:03.280 align:middle line:84% For example, if f is a constant function, 00:04:03.280 --> 00:04:07.200 align:middle line:84% then the image would just be a single point. 00:04:07.200 --> 00:04:11.970 align:middle line:84% So this is a theme of this lecture and the next lecture. 00:04:11.970 --> 00:04:18.230 align:middle line:90% 00:04:18.230 --> 00:04:21.399 align:middle line:90% And so, let's start. 00:04:21.399 --> 00:04:24.280 align:middle line:84% Basically, we're going to prove two theorems. 00:04:24.280 --> 00:04:26.830 align:middle line:84% One is called the min-max theorem, 00:04:26.830 --> 00:04:31.360 align:middle line:84% and the other is called Bolzano's intermediate value 00:04:31.360 --> 00:04:32.170 align:middle line:90% theorem. 00:04:32.170 --> 00:04:37.480 align:middle line:84% The min-max theorem will tell you that f of an interval 00:04:37.480 --> 00:04:41.905 align:middle line:84% is always contained in an interval of this type. 00:04:41.905 --> 00:04:43.530 align:middle line:84% And then the intermediate value theorem 00:04:43.530 --> 00:04:47.880 align:middle line:84% will tell you everything in between two certain bounds 00:04:47.880 --> 00:04:49.020 align:middle line:90% is attained by f. 00:04:49.020 --> 00:04:57.810 align:middle line:84% So let's start off with showing that f of a closed and bounded 00:04:57.810 --> 00:04:59.052 align:middle line:90% interval is a bounded set. 00:04:59.052 --> 00:05:00.510 align:middle line:84% So let me just recall what it means 00:05:00.510 --> 00:05:04.080 align:middle line:90% for a function to be bounded. 00:05:04.080 --> 00:05:09.070 align:middle line:84% We say that a function from a subset S to R is bounded. 00:05:09.070 --> 00:05:12.810 align:middle line:90% So here, S is a subset of R-- 00:05:12.810 --> 00:05:18.090 align:middle line:84% is bounded if there exists a non-negative B such 00:05:18.090 --> 00:05:29.610 align:middle line:84% that for all x and S, f of x is bigger than or equal to B. 00:05:29.610 --> 00:05:41.660 align:middle line:84% So for example, if I take f of x to be 3x plus 1. 00:05:41.660 --> 00:05:50.180 align:middle line:84% And here, S is closed and bounded interval 0, 1, 00:05:50.180 --> 00:05:51.245 align:middle line:90% then f is bounded. 00:05:51.245 --> 00:05:58.320 align:middle line:90% 00:05:58.320 --> 00:06:00.510 align:middle line:84% So pictorially, what does this mean? 00:06:00.510 --> 00:06:02.960 align:middle line:90% Well, f of x equals 3x plus 1. 00:06:02.960 --> 00:06:08.410 align:middle line:90% So at 1, it's going to be 4. 00:06:08.410 --> 00:06:10.930 align:middle line:84% And pictorially, a function is bounded 00:06:10.930 --> 00:06:17.270 align:middle line:84% if the graph is always bounded between two real numbers. 00:06:17.270 --> 00:06:21.200 align:middle line:84% And for us, for f of x, it's always bounded between 0 and 4 00:06:21.200 --> 00:06:22.850 align:middle line:90% in terms of this absolute value. 00:06:22.850 --> 00:06:26.340 align:middle line:90% 00:06:26.340 --> 00:06:31.610 align:middle line:90% So f of x equals 3x plus 1. 00:06:31.610 --> 00:06:33.500 align:middle line:84% This is less than or equal to by the triangle 00:06:33.500 --> 00:06:37.100 align:middle line:84% inequality 3 times the absolute value of x plus 1. 00:06:37.100 --> 00:06:38.840 align:middle line:90% X is between 0 and 1. 00:06:38.840 --> 00:06:40.910 align:middle line:84% So it's absolute value is bounded by 1. 00:06:40.910 --> 00:06:45.660 align:middle line:90% 00:06:45.660 --> 00:06:46.845 align:middle line:90% So this function is bounded. 00:06:46.845 --> 00:06:51.160 align:middle line:90% 00:06:51.160 --> 00:06:57.430 align:middle line:84% And so, how about a function which is not bounded? 00:06:57.430 --> 00:06:59.470 align:middle line:90% What does that mean? 00:06:59.470 --> 00:07:03.040 align:middle line:84% Remember, whenever we see a definition, 00:07:03.040 --> 00:07:05.540 align:middle line:84% we should try to negate it to understand it better. 00:07:05.540 --> 00:07:21.310 align:middle line:84% So f is unbounded if for all B bigger than or equal to 0, 00:07:21.310 --> 00:07:31.520 align:middle line:84% there exists some bad x such that f of x in absolute value 00:07:31.520 --> 00:07:36.650 align:middle line:90% is bigger than or equal to B. 00:07:36.650 --> 00:07:39.410 align:middle line:84% So if a bounded function is supposed 00:07:39.410 --> 00:07:42.470 align:middle line:84% to have a graph bounded between two real numbers, 00:07:42.470 --> 00:07:45.710 align:middle line:84% an unbounded one is one in which at least part of the graph 00:07:45.710 --> 00:07:51.380 align:middle line:84% is going off to plus infinity or minus infinity. 00:07:51.380 --> 00:07:56.530 align:middle line:84% So most basic example we could think of 00:07:56.530 --> 00:08:02.200 align:middle line:84% is, let's say if we wanted to go from 0, 1 to R. Again, 00:08:02.200 --> 00:08:12.370 align:middle line:84% we could take f of x to be well, 0 at x equals 0, 1 over x 00:08:12.370 --> 00:08:17.060 align:middle line:90% if x is not equal to 0. 00:08:17.060 --> 00:08:29.460 align:middle line:84% So this function looks like 1, 1, 00:08:29.460 --> 00:08:35.010 align:middle line:84% and goes up to infinity as you approach x equals 0. 00:08:35.010 --> 00:08:36.360 align:middle line:90% So why is this unbounded? 00:08:36.360 --> 00:08:39.159 align:middle line:84% We have to verify the negation of the definition. 00:08:39.159 --> 00:08:43.830 align:middle line:90% So claim f is unbounded. 00:08:43.830 --> 00:08:47.380 align:middle line:90% 00:08:47.380 --> 00:08:49.060 align:middle line:84% Let B be bigger than or equal to 0. 00:08:49.060 --> 00:08:53.560 align:middle line:84% We now have to find an x in 0, 1, so that 1 over x-- 00:08:53.560 --> 00:09:01.000 align:middle line:84% or so that f of x is bigger than B. And this is pretty easy. 00:09:01.000 --> 00:09:24.230 align:middle line:84% So let x be in R such that 1 over x is bigger than B. 00:09:24.230 --> 00:09:28.940 align:middle line:84% So if B is equal to 0, we take x to be anything in 0, 1. 00:09:28.940 --> 00:09:36.000 align:middle line:90% That's fair enough-- i.e. 00:09:36.000 --> 00:09:40.920 align:middle line:84% x [? less than ?] 1 over B. So if we like, 00:09:40.920 --> 00:09:43.700 align:middle line:90% we could make this explicit. 00:09:43.700 --> 00:09:45.710 align:middle line:90% So I'll just keep it here. 00:09:45.710 --> 00:09:54.850 align:middle line:90% 00:09:54.850 --> 00:09:59.990 align:middle line:84% And so, now if we take x here and look at the graph, 00:09:59.990 --> 00:10:14.370 align:middle line:84% it's going to be bigger than B. And that's it. 00:10:14.370 --> 00:10:24.210 align:middle line:90% 00:10:24.210 --> 00:10:40.220 align:middle line:84% So several of these proofs will have the same flavor of what 00:10:40.220 --> 00:10:41.990 align:middle line:90% we're going to try and prove. 00:10:41.990 --> 00:10:47.690 align:middle line:90% 00:10:47.690 --> 00:10:51.150 align:middle line:84% And this one is a nice one to start off with, and necessary. 00:10:51.150 --> 00:10:56.690 align:middle line:84% So my first theorem that I want to prove 00:10:56.690 --> 00:11:00.260 align:middle line:84% about continuous functions on a bounded interval-- 00:11:00.260 --> 00:11:03.740 align:middle line:84% closed and bounded interval-- is that they're bounded. 00:11:03.740 --> 00:11:18.620 align:middle line:84% So if I have a function from a, b to R is continuous, 00:11:18.620 --> 00:11:20.780 align:middle line:84% meaning it's continuous at every point in a, b, 00:11:20.780 --> 00:11:26.945 align:middle line:90% then f is bounded. 00:11:26.945 --> 00:11:35.470 align:middle line:90% 00:11:35.470 --> 00:11:41.320 align:middle line:84% So the proof of this theorem, and several of these theorems, 00:11:41.320 --> 00:11:45.260 align:middle line:90% will be by contradiction. 00:11:45.260 --> 00:11:50.170 align:middle line:84% And the fact that we're on the closed and bounded interval 00:11:50.170 --> 00:11:52.120 align:middle line:90% is-- 00:11:52.120 --> 00:11:54.190 align:middle line:84% or that we're in this setting is what 00:11:54.190 --> 00:11:56.200 align:middle line:84% allows us to use Bolzano-Weierstrass, 00:11:56.200 --> 00:11:58.410 align:middle line:90% this powerful tool. 00:11:58.410 --> 00:12:13.950 align:middle line:84% So this is-- so assume and, so like I said, 00:12:13.950 --> 00:12:24.245 align:middle line:84% this proof is by contradiction, that the conclusion is false. 00:12:24.245 --> 00:12:33.850 align:middle line:90% 00:12:33.850 --> 00:12:38.120 align:middle line:84% Which the precise meaning is right here. 00:12:38.120 --> 00:12:41.170 align:middle line:84% For all b, there exists an x and S such that that. 00:12:41.170 --> 00:12:44.380 align:middle line:84% So for every b, I can find an x and S 00:12:44.380 --> 00:12:46.180 align:middle line:84% so that the absolute value of f of x 00:12:46.180 --> 00:12:48.930 align:middle line:90% is bigger than or equal to b. 00:12:48.930 --> 00:12:55.830 align:middle line:84% So I could take b to be N, for example, 00:12:55.830 --> 00:12:57.045 align:middle line:90% where N as a natural number. 00:12:57.045 --> 00:13:00.730 align:middle line:90% 00:13:00.730 --> 00:13:04.210 align:middle line:84% Then for all N, a natural number, 00:13:04.210 --> 00:13:12.400 align:middle line:84% so I'm taking b to be N, there exists x sub n in a, b, such 00:13:12.400 --> 00:13:17.710 align:middle line:84% that f of x sub n in absolute value 00:13:17.710 --> 00:13:18.970 align:middle line:90% is bigger than or equal to n. 00:13:18.970 --> 00:13:21.948 align:middle line:90% 00:13:21.948 --> 00:13:23.660 align:middle line:84% So now I have this sequence, which 00:13:23.660 --> 00:13:27.815 align:middle line:84% is bounded because it's in this closed and bounded interval. 00:13:27.815 --> 00:13:35.080 align:middle line:90% 00:13:35.080 --> 00:13:42.690 align:middle line:84% Then xn is bounded, because it's in this closed and bounded 00:13:42.690 --> 00:13:43.710 align:middle line:90% interval a, b. 00:13:43.710 --> 00:13:54.908 align:middle line:84% So by Bolzano-Weierstrass, there exists a subsequence 00:13:54.908 --> 00:14:08.310 align:middle line:84% x in sub k, of the sequence x sub n and x in R, such 00:14:08.310 --> 00:14:15.150 align:middle line:84% that limit as k goes to infinity of x of n sub k equals x. 00:14:15.150 --> 00:14:20.880 align:middle line:90% 00:14:20.880 --> 00:14:24.330 align:middle line:84% Now, I claim that this x is actually in the interval a, b. 00:14:24.330 --> 00:14:28.640 align:middle line:90% 00:14:28.640 --> 00:14:32.300 align:middle line:84% Since x of n sub k is a subsequence 00:14:32.300 --> 00:14:35.840 align:middle line:84% of x sub n, which are in a sub b, or a comma 00:14:35.840 --> 00:14:47.870 align:middle line:84% b, so I have for all k x sub n sub k is between a and b. 00:14:47.870 --> 00:14:50.930 align:middle line:84% This implies by what we know about limits that 00:14:50.930 --> 00:14:53.900 align:middle line:90% limits respect inequalities. 00:14:53.900 --> 00:15:01.890 align:middle line:84% And therefore, a is less than or equal to limit 00:15:01.890 --> 00:15:05.250 align:middle line:84% as k goes to infinity of x sub n sub k, which is less than 00:15:05.250 --> 00:15:05.970 align:middle line:90% or equal to b. 00:15:05.970 --> 00:15:07.980 align:middle line:90% And this is x. 00:15:07.980 --> 00:15:10.100 align:middle line:90% So i.e. 00:15:10.100 --> 00:15:13.550 align:middle line:90% 00:15:13.550 --> 00:15:15.070 align:middle line:90% x is between a and b. 00:15:15.070 --> 00:15:21.050 align:middle line:90% 00:15:21.050 --> 00:15:25.682 align:middle line:84% Another way, instead of going through this is, 00:15:25.682 --> 00:15:27.140 align:middle line:84% I believe we did in the assignment, 00:15:27.140 --> 00:15:32.670 align:middle line:84% that an interval of this form is closed. 00:15:32.670 --> 00:15:37.600 align:middle line:84% And you proved that for a closed set-- 00:15:37.600 --> 00:15:39.750 align:middle line:84% so when I say an interval of this form is closed, 00:15:39.750 --> 00:15:40.860 align:middle line:90% I mean it's a closed set. 00:15:40.860 --> 00:15:43.470 align:middle line:84% And you proved on the assignment that for a closed 00:15:43.470 --> 00:15:46.495 align:middle line:84% set, if I have a convergent sequence, 00:15:46.495 --> 00:15:48.870 align:middle line:84% then the limit of that sequence has to belong to the set. 00:15:48.870 --> 00:15:55.270 align:middle line:84% So this is really a consequence of the fact 00:15:55.270 --> 00:15:59.230 align:middle line:84% that this interval is a closed set. 00:15:59.230 --> 00:16:12.860 align:middle line:84% So I have this subsequence x sub n sub k. 00:16:12.860 --> 00:16:15.500 align:middle line:84% It's converging to x, which is in a, b. 00:16:15.500 --> 00:16:16.760 align:middle line:90% And I know the function. 00:16:16.760 --> 00:16:19.640 align:middle line:84% I haven't used anything about f yet. 00:16:19.640 --> 00:16:21.090 align:middle line:84% And I'm assuming it's continuous. 00:16:21.090 --> 00:16:23.690 align:middle line:90% So let's look at f of x. 00:16:23.690 --> 00:16:30.700 align:middle line:90% 00:16:30.700 --> 00:16:44.737 align:middle line:84% Since f is continuous, and the x sub n sub 00:16:44.737 --> 00:16:52.080 align:middle line:84% k is converged to x as k goes to infinity, I know by the theorem 00:16:52.080 --> 00:16:57.990 align:middle line:84% up top over there, that f of x equals the limit of f 00:16:57.990 --> 00:17:02.550 align:middle line:90% of x sub n sub k. 00:17:02.550 --> 00:17:08.150 align:middle line:84% And therefore, the absolute value of x 00:17:08.150 --> 00:17:10.940 align:middle line:84% is equal to the limit as k goes to infinity 00:17:10.940 --> 00:17:12.035 align:middle line:90% of the absolute values. 00:17:12.035 --> 00:17:16.750 align:middle line:90% 00:17:16.750 --> 00:17:21.430 align:middle line:84% But now, f of x sub n is always bigger than or equal to n. 00:17:21.430 --> 00:17:23.619 align:middle line:84% That's how the x of n's were chosen. 00:17:23.619 --> 00:17:26.800 align:middle line:84% We assume that f is unbounded, and therefore, 00:17:26.800 --> 00:17:30.220 align:middle line:84% all of these x sub n's in a, b were chosen so that f of x sub 00:17:30.220 --> 00:17:32.320 align:middle line:90% n is bigger than or equal to n. 00:17:32.320 --> 00:17:34.270 align:middle line:84% So this is bigger than or equal to-- 00:17:34.270 --> 00:17:48.870 align:middle line:90% 00:17:48.870 --> 00:17:56.100 align:middle line:84% so each of these is bigger than or equal to n sub k. 00:17:56.100 --> 00:18:02.082 align:middle line:90% 00:18:02.082 --> 00:18:03.665 align:middle line:84% Let's see, how to write this without-- 00:18:03.665 --> 00:18:10.330 align:middle line:90% 00:18:10.330 --> 00:18:10.870 align:middle line:90% OK. 00:18:10.870 --> 00:18:13.067 align:middle line:90% So we'll write it this way. 00:18:13.067 --> 00:18:14.900 align:middle line:84% I don't want to write the limit of something 00:18:14.900 --> 00:18:16.358 align:middle line:84% equals infinity, because we haven't 00:18:16.358 --> 00:18:19.970 align:middle line:84% said what it means for the limit of something to equal infinity. 00:18:19.970 --> 00:18:24.420 align:middle line:90% 00:18:24.420 --> 00:18:27.570 align:middle line:84% So then, the limit as k goes to infinity 00:18:27.570 --> 00:18:32.010 align:middle line:84% of f of x sub n sub k exists, which implies 00:18:32.010 --> 00:18:42.570 align:middle line:90% that the sequence is bounded. 00:18:42.570 --> 00:18:44.355 align:middle line:84% A convergent sequence is always bounded. 00:18:44.355 --> 00:18:49.320 align:middle line:90% 00:18:49.320 --> 00:18:54.210 align:middle line:84% And since n sub k is always less than or equal to f of x sub k, 00:18:54.210 --> 00:18:57.750 align:middle line:84% and this is a bounded sequence, this 00:18:57.750 --> 00:19:05.400 align:middle line:84% tells me that the sequence n sub k is bounded. 00:19:05.400 --> 00:19:07.170 align:middle line:90% But this is impossible. 00:19:07.170 --> 00:19:09.980 align:middle line:84% Remember, to form a subsequence, the n sub k's 00:19:09.980 --> 00:19:15.240 align:middle line:84% are increasing integers, which is a contradiction. 00:19:15.240 --> 00:19:18.840 align:middle line:84% And this is always because n sub 1, is always less than n sub 2, 00:19:18.840 --> 00:19:21.340 align:middle line:90% is less than n 3, and so on. 00:19:21.340 --> 00:19:23.400 align:middle line:84% So these are always getting bigger without bound. 00:19:23.400 --> 00:19:27.470 align:middle line:90% 00:19:27.470 --> 00:19:32.290 align:middle line:84% And therefore, we have our contradiction. 00:19:32.290 --> 00:19:34.240 align:middle line:84% So again, here's the structure of the proof. 00:19:34.240 --> 00:19:41.310 align:middle line:84% We want to assume some property of f. 00:19:41.310 --> 00:19:42.210 align:middle line:90% So we assume not. 00:19:42.210 --> 00:19:48.900 align:middle line:84% We get this kind of bad sequence of numbers in this interval. 00:19:48.900 --> 00:19:53.280 align:middle line:84% And using Bolzano-Weierstrass, we get to pass to some limit x. 00:19:53.280 --> 00:20:01.140 align:middle line:84% And the continuity of f at x essentially breaks the badness 00:20:01.140 --> 00:20:04.450 align:middle line:90% of this sequence x sub n. 00:20:04.450 --> 00:20:07.030 align:middle line:84% And we'll see another argument where 00:20:07.030 --> 00:20:08.890 align:middle line:90% it's kind of the same flavor. 00:20:08.890 --> 00:20:14.316 align:middle line:90% 00:20:14.316 --> 00:20:21.830 align:middle line:84% So we'll soon state that f always 00:20:21.830 --> 00:20:25.050 align:middle line:84% achieves a maximum value and minimum value. 00:20:25.050 --> 00:20:27.710 align:middle line:84% So let me first on the closed and bounded intervals-- so 00:20:27.710 --> 00:20:30.590 align:middle line:84% let me precisely define what these absolute mins 00:20:30.590 --> 00:20:34.340 align:middle line:90% and absolute maxes are. 00:20:34.340 --> 00:20:37.910 align:middle line:84% So let f be a function from S to R. S 00:20:37.910 --> 00:20:41.400 align:middle line:90% is a non-empty subset of R. 00:20:41.400 --> 00:20:53.220 align:middle line:84% We say, f achieves an absolute max. 00:20:53.220 --> 00:21:00.790 align:middle line:90% 00:21:00.790 --> 00:21:03.046 align:middle line:90% Let me write it this way-- 00:21:03.046 --> 00:21:09.430 align:middle line:84% an absolute min at c in S if f of c 00:21:09.430 --> 00:21:20.120 align:middle line:84% sits below f of x for all x in S. If for all x in S, f of c 00:21:20.120 --> 00:21:23.170 align:middle line:90% is less than or equal to f of x. 00:21:23.170 --> 00:21:25.660 align:middle line:90% So this is an absolute min. 00:21:25.660 --> 00:21:28.320 align:middle line:90% 00:21:28.320 --> 00:21:48.770 align:middle line:84% f achieves absolute max at d and S if f of d 00:21:48.770 --> 00:21:52.140 align:middle line:90% sits above everything. 00:21:52.140 --> 00:21:58.970 align:middle line:84% Every x you stick into f, if for all x in S, 00:21:58.970 --> 00:22:01.592 align:middle line:84% f of x is less than or equal to f of d. 00:22:01.592 --> 00:22:09.020 align:middle line:90% 00:22:09.020 --> 00:22:10.050 align:middle line:90% So for example. 00:22:10.050 --> 00:22:12.095 align:middle line:84% So what's the picture that goes with this? 00:22:12.095 --> 00:22:16.217 align:middle line:90% 00:22:16.217 --> 00:22:18.425 align:middle line:84% Let's imagine we're on a closed and bounded interval. 00:22:18.425 --> 00:22:27.440 align:middle line:90% 00:22:27.440 --> 00:22:41.750 align:middle line:84% Then d-- so f achieves absolute max at d. 00:22:41.750 --> 00:22:48.320 align:middle line:84% The graph of f sits below the value f of d. 00:22:48.320 --> 00:22:55.190 align:middle line:90% And sits above the value f of c. 00:22:55.190 --> 00:23:00.220 align:middle line:90% 00:23:00.220 --> 00:23:02.650 align:middle line:84% Now, absolute max and mins at points 00:23:02.650 --> 00:23:06.150 align:middle line:84% are so that the function is always 00:23:06.150 --> 00:23:09.690 align:middle line:84% sitting above f evaluated at that point. 00:23:09.690 --> 00:23:15.002 align:middle line:90% Just a quick Warning. 00:23:15.002 --> 00:23:16.210 align:middle line:90% Let's say, we're looking at-- 00:23:16.210 --> 00:23:21.280 align:middle line:90% 00:23:21.280 --> 00:23:22.900 align:middle line:84% all right, so this is not to scale. 00:23:22.900 --> 00:23:25.750 align:middle line:84% But let's say our function looks like this. 00:23:25.750 --> 00:23:38.370 align:middle line:90% 00:23:38.370 --> 00:23:44.700 align:middle line:84% So this is one, two, three halves. 00:23:44.700 --> 00:23:47.760 align:middle line:84% Then, this function does not have an absolute max, 00:23:47.760 --> 00:23:49.530 align:middle line:84% does not achieve an absolute max, 00:23:49.530 --> 00:23:53.795 align:middle line:84% or achieve an absolute min on the set 1, 2. 00:23:53.795 --> 00:23:56.600 align:middle line:90% 00:23:56.600 --> 00:24:01.820 align:middle line:84% What you would like to say is that f achieves an absolute min 00:24:01.820 --> 00:24:02.320 align:middle line:90% at 1. 00:24:02.320 --> 00:24:10.150 align:middle line:84% But the graph does not sit above f evaluated at 1. 00:24:10.150 --> 00:24:11.770 align:middle line:84% Just like over here, the graph of f 00:24:11.770 --> 00:24:15.290 align:middle line:84% does not sit below f evaluated at 2. 00:24:15.290 --> 00:24:20.500 align:middle line:84% So absolute f achieves an absolute min at a point 00:24:20.500 --> 00:24:24.680 align:middle line:84% if the whole graph sits above f evaluated at that point, 00:24:24.680 --> 00:24:28.570 align:middle line:84% not necessarily some number, it sits below the whole graph. 00:24:28.570 --> 00:24:29.890 align:middle line:90% That's what boundedness means. 00:24:29.890 --> 00:24:33.376 align:middle line:90% 00:24:33.376 --> 00:24:38.250 align:middle line:84% So I just want to clarify that distinction between the graph 00:24:38.250 --> 00:24:40.860 align:middle line:84% of a function being bounded below by a number and f 00:24:40.860 --> 00:24:43.480 align:middle line:84% achieving a min at a certain point. 00:24:43.480 --> 00:24:45.300 align:middle line:84% So for example, the graph of this function 00:24:45.300 --> 00:24:47.820 align:middle line:90% is bounded below by 1. 00:24:47.820 --> 00:24:53.190 align:middle line:84% But f does not achieve an absolute min at c equals 1. 00:24:53.190 --> 00:25:11.530 align:middle line:90% 00:25:11.530 --> 00:25:19.730 align:middle line:84% So our next theorem is the min-max theorem 00:25:19.730 --> 00:25:24.860 align:middle line:84% for continuous functions, which is the following. 00:25:24.860 --> 00:25:31.200 align:middle line:90% 00:25:31.200 --> 00:25:40.410 align:middle line:84% Let f be a function from a, b to R. If f is continuous, 00:25:40.410 --> 00:25:46.520 align:middle line:84% meaning it's continuous at every point in a,b, 00:25:46.520 --> 00:26:07.410 align:middle line:84% then f achieves absolute min and absolute max on a, b. 00:26:07.410 --> 00:26:13.290 align:middle line:84% So another way of stating this, so let's just 00:26:13.290 --> 00:26:19.215 align:middle line:84% make this a remark, it says that they there exists c, d, and a, 00:26:19.215 --> 00:26:34.100 align:middle line:84% b, such that f of a, b is contained in f of c, f of d. 00:26:34.100 --> 00:26:43.710 align:middle line:90% 00:26:43.710 --> 00:26:51.720 align:middle line:84% So we'll do the proof of an absolute max. 00:26:51.720 --> 00:26:56.790 align:middle line:84% And the proof of the absolute min, I'll leave to you. 00:26:56.790 --> 00:27:02.040 align:middle line:84% It's a simple change in the argument that's 00:27:02.040 --> 00:27:07.170 align:middle line:84% not too difficult. And again, we're 00:27:07.170 --> 00:27:09.540 align:middle line:84% going to use this powerful tool, Bolzano-Weierstrass, 00:27:09.540 --> 00:27:14.310 align:middle line:84% that allows us to go from an arbitrary sequence in a, b, 00:27:14.310 --> 00:27:16.800 align:middle line:90% to a convergent subsequence. 00:27:16.800 --> 00:27:26.388 align:middle line:90% So this equals f from a to b-- 00:27:26.388 --> 00:27:27.430 align:middle line:90% let me write it this way. 00:27:27.430 --> 00:27:37.700 align:middle line:84% If this function is continuous, then by what we've proven-- 00:27:37.700 --> 00:27:39.030 align:middle line:90% where? 00:27:39.030 --> 00:27:41.950 align:middle line:90% Over there. 00:27:41.950 --> 00:27:44.770 align:middle line:90% This implies f is bounded. 00:27:44.770 --> 00:27:48.380 align:middle line:90% 00:27:48.380 --> 00:27:58.720 align:middle line:84% Then the set E given by the range of f, 00:27:58.720 --> 00:28:11.350 align:middle line:84% so f of x, where x is in a, b, this set is bounded above. 00:28:11.350 --> 00:28:13.180 align:middle line:84% If the absolute value of f of x is always 00:28:13.180 --> 00:28:15.040 align:middle line:84% less than or equal to some b, then 00:28:15.040 --> 00:28:18.160 align:middle line:84% f of x is always bounded above by b for all x and a, b. 00:28:18.160 --> 00:28:19.930 align:middle line:90% So this set is bounded above. 00:28:19.930 --> 00:28:25.840 align:middle line:90% 00:28:25.840 --> 00:28:32.950 align:middle line:84% Now, let L be the supremum of E, which 00:28:32.950 --> 00:28:36.490 align:middle line:84% exists, because the real numbers have the least upper 00:28:36.490 --> 00:28:37.300 align:middle line:90% bound property. 00:28:37.300 --> 00:28:40.850 align:middle line:84% Whenever we have a non-empty subset which is bounded above, 00:28:40.850 --> 00:28:44.290 align:middle line:90% we can always find a supremum. 00:28:44.290 --> 00:28:49.690 align:middle line:84% Now, since L is a supremum of this set, we did this, I think, 00:28:49.690 --> 00:28:52.840 align:middle line:90% in an assignment. 00:28:52.840 --> 00:28:55.270 align:middle line:84% Yes, I definitely did an assignment. 00:28:55.270 --> 00:29:02.590 align:middle line:84% Then there exists a sequence of elements 00:29:02.590 --> 00:29:08.640 align:middle line:84% of this set E converging to L. And to express 00:29:08.640 --> 00:29:11.040 align:middle line:84% that, that means there exists some sequence of elements. 00:29:11.040 --> 00:29:13.290 align:middle line:84% So there exists a sequence of the form 00:29:13.290 --> 00:29:25.460 align:middle line:84% f of x sub n such that limit as n goes to infinity 00:29:25.460 --> 00:29:31.120 align:middle line:90% of f of x sub n equals L. 00:29:31.120 --> 00:29:35.190 align:middle line:84% Now, we would like to show that L is equal to f 00:29:35.190 --> 00:29:37.845 align:middle line:90% of d for some number d. 00:29:37.845 --> 00:29:40.970 align:middle line:90% 00:29:40.970 --> 00:29:44.750 align:middle line:84% And how we're going to do that is, well, this is-- 00:29:44.750 --> 00:29:48.260 align:middle line:84% the x sub n's are just some sequence in a, b. 00:29:48.260 --> 00:29:51.620 align:middle line:84% So we can pass to a subsequence, which converges to some number, 00:29:51.620 --> 00:29:53.270 align:middle line:90% call it d. 00:29:53.270 --> 00:29:55.550 align:middle line:84% And then we need to show that f of d equals L. 00:29:55.550 --> 00:29:57.140 align:middle line:84% And that's where we use continuity. 00:29:57.140 --> 00:30:00.350 align:middle line:84% And then, that's the whole proof. 00:30:00.350 --> 00:30:02.390 align:middle line:84% Which we saw a little bit over here. 00:30:02.390 --> 00:30:03.620 align:middle line:90% We passed a subsequence. 00:30:03.620 --> 00:30:07.640 align:middle line:84% And then we get to say that the limit of this subsequence 00:30:07.640 --> 00:30:11.015 align:middle line:84% has the same property as the original sequence. 00:30:11.015 --> 00:30:17.300 align:middle line:90% 00:30:17.300 --> 00:30:28.230 align:middle line:84% So by Bolzano-Weierstrass, there exists a subsequence x sub n 00:30:28.230 --> 00:30:29.040 align:middle line:90% sub k. 00:30:29.040 --> 00:30:32.730 align:middle line:84% So this is a subsequence of the sequence x sub n. 00:30:32.730 --> 00:30:39.960 align:middle line:90% 00:30:39.960 --> 00:30:46.110 align:middle line:84% And as before, the same argument and limit 00:30:46.110 --> 00:31:00.820 align:middle line:84% d, but as before, d will be in the set a, b, 00:31:00.820 --> 00:31:06.970 align:middle line:84% such that limit as k goes to infinity of x of n sub k 00:31:06.970 --> 00:31:10.118 align:middle line:90% equals d. 00:31:10.118 --> 00:31:11.660 align:middle line:84% So we use Bolzano-Weierstrass to pass 00:31:11.660 --> 00:31:14.120 align:middle line:84% to a subsequence of the x sub n's, 00:31:14.120 --> 00:31:17.930 align:middle line:90% and a convergent subsequence. 00:31:17.930 --> 00:31:23.180 align:middle line:84% And yeah, so, all of these x sub n sub k's are between a and b. 00:31:23.180 --> 00:31:25.580 align:middle line:84% So their limit d will be between a and b. 00:31:25.580 --> 00:31:30.690 align:middle line:90% 00:31:30.690 --> 00:31:42.610 align:middle line:84% But now, since f is continuous at d, 00:31:42.610 --> 00:31:44.380 align:middle line:84% it's continuous at every point in a, b. 00:31:44.380 --> 00:31:47.530 align:middle line:84% So it's certainly continuous at d-- f of d, this 00:31:47.530 --> 00:31:52.150 align:middle line:84% is equal to the limit as k goes to infinity 00:31:52.150 --> 00:31:55.100 align:middle line:84% of f of x of n sub k, since the x of n 00:31:55.100 --> 00:31:58.350 align:middle line:90% sub k's are converging to d. 00:31:58.350 --> 00:32:01.380 align:middle line:84% And now, f of x sub n sub k, this 00:32:01.380 --> 00:32:06.240 align:middle line:84% is also a subsequence of the sequence f of x sub n. 00:32:06.240 --> 00:32:08.700 align:middle line:84% And f of x sub n is a convergent sequence. 00:32:08.700 --> 00:32:11.640 align:middle line:84% So any subsequence has to converge to the same thing. 00:32:11.640 --> 00:32:18.450 align:middle line:90% 00:32:18.450 --> 00:32:26.330 align:middle line:84% Since f of x sub n converges to L, and f of n sub k, 00:32:26.330 --> 00:32:37.540 align:middle line:84% k is a subsequence of f of x sub n. 00:32:37.540 --> 00:32:41.890 align:middle line:90% 00:32:41.890 --> 00:32:44.260 align:middle line:90% And remember, what was L? 00:32:44.260 --> 00:32:56.960 align:middle line:84% L was the sup of V. And therefore, for all x in a, 00:32:56.960 --> 00:33:04.200 align:middle line:84% b, f of x is less than or equal to f of d, which means f 00:33:04.200 --> 00:33:14.980 align:middle line:90% achieves an absolute max at d. 00:33:14.980 --> 00:33:16.545 align:middle line:90% And the absolute min is similar. 00:33:16.545 --> 00:33:22.608 align:middle line:90% 00:33:22.608 --> 00:33:23.650 align:middle line:90% And I'll leave it to you. 00:33:23.650 --> 00:33:38.450 align:middle line:90% 00:33:38.450 --> 00:33:41.630 align:middle line:84% So just rerunning through the proof. 00:33:41.630 --> 00:33:46.670 align:middle line:84% We did use the fact that a continuous function is bounded. 00:33:46.670 --> 00:33:51.830 align:middle line:84% And we extracted the sup, if you like, of the range of f. 00:33:51.830 --> 00:33:57.220 align:middle line:84% And so, which is by this exercise we did in one 00:33:57.220 --> 00:34:00.100 align:middle line:84% of the assignments, means that this supremum is equal 00:34:00.100 --> 00:34:02.560 align:middle line:90% to the limit of f of x sub n's. 00:34:02.560 --> 00:34:05.740 align:middle line:84% We would like to show L is equal to f of d for some d. 00:34:05.740 --> 00:34:08.139 align:middle line:84% So we passed a subsequence of the x sub n's. 00:34:08.139 --> 00:34:12.550 align:middle line:84% That does converge to something in the interval a, b. 00:34:12.550 --> 00:34:15.610 align:middle line:84% And we can show using the continuity of f. 00:34:15.610 --> 00:34:22.840 align:middle line:84% And how this original sequence was chosen, that f at that 00:34:22.840 --> 00:34:27.100 align:middle line:84% point is actually equal to the supremum of that set. 00:34:27.100 --> 00:34:31.840 align:middle line:84% And therefore, f achieves an absolute max at that set. 00:34:31.840 --> 00:34:42.199 align:middle line:84% As beginning students of math, one 00:34:42.199 --> 00:34:44.060 align:middle line:84% of the things we should be curious about 00:34:44.060 --> 00:34:50.600 align:middle line:84% is what hypotheses are needed and what are not 00:34:50.600 --> 00:34:53.719 align:middle line:90% in the statements of theorems. 00:34:53.719 --> 00:34:56.639 align:middle line:90% So our main hypothesis-- 00:34:56.639 --> 00:35:01.270 align:middle line:84% so we had two hypotheses really in the theorem, 00:35:01.270 --> 00:35:05.950 align:middle line:84% that f is going from this closed and bounded interval to R, 00:35:05.950 --> 00:35:07.460 align:middle line:90% and that f is continuous. 00:35:07.460 --> 00:35:10.660 align:middle line:84% So what happens if we drop one of those hypotheses? 00:35:10.660 --> 00:35:13.840 align:middle line:84% Is the conclusion still true, that f achieves an absolute max 00:35:13.840 --> 00:35:15.280 align:middle line:90% and absolute min? 00:35:15.280 --> 00:35:17.230 align:middle line:84% So this example that I drew right here 00:35:17.230 --> 00:35:21.100 align:middle line:84% shows that we need to have the function be continuous in order 00:35:21.100 --> 00:35:23.170 align:middle line:84% for it to achieve an absolute max 00:35:23.170 --> 00:35:27.750 align:middle line:84% and absolute min on the closed and bounded interval. 00:35:27.750 --> 00:35:38.930 align:middle line:84% But it also needs to be on a closed and bounded interval. 00:35:38.930 --> 00:35:58.200 align:middle line:90% 00:35:58.200 --> 00:36:00.810 align:middle line:84% So we have some continuous function from S 00:36:00.810 --> 00:36:05.820 align:middle line:84% to R. Can we drop the assumption that S 00:36:05.820 --> 00:36:07.920 align:middle line:84% is equal to this closed and bounded interval? 00:36:07.920 --> 00:36:10.860 align:middle line:84% Meaning could it be, say, an open interval? 00:36:10.860 --> 00:36:12.195 align:middle line:90% Does the same conclusion hold? 00:36:12.195 --> 00:36:16.400 align:middle line:90% 00:36:16.400 --> 00:36:17.000 align:middle line:90% I.e. 00:36:17.000 --> 00:36:19.025 align:middle line:84% let's say f is from an open interval. 00:36:19.025 --> 00:36:37.620 align:middle line:90% 00:36:37.620 --> 00:36:39.900 align:middle line:90% And the answer is, no. 00:36:39.900 --> 00:36:44.610 align:middle line:84% Simple example is, if I take the function f of x equals, 00:36:44.610 --> 00:36:52.920 align:middle line:84% for example, 1 over x, minus 1 over 1 minus x. 00:36:52.920 --> 00:37:00.180 align:middle line:84% And on the open interval 0, 1, so what 00:37:00.180 --> 00:37:03.340 align:middle line:90% does this guy look like? 00:37:03.340 --> 00:37:04.170 align:middle line:90% Here's 0. 00:37:04.170 --> 00:37:08.730 align:middle line:90% Let me draw a dotted line there. 00:37:08.730 --> 00:37:09.330 align:middle line:90% Here's 1. 00:37:09.330 --> 00:37:12.030 align:middle line:90% 00:37:12.030 --> 00:37:16.350 align:middle line:84% And the graph shoots up to positive infinity 00:37:16.350 --> 00:37:20.145 align:middle line:84% as you approach 0 and as you approach 1. 00:37:20.145 --> 00:37:23.630 align:middle line:90% 00:37:23.630 --> 00:37:25.190 align:middle line:90% So where does it equal 0? 00:37:25.190 --> 00:37:30.440 align:middle line:90% I guess there's 1/2. 00:37:30.440 --> 00:37:36.060 align:middle line:90% 00:37:36.060 --> 00:37:37.530 align:middle line:84% So this function does not achieve 00:37:37.530 --> 00:37:40.320 align:middle line:90% a absolute min or absolute max. 00:37:40.320 --> 00:37:43.410 align:middle line:84% But it is continuous on this interval. 00:37:43.410 --> 00:37:46.740 align:middle line:84% But because the interval is not closed and bounded, 00:37:46.740 --> 00:37:48.750 align:middle line:84% this function does not achieve an absolute min 00:37:48.750 --> 00:37:49.890 align:middle line:90% or absolute max. 00:37:49.890 --> 00:37:53.708 align:middle line:84% So what I'm trying to say is that the assumptions that two-- 00:37:53.708 --> 00:37:55.500 align:middle line:84% so there's two assumptions here, that we're 00:37:55.500 --> 00:37:57.690 align:middle line:84% working on a closed and bounded interval, 00:37:57.690 --> 00:37:59.220 align:middle line:90% and that f is continuous. 00:37:59.220 --> 00:38:01.470 align:middle line:84% These two assumptions are necessary for the theorem 00:38:01.470 --> 00:38:03.240 align:middle line:90% to be true. 00:38:03.240 --> 00:38:06.300 align:middle line:84% Now, they are not the most precise way 00:38:06.300 --> 00:38:08.070 align:middle line:90% of stating this theorem. 00:38:08.070 --> 00:38:12.360 align:middle line:84% You could replace a closed and bounded interval 00:38:12.360 --> 00:38:15.000 align:middle line:84% with what's called a compact set, which maybe I'll 00:38:15.000 --> 00:38:18.810 align:middle line:84% put in the assignment or on the midterm, 00:38:18.810 --> 00:38:22.980 align:middle line:90% depending on where there's room. 00:38:22.980 --> 00:38:27.980 align:middle line:84% But just in the setting of intervals, 00:38:27.980 --> 00:38:29.960 align:middle line:84% the interval has to be closed and bounded. 00:38:29.960 --> 00:38:33.890 align:middle line:84% And the function has to be continuous for this theorem 00:38:33.890 --> 00:38:35.390 align:middle line:90% to hold true. 00:38:35.390 --> 00:38:37.280 align:middle line:84% So that's what I wanted to get at. 00:38:37.280 --> 00:38:54.790 align:middle line:90% 00:38:54.790 --> 00:39:03.610 align:middle line:84% So what we've proven, as I said in this remark here, 00:39:03.610 --> 00:39:07.840 align:middle line:84% is that there exists two numbers c and d in a, b, 00:39:07.840 --> 00:39:12.595 align:middle line:84% so that f of a, b is contained in f of c, f of d. 00:39:12.595 --> 00:39:15.610 align:middle line:84% c would be where f achieves an absolute min. 00:39:15.610 --> 00:39:18.955 align:middle line:84% d would be where if f achieves an absolute max. 00:39:18.955 --> 00:39:26.420 align:middle line:90% 00:39:26.420 --> 00:39:32.600 align:middle line:90% And this is absolute max. 00:39:32.600 --> 00:39:37.690 align:middle line:84% So now the question becomes, do I hit everything in between? 00:39:37.690 --> 00:39:42.340 align:middle line:84% Does this inclusion become equality? 00:39:42.340 --> 00:39:46.060 align:middle line:84% And I gave the game away at the beginning of the lecture 00:39:46.060 --> 00:39:47.860 align:middle line:90% by saying, yes, it will. 00:39:47.860 --> 00:39:50.740 align:middle line:90% But that's actually a theorem. 00:39:50.740 --> 00:39:58.927 align:middle line:84% So this was the min-max theorem, which I didn't call it that. 00:39:58.927 --> 00:39:59.510 align:middle line:90% I should have. 00:39:59.510 --> 00:40:01.280 align:middle line:84% It's actually called that in the notes. 00:40:01.280 --> 00:40:04.130 align:middle line:84% But now we're going to move on to the intermediate value 00:40:04.130 --> 00:40:06.840 align:middle line:90% theorem. 00:40:06.840 --> 00:40:10.080 align:middle line:84% And first we're going to do what looks like a special case, 00:40:10.080 --> 00:40:14.515 align:middle line:84% and then we'll prove the general case. 00:40:14.515 --> 00:40:18.630 align:middle line:90% 00:40:18.630 --> 00:40:26.190 align:middle line:84% So this theorem, which I'm actually-- 00:40:26.190 --> 00:40:32.670 align:middle line:84% it's not called this in the textbook. 00:40:32.670 --> 00:40:36.690 align:middle line:84% But I'm going to call it this, which I'll call the bisection 00:40:36.690 --> 00:40:40.600 align:middle line:90% method, is the following. 00:40:40.600 --> 00:40:57.800 align:middle line:84% Let f be a function from a, b to R if f of a is less than 0-- 00:40:57.800 --> 00:41:00.170 align:middle line:90% and so, I need one more-- 00:41:00.170 --> 00:41:05.640 align:middle line:90% 00:41:05.640 --> 00:41:06.520 align:middle line:90% be continuous. 00:41:06.520 --> 00:41:08.430 align:middle line:84% So we're always in the continuous setting 00:41:08.430 --> 00:41:14.530 align:middle line:90% for this section. 00:41:14.530 --> 00:41:25.650 align:middle line:84% So if f of a is less than 0 and f of b is bigger than 0, 00:41:25.650 --> 00:41:38.110 align:middle line:84% then there exists c and E in the interval a, b such that f of c 00:41:38.110 --> 00:41:40.660 align:middle line:90% equals 0. 00:41:40.660 --> 00:41:43.600 align:middle line:84% So the picture that goes on with goes along with this is, 00:41:43.600 --> 00:41:46.960 align:middle line:90% here's a, b. 00:41:46.960 --> 00:41:49.750 align:middle line:90% Here's f of a. 00:41:49.750 --> 00:41:52.270 align:middle line:90% f of b is positive. 00:41:52.270 --> 00:41:53.830 align:middle line:84% And therefore, if it's continuous, 00:41:53.830 --> 00:41:57.340 align:middle line:84% and you believe the definition of a continuity 00:41:57.340 --> 00:41:59.950 align:middle line:84% is the fact that is that I don't have 00:41:59.950 --> 00:42:02.380 align:middle line:90% to pick up my piece of chalk. 00:42:02.380 --> 00:42:03.520 align:middle line:90% It pains me to say this. 00:42:03.520 --> 00:42:07.240 align:middle line:84% But that's, again, not the official definition, 00:42:07.240 --> 00:42:08.800 align:middle line:90% but one you keep in your head. 00:42:08.800 --> 00:42:11.510 align:middle line:84% Since somebody probably told you that at some point, 00:42:11.510 --> 00:42:13.630 align:middle line:84% if I don't have to pick up my piece of chalk, 00:42:13.630 --> 00:42:17.170 align:middle line:84% then eventually, I have to cross the x-axis. 00:42:17.170 --> 00:42:23.300 align:middle line:84% And therefore, at this point c, f of c will be 0. 00:42:23.300 --> 00:42:26.420 align:middle line:84% Now, why do I call this theorem the bisection method? 00:42:26.420 --> 00:42:32.540 align:middle line:84% As we'll see, the way you determine this c is by what 00:42:32.540 --> 00:42:35.300 align:middle line:84% in calculus books is called the bisection method. 00:42:35.300 --> 00:42:48.110 align:middle line:90% 00:42:48.110 --> 00:42:58.610 align:middle line:84% So for the proof, let-- we're going 00:42:58.610 --> 00:43:01.970 align:middle line:84% to define a sequence of numbers a in b 00:43:01.970 --> 00:43:03.600 align:middle line:90% and with special properties. 00:43:03.600 --> 00:43:06.350 align:middle line:84% So I'm first going to tell you what a1 and b1 are. 00:43:06.350 --> 00:43:10.780 align:middle line:84% a1 is just going to be a, b1 is going to be b. 00:43:10.780 --> 00:43:14.380 align:middle line:90% 00:43:14.380 --> 00:43:16.600 align:middle line:84% And so, now I'm going to tell you 00:43:16.600 --> 00:43:20.050 align:middle line:84% how to choose the next element in the sequence knowing 00:43:20.050 --> 00:43:23.318 align:middle line:84% the element in the sequence before. 00:43:23.318 --> 00:43:24.110 align:middle line:90% So we're going to-- 00:43:24.110 --> 00:43:37.670 align:middle line:90% 00:43:37.670 --> 00:43:44.570 align:middle line:84% first to find two sequences a n and bn. 00:43:44.570 --> 00:43:46.260 align:middle line:84% So let me tell you how to do that. 00:43:46.260 --> 00:43:48.110 align:middle line:84% And the way we're going to choose this 00:43:48.110 --> 00:43:51.260 align:middle line:84% is so that f of a sub n is always less than 0, 00:43:51.260 --> 00:43:55.040 align:middle line:84% and f of b sub n is always bigger than or equal to 0. 00:43:55.040 --> 00:43:58.970 align:middle line:84% And there obtained from the previous two guys 00:43:58.970 --> 00:44:04.050 align:middle line:84% by taking the midpoint, so bisecting the ones before. 00:44:04.050 --> 00:44:11.846 align:middle line:84% So to get this started, a1 will be just a, b1 will be b. 00:44:11.846 --> 00:44:15.540 align:middle line:90% 00:44:15.540 --> 00:44:20.370 align:middle line:84% Now, for in a natural number, we're going to define-- 00:44:20.370 --> 00:44:21.450 align:middle line:90% so we know a1. 00:44:21.450 --> 00:44:24.180 align:middle line:84% So now, I'm going to tell you how to define a2 00:44:24.180 --> 00:44:26.370 align:middle line:90% based on now you know a1. 00:44:26.370 --> 00:44:29.080 align:middle line:84% But what I'm about to write down will also 00:44:29.080 --> 00:44:32.320 align:middle line:84% tell you how to define a3, since you now know how to do a2. 00:44:32.320 --> 00:44:33.910 align:middle line:84% So I'm going to write it this way. 00:44:33.910 --> 00:44:42.895 align:middle line:84% For in a natural number, knowing a sub n, 00:44:42.895 --> 00:44:52.265 align:middle line:84% b sub n, we define a sub n plus 1 and b sub n plus 1 as 00:44:52.265 --> 00:44:52.765 align:middle line:90% follows. 00:44:52.765 --> 00:45:01.290 align:middle line:90% 00:45:01.290 --> 00:45:08.990 align:middle line:84% So if f of a sub n plus b sub n-- so the midpoint 00:45:08.990 --> 00:45:11.900 align:middle line:84% between the two guys that I already 00:45:11.900 --> 00:45:18.910 align:middle line:84% know-- so if you like, take n to be 1 for when you first read 00:45:18.910 --> 00:45:22.390 align:middle line:90% how to define a sub 2, say. 00:45:22.390 --> 00:45:25.210 align:middle line:84% But what I'm writing down applies to every in. 00:45:25.210 --> 00:45:31.320 align:middle line:90% 00:45:31.320 --> 00:45:35.510 align:middle line:84% So we take f of a sub n plus b sub n over 2 to be, 00:45:35.510 --> 00:45:39.770 align:middle line:84% if this is bigger than or equal to 0, 00:45:39.770 --> 00:45:47.160 align:middle line:84% then we define a sub n plus 1 to be a sub n, 00:45:47.160 --> 00:45:52.290 align:middle line:84% and b sub n plus 1 to be the midpoint. 00:45:52.290 --> 00:45:59.300 align:middle line:90% 00:45:59.300 --> 00:46:06.650 align:middle line:84% And if of the midpoint is less than 0, 00:46:06.650 --> 00:46:16.310 align:middle line:84% then a sub n plus 1 is going to be the midpoint, and b sub n 00:46:16.310 --> 00:46:19.370 align:middle line:84% plus 1 will be the previous point. 00:46:19.370 --> 00:46:23.778 align:middle line:84% So let me-- in fact, so this is how you 00:46:23.778 --> 00:46:25.070 align:middle line:90% define the sequence in general. 00:46:25.070 --> 00:46:27.500 align:middle line:84% Let's walk through what this means just for n 00:46:27.500 --> 00:46:31.460 align:middle line:90% equals 2 so that you see-- 00:46:31.460 --> 00:46:32.795 align:middle line:90% you get the idea. 00:46:32.795 --> 00:46:34.295 align:middle line:84% Maybe I should have done this first. 00:46:34.295 --> 00:46:37.450 align:middle line:90% 00:46:37.450 --> 00:46:39.310 align:middle line:90% So let's draw a picture to go-- 00:46:39.310 --> 00:46:42.950 align:middle line:90% 00:46:42.950 --> 00:46:45.270 align:middle line:90% in fact, I don't need this axis. 00:46:45.270 --> 00:46:50.940 align:middle line:90% I'll just-- b. 00:46:50.940 --> 00:46:55.110 align:middle line:84% So and this is a sub 1, this is b sub 1. 00:46:55.110 --> 00:46:56.820 align:middle line:84% And then, we look at the midpoint. 00:46:56.820 --> 00:47:03.450 align:middle line:90% 00:47:03.450 --> 00:47:09.920 align:middle line:84% So we know that f of a sub 1 is less than 0. 00:47:09.920 --> 00:47:14.760 align:middle line:84% f of b sub 1 is bigger than 0, which 00:47:14.760 --> 00:47:16.440 align:middle line:84% is certainly bigger than or equal to 0 00:47:16.440 --> 00:47:17.850 align:middle line:90% if it's bigger than 0. 00:47:17.850 --> 00:47:20.780 align:middle line:90% Now we look at the midpoint. 00:47:20.780 --> 00:47:26.000 align:middle line:84% And based on the sign of this guy 00:47:26.000 --> 00:47:27.860 align:middle line:90% will be how we define a2 and b2. 00:47:27.860 --> 00:47:30.060 align:middle line:84% So for the sake of me going through this, 00:47:30.060 --> 00:47:36.950 align:middle line:84% Let's suppose that f of this thing is less than 0. 00:47:36.950 --> 00:47:40.590 align:middle line:90% 00:47:40.590 --> 00:47:46.680 align:middle line:84% Then I take, a2 will be a1 plus b1 over 2. 00:47:46.680 --> 00:47:52.120 align:middle line:90% And b2 will be b1. 00:47:52.120 --> 00:47:54.520 align:middle line:84% And now, I look at the midpoint of these two guys. 00:47:54.520 --> 00:48:02.020 align:middle line:84% So now, I'm at this stage, f of a2 is less than 0, f of b2 00:48:02.020 --> 00:48:03.010 align:middle line:90% is bigger than 0. 00:48:03.010 --> 00:48:06.160 align:middle line:84% So I'm in the picture before, except now I'm 00:48:06.160 --> 00:48:09.790 align:middle line:84% at half the distance between the two endpoints. 00:48:09.790 --> 00:48:18.860 align:middle line:84% And so, I look at this point now, a2 plus V2 over 2. 00:48:18.860 --> 00:48:20.580 align:middle line:90% And I look at the sign of that. 00:48:20.580 --> 00:48:22.490 align:middle line:84% And let's suppose f of this thing 00:48:22.490 --> 00:48:23.840 align:middle line:90% is bigger than or equal to 0. 00:48:23.840 --> 00:48:27.150 align:middle line:90% 00:48:27.150 --> 00:48:32.400 align:middle line:84% Then I will take this point to be a 3, 00:48:32.400 --> 00:48:36.630 align:middle line:90% and this point will now be b3. 00:48:36.630 --> 00:48:43.500 align:middle line:84% And what I have is f of a3 is less than 0, f of b3 00:48:43.500 --> 00:48:44.745 align:middle line:90% is bigger than or equal to 0. 00:48:44.745 --> 00:48:49.290 align:middle line:90% 00:48:49.290 --> 00:48:52.560 align:middle line:84% So the point is that the left endpoint is always negative 00:48:52.560 --> 00:48:54.120 align:middle line:90% when I stick it into f. 00:48:54.120 --> 00:48:56.160 align:middle line:84% The right endpoint is always non-negative 00:48:56.160 --> 00:48:58.490 align:middle line:90% when I stick it into f. 00:48:58.490 --> 00:49:01.520 align:middle line:84% And that the distance between the two midpoints 00:49:01.520 --> 00:49:04.745 align:middle line:84% is always getting cut in half by the previous distance before. 00:49:04.745 --> 00:49:08.330 align:middle line:90% 00:49:08.330 --> 00:49:12.290 align:middle line:84% But we're always in the picture, but kind of in the setting 00:49:12.290 --> 00:49:13.760 align:middle line:90% we were in the step before. 00:49:13.760 --> 00:49:22.800 align:middle line:90% 00:49:22.800 --> 00:49:36.450 align:middle line:84% Now, we have three properties of this sequence of a n's and bn's 00:49:36.450 --> 00:49:38.700 align:middle line:84% So for all n, they're always bounded 00:49:38.700 --> 00:49:42.510 align:middle line:84% between the original two endpoints. 00:49:42.510 --> 00:49:48.070 align:middle line:84% Not only that, the a sub n's are always moving to the right. 00:49:48.070 --> 00:49:52.310 align:middle line:84% Remember, a sub n is always getting 00:49:52.310 --> 00:49:59.610 align:middle line:84% replaced by the midpoint between a sub n and b sub n, possibly, 00:49:59.610 --> 00:50:00.970 align:middle line:90% or staying the same. 00:50:00.970 --> 00:50:05.100 align:middle line:84% So a sub n is always less than or equal to a sub n plus 1. 00:50:05.100 --> 00:50:13.180 align:middle line:90% 00:50:13.180 --> 00:50:15.850 align:middle line:84% So in fact, let me write it this way. 00:50:15.850 --> 00:50:18.920 align:middle line:90% 00:50:18.920 --> 00:50:21.860 align:middle line:84% We always have a sub n is less than or equal to b sub n. 00:50:21.860 --> 00:50:26.390 align:middle line:84% We always have a is less than or equal to a n plus 1, which 00:50:26.390 --> 00:50:34.630 align:middle line:90% is less than or equal to a n. 00:50:34.630 --> 00:50:35.800 align:middle line:90% And then, we also have-- 00:50:35.800 --> 00:50:40.170 align:middle line:90% 00:50:40.170 --> 00:50:46.500 align:middle line:90% so for all in b. 00:50:46.500 --> 00:50:48.690 align:middle line:84% I'm making a mess of what I'm writing down. 00:50:48.690 --> 00:50:51.010 align:middle line:90% So let's start this over. 00:50:51.010 --> 00:50:54.000 align:middle line:84% So the first property I want to write, 00:50:54.000 --> 00:51:08.710 align:middle line:84% for all N natural numbers, we have b, a sub n 00:51:08.710 --> 00:51:11.410 align:middle line:84% is always less than or equal to a sub n plus 1. 00:51:11.410 --> 00:51:14.950 align:middle line:90% 00:51:14.950 --> 00:51:19.210 align:middle line:84% And b sub n plus 1, remember, whenever-- 00:51:19.210 --> 00:51:21.520 align:middle line:84% if we're going to change b, it's always 00:51:21.520 --> 00:51:25.480 align:middle line:84% to change it to the midpoint between the previous guy 00:51:25.480 --> 00:51:27.600 align:middle line:84% and the a from the previous step. 00:51:27.600 --> 00:51:29.650 align:middle line:84% So this is always less than or equal to b sub n. 00:51:29.650 --> 00:51:35.800 align:middle line:90% 00:51:35.800 --> 00:51:52.270 align:middle line:84% And now, two, if I look at the difference, 00:51:52.270 --> 00:51:57.780 align:middle line:84% this is always equal to 1/2 of the previous distance. 00:51:57.780 --> 00:52:01.080 align:middle line:84% Because either one of these are getting 00:52:01.080 --> 00:52:02.250 align:middle line:90% changed to the midpoint. 00:52:02.250 --> 00:52:05.190 align:middle line:84% So if it's a sub n plus 1, then b 00:52:05.190 --> 00:52:09.510 align:middle line:84% sub n plus 1 minus a sub n plus 1, is b sub n minus a sub n 00:52:09.510 --> 00:52:11.955 align:middle line:84% plus b sub n over 2, which gives me this. 00:52:11.955 --> 00:52:16.000 align:middle line:90% 00:52:16.000 --> 00:52:23.530 align:middle line:84% Three, for all n, a natural number-- 00:52:23.530 --> 00:52:28.240 align:middle line:84% and this is just based on how we are choosing-- 00:52:28.240 --> 00:52:34.060 align:middle line:84% we are doing this construction, f of a sub n is less than 0, 00:52:34.060 --> 00:52:40.810 align:middle line:84% and f of b sub n is bigger than or equal to 0. 00:52:40.810 --> 00:52:42.700 align:middle line:90% So I hope this is clear. 00:52:42.700 --> 00:52:45.720 align:middle line:90% 00:52:45.720 --> 00:52:50.020 align:middle line:84% All of these, if you like, you could prove by induction. 00:52:50.020 --> 00:52:54.380 align:middle line:84% For equals 1, this is certainly true. 00:52:54.380 --> 00:52:56.320 align:middle line:84% Assume it holds for n equals m, and then 00:52:56.320 --> 00:53:00.460 align:middle line:84% prove all of these statements here for n equals m plus 1. 00:53:00.460 --> 00:53:04.000 align:middle line:90% 00:53:04.000 --> 00:53:08.170 align:middle line:84% But I'm just stating it as clear from the construction, 00:53:08.170 --> 00:53:09.070 align:middle line:90% and hopefully it is. 00:53:09.070 --> 00:53:12.070 align:middle line:90% 00:53:12.070 --> 00:53:15.260 align:middle line:90% So what does this give us? 00:53:15.260 --> 00:53:24.790 align:middle line:84% Well, by one, the sequence is a sub n 00:53:24.790 --> 00:53:31.740 align:middle line:84% and b sub n are bounded, because they're always 00:53:31.740 --> 00:53:42.170 align:middle line:84% bounded between a and b monotone sequences. 00:53:42.170 --> 00:53:46.190 align:middle line:84% a sub n is increasing, b sub n is decreasing. 00:53:46.190 --> 00:53:52.280 align:middle line:90% 00:53:52.280 --> 00:53:58.610 align:middle line:90% Thus, there exists limits. 00:53:58.610 --> 00:54:01.070 align:middle line:84% So I'm going to call them c and d. 00:54:01.070 --> 00:54:06.120 align:middle line:84% But it has nothing to do with the c and d from over here. 00:54:06.120 --> 00:54:14.095 align:middle line:90% Thus these sequences converge. 00:54:14.095 --> 00:54:19.570 align:middle line:90% 00:54:19.570 --> 00:54:25.530 align:middle line:84% I.e, there exists element c and d in a, b, 00:54:25.530 --> 00:54:31.380 align:middle line:84% such that limit as n goes to infinity of a sub n equals c-- 00:54:31.380 --> 00:54:34.560 align:middle line:84% again, the a sub n's and b sub n's are between a and b. 00:54:34.560 --> 00:54:36.990 align:middle line:84% So their limits will also be between a and b-- 00:54:36.990 --> 00:54:41.960 align:middle line:90% 00:54:41.960 --> 00:54:45.060 align:middle line:90% b sub n equals d. 00:54:45.060 --> 00:54:55.370 align:middle line:90% 00:54:55.370 --> 00:54:57.680 align:middle line:90% Now, I claim c equals d. 00:54:57.680 --> 00:55:00.080 align:middle line:84% Why should this not come as a surprise? 00:55:00.080 --> 00:55:02.310 align:middle line:84% Well, the a n's and bn's are getting very close. 00:55:02.310 --> 00:55:04.738 align:middle line:84% They're always-- the distance between any two of them 00:55:04.738 --> 00:55:05.780 align:middle line:90% is always getting halved. 00:55:05.780 --> 00:55:12.170 align:middle line:90% 00:55:12.170 --> 00:55:18.230 align:middle line:84% Now, a sub m minus b sub n, this equals a sub n minus 1, 00:55:18.230 --> 00:55:26.180 align:middle line:84% minus b sub n minus 2 over 2, which equals 1 over 2 squared, 00:55:26.180 --> 00:55:32.560 align:middle line:84% a sub n minus 2, minus b sub n minus 2, dot, dot, dot. 00:55:32.560 --> 00:55:39.790 align:middle line:84% And therefore, this equals 1 over 2n minus 1 b sub 1 00:55:39.790 --> 00:55:47.680 align:middle line:84% minus a sub 1, equals 1 over 2 to the n minus 1, b minus a. 00:55:47.680 --> 00:55:54.240 align:middle line:90% 00:55:54.240 --> 00:56:00.690 align:middle line:84% And therefore, c minus d, which is equal to the limit 00:56:00.690 --> 00:56:04.560 align:middle line:84% as n goes to infinity of a sub n minus 00:56:04.560 --> 00:56:11.100 align:middle line:84% b sub n equals the limit as n goes to infinity of 1 over 2 00:56:11.100 --> 00:56:14.740 align:middle line:90% to the n minus 1, b minus a. 00:56:14.740 --> 00:56:18.270 align:middle line:84% 1 over 2 to the n minus 1 is converging to 0, 00:56:18.270 --> 00:56:21.960 align:middle line:84% times this fixed number, b minus a, equals 0. 00:56:21.960 --> 00:56:24.720 align:middle line:84% And therefore, these two numbers, these two limits 00:56:24.720 --> 00:56:26.160 align:middle line:90% c and d equal each other. 00:56:26.160 --> 00:56:34.460 align:middle line:90% 00:56:34.460 --> 00:56:39.320 align:middle line:84% So this step here is by the second property here. 00:56:39.320 --> 00:56:41.912 align:middle line:90% 00:56:41.912 --> 00:56:43.245 align:middle line:90% And now, we're essentially done. 00:56:43.245 --> 00:56:46.530 align:middle line:90% 00:56:46.530 --> 00:56:58.312 align:middle line:84% Using the third property, by three and continuity, 00:56:58.312 --> 00:57:03.950 align:middle line:84% if I look at f of c, since the a sub n's are converging to c, 00:57:03.950 --> 00:57:04.880 align:middle line:90% this is equal to-- 00:57:04.880 --> 00:57:09.450 align:middle line:90% 00:57:09.450 --> 00:57:12.480 align:middle line:84% and f of a sub n is always less than 0. 00:57:12.480 --> 00:57:16.740 align:middle line:84% So its limit is less than or equal to 0. 00:57:16.740 --> 00:57:23.255 align:middle line:84% And now, remember, c is also equal to the limit of the b sub 00:57:23.255 --> 00:57:24.050 align:middle line:90% n's. 00:57:24.050 --> 00:57:31.670 align:middle line:84% And therefore, by continuity, f of c is equal to f of b sub n. 00:57:31.670 --> 00:57:34.250 align:middle line:84% And the f of b sub n's are always non-negative. 00:57:34.250 --> 00:57:37.370 align:middle line:84% So their limit is also bigger than or equal to 0. 00:57:37.370 --> 00:57:39.830 align:middle line:84% So I've shown that this number f of c 00:57:39.830 --> 00:57:42.380 align:middle line:84% is less than or equal to 0, and that it's also 00:57:42.380 --> 00:57:44.060 align:middle line:90% bigger than or equal to 0. 00:57:44.060 --> 00:57:48.260 align:middle line:90% And therefore, f of c equals 0. 00:57:48.260 --> 00:58:04.050 align:middle line:90% 00:58:04.050 --> 00:58:06.480 align:middle line:90% So we use the-- 00:58:06.480 --> 00:58:08.010 align:middle line:84% why I put the bisection method here 00:58:08.010 --> 00:58:09.570 align:middle line:84% is because we used a bisection method 00:58:09.570 --> 00:58:11.490 align:middle line:84% to prove this theorem, which looks 00:58:11.490 --> 00:58:20.320 align:middle line:84% like a kind of a special case of a certain intermediate value 00:58:20.320 --> 00:58:20.870 align:middle line:90% property. 00:58:20.870 --> 00:58:27.420 align:middle line:84% Meaning that I can always find something, that if I have-- 00:58:27.420 --> 00:58:30.540 align:middle line:84% that I can always find something in the interval that 00:58:30.540 --> 00:58:32.910 align:middle line:84% achieves something in between f of a and f of b. 00:58:32.910 --> 00:58:34.530 align:middle line:90% In this case, it's just 0. 00:58:34.530 --> 00:58:36.715 align:middle line:90% But I can in fact upgrade this. 00:58:36.715 --> 00:58:39.385 align:middle line:90% 00:58:39.385 --> 00:58:46.820 align:middle line:84% This is the following theorem due to Bolzano 00:58:46.820 --> 00:58:49.730 align:middle line:84% so Bolzano's intermediate value theorem. 00:58:49.730 --> 00:58:58.540 align:middle line:90% 00:58:58.540 --> 00:59:02.790 align:middle line:84% Let f from a to b, R be continuous. 00:59:02.790 --> 00:59:13.830 align:middle line:90% 00:59:13.830 --> 00:59:20.210 align:middle line:90% If y is between f of b-- 00:59:20.210 --> 00:59:23.330 align:middle line:84% so meaning, let's say, so we're in one of two cases. 00:59:23.330 --> 00:59:26.120 align:middle line:84% Let's say f of a is less than f of b. 00:59:26.120 --> 00:59:35.930 align:middle line:84% If y is between f of a and f of b, 00:59:35.930 --> 00:59:45.950 align:middle line:84% then there exists a c in a, b, such that f of c equals y. 00:59:45.950 --> 00:59:49.840 align:middle line:90% 00:59:49.840 --> 00:59:53.740 align:middle line:84% If f of a is, say, bigger than f of b, 00:59:53.740 --> 00:59:56.980 align:middle line:84% and we take something in between, 00:59:56.980 --> 00:59:58.090 align:middle line:90% the same conclusion holds. 00:59:58.090 --> 01:00:07.450 align:middle line:90% 01:00:07.450 --> 01:00:10.810 align:middle line:84% So in short, if I take anything between the value 01:00:10.810 --> 01:00:14.290 align:middle line:84% of the function evaluated at the two endpoints a and b, 01:00:14.290 --> 01:00:20.300 align:middle line:84% then the function achieves that value at some point. 01:00:20.300 --> 01:00:22.930 align:middle line:90% 01:00:22.930 --> 01:00:41.140 align:middle line:84% So if I take any y between here, f 01:00:41.140 --> 01:00:43.510 align:middle line:84% has to cross this horizontal line. 01:00:43.510 --> 01:00:47.590 align:middle line:84% And it's achieved at some point c. 01:00:47.590 --> 01:00:52.370 align:middle line:90% 01:00:52.370 --> 01:00:54.690 align:middle line:90% And we'll deduce this from-- 01:00:54.690 --> 01:00:59.710 align:middle line:84% so this looks like a special case of this intermediate value 01:00:59.710 --> 01:01:04.690 align:middle line:84% theorem, if f of a is less than 0, f of b is bigger than 0, 01:01:04.690 --> 01:01:06.340 align:middle line:90% and y is equal to 0. 01:01:06.340 --> 01:01:08.680 align:middle line:84% But in fact, you can reduce this general statement 01:01:08.680 --> 01:01:13.270 align:middle line:84% to that special case just by a simple trick. 01:01:13.270 --> 01:01:20.220 align:middle line:84% So proof, suppose f of a is less than f of b. 01:01:20.220 --> 01:01:23.040 align:middle line:90% So we're in this first case. 01:01:23.040 --> 01:01:41.740 align:middle line:84% And y is in f of a, f of b, let g of x be f of x minus y. 01:01:41.740 --> 01:01:43.960 align:middle line:84% This is not a function of x and y. y 01:01:43.960 --> 01:01:47.020 align:middle line:84% is a fixed number between f of a and f of b. 01:01:47.020 --> 01:01:49.800 align:middle line:90% 01:01:49.800 --> 01:01:55.910 align:middle line:84% And this is a function going from f to b to R, 01:01:55.910 --> 01:02:02.040 align:middle line:90% and which is continuous. 01:02:02.040 --> 01:02:04.980 align:middle line:84% So sum of two continuous functions is continuous, 01:02:04.980 --> 01:02:07.620 align:middle line:84% f is a continuous function by assumption. 01:02:07.620 --> 01:02:09.660 align:middle line:90% This is just a constant minus y. 01:02:09.660 --> 01:02:11.340 align:middle line:90% So their sum is also continuous. 01:02:11.340 --> 01:02:14.910 align:middle line:90% 01:02:14.910 --> 01:02:18.030 align:middle line:84% But now, if we look at g of a, this 01:02:18.030 --> 01:02:20.955 align:middle line:90% is equal to f of a minus y. 01:02:20.955 --> 01:02:24.110 align:middle line:90% 01:02:24.110 --> 01:02:27.890 align:middle line:84% Now, y is bigger than f of a, since y is in f of a, f of b. 01:02:27.890 --> 01:02:31.460 align:middle line:90% So this is less than 0. 01:02:31.460 --> 01:02:38.000 align:middle line:84% And g of b is equal to f of b minus y is bigger than 0. 01:02:38.000 --> 01:02:45.560 align:middle line:84% And therefore, by the previous theorem, 01:02:45.560 --> 01:02:53.430 align:middle line:84% there exists a c in a, b, such that g of c 01:02:53.430 --> 01:02:58.250 align:middle line:90% equals 0, which means i.e. 01:02:58.250 --> 01:02:59.960 align:middle line:90% f of c equals y. 01:02:59.960 --> 01:03:04.480 align:middle line:90% 01:03:04.480 --> 01:03:08.750 align:middle line:84% And the proof, assuming f of a is bigger than f of b 01:03:08.750 --> 01:03:12.950 align:middle line:84% is similar, you just now put y here minus f of x. 01:03:12.950 --> 01:03:16.520 align:middle line:84% Or you put-- now you define g of x would be y minus f of x. 01:03:16.520 --> 01:03:20.390 align:middle line:84% So let me just make a remark about that. 01:03:20.390 --> 01:03:29.420 align:middle line:90% 01:03:29.420 --> 01:03:45.740 align:middle line:84% Is similar, look at g of x equals y minus f of x. 01:03:45.740 --> 01:03:53.190 align:middle line:90% 01:03:53.190 --> 01:04:00.320 align:middle line:84% So Bolzano's intermediate value theorem tells us 01:04:00.320 --> 01:04:10.690 align:middle line:84% that if I take any value between the function evaluated 01:04:10.690 --> 01:04:15.760 align:middle line:84% at the two endpoints, then that value is achieved by something 01:04:15.760 --> 01:04:17.005 align:middle line:90% in between a and b. 01:04:17.005 --> 01:04:20.050 align:middle line:90% 01:04:20.050 --> 01:04:24.610 align:middle line:84% And therefore, this will, in fact, give us that f of a, b 01:04:24.610 --> 01:04:28.330 align:middle line:84% is equal to f of c, f of d for some c and d. 01:04:28.330 --> 01:04:30.610 align:middle line:84% So this is a simple theorem from this. 01:04:30.610 --> 01:04:41.544 align:middle line:90% 01:04:41.544 --> 01:04:57.890 align:middle line:84% Is continuous if f local n at c-- 01:04:57.890 --> 01:05:01.520 align:middle line:90% 01:05:01.520 --> 01:05:15.100 align:middle line:84% not local, absolute min at c and absolute max 01:05:15.100 --> 01:05:34.365 align:middle line:84% at d, then the image by f of a, b is equal to f of c, f of d. 01:05:34.365 --> 01:05:35.240 align:middle line:90% And what's the proof? 01:05:35.240 --> 01:05:46.740 align:middle line:84% Well, apply Bolzano's intermediate value theorem 01:05:46.740 --> 01:05:53.160 align:middle line:84% to f now going from the possibly smaller interval, c, d. 01:05:53.160 --> 01:05:59.120 align:middle line:84% It could be d, c or c, d, depending on 01:05:59.120 --> 01:06:03.200 align:middle line:84% if the thing that I have to stick in to f to get the min 01:06:03.200 --> 01:06:05.502 align:middle line:84% and the thing I have to stick into f to get the max 01:06:05.502 --> 01:06:06.965 align:middle line:90% is smaller than the other. 01:06:06.965 --> 01:06:13.680 align:middle line:90% 01:06:13.680 --> 01:06:17.600 align:middle line:84% So apply the Bolzano's intermediate value theorem 01:06:17.600 --> 01:06:19.190 align:middle line:84% to this function now just restricted 01:06:19.190 --> 01:06:20.840 align:middle line:90% to the smaller closed interval. 01:06:20.840 --> 01:06:23.340 align:middle line:90% It's still continuous. 01:06:23.340 --> 01:06:24.750 align:middle line:90% And that's the proof. 01:06:24.750 --> 01:06:39.600 align:middle line:90% 01:06:39.600 --> 01:06:41.567 align:middle line:84% So maybe I was a little too fast with that. 01:06:41.567 --> 01:06:47.745 align:middle line:84% So let me just say a couple of more words about this. 01:06:47.745 --> 01:06:53.520 align:middle line:90% 01:06:53.520 --> 01:07:04.840 align:middle line:84% So this tells me that everything between the two values 01:07:04.840 --> 01:07:11.770 align:middle line:84% f of c and f of d is achieved by f on this interval. 01:07:11.770 --> 01:07:15.310 align:middle line:90% 01:07:15.310 --> 01:07:18.790 align:middle line:84% But this is a possibly smaller interval than a, b. 01:07:18.790 --> 01:07:29.570 align:middle line:84% So this is contained in the range of f now 01:07:29.570 --> 01:07:31.685 align:middle line:90% of f of this larger interval. 01:07:31.685 --> 01:07:34.460 align:middle line:90% 01:07:34.460 --> 01:07:37.070 align:middle line:84% And since we already know that f of a, b 01:07:37.070 --> 01:07:41.060 align:middle line:84% is contained in f of c, f of d, and now we've 01:07:41.060 --> 01:07:43.490 align:middle line:84% proven the reverse inclusion, this 01:07:43.490 --> 01:07:54.285 align:middle line:84% implies that f of c, f of d equals f of a, b. 01:07:54.285 --> 01:07:55.160 align:middle line:90% And that's the proof. 01:07:55.160 --> 01:07:59.040 align:middle line:90% 01:07:59.040 --> 01:08:00.850 align:middle line:90% So I think we'll stop there. 01:08:00.850 --> 01:08:08.160 align:middle line:84% Next time, we'll finish with a few remarks 01:08:08.160 --> 01:08:13.710 align:middle line:84% about an application of the intermediate value theorem. 01:08:13.710 --> 01:08:17.270 align:middle line:84% And then we'll move on to uniform continuity. 01:08:17.270 --> 01:08:24.000 align:middle line:90%