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YUFEI ZHAO: Let us
look at an application

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of the probabilistic method
to extremal set theory.

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The extremal set theory
concerns the study

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of families of sets that have
certain desirable properties,

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and asking questions such
as what is the largest set

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family that you can have
with such families--

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with such properties?

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And in this video, we'll focus
on the following property

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of being an intersecting family.

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This means a collection
of sets, A1 through Al,

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such that these sets
pairwise intersect

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in a nonempty intersection.

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So we want sets A1 through
Al, such that whenever

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you take any two of these sets
and take their intersection,

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it is never the empty set.

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So here are two basic
questions that one can ask.

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The first one, which will turn
out to be an easy warm up,

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is what is the largest
intersecting family

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of subsets of 1 through n.

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So here, there are, in total, 2
to the n subsets of 1 through n.

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But if I want to only keep
a collection of sets that's

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an intersecting family, what is
the largest family I can get?

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And the second question,
which will turn out

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to be far more interesting,
will involve an application

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of the probabilistic
method, is a question,

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what is the largest intersecting
family of K element subsets of 1

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through n?

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So here, n and K are
inputs to the question.

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So they are given.

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And we're asked to find what
is the largest intersecting

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family of K element
subsets of 1 through n.

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All right.

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So the first question,
as I mentioned earlier,

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turns out to be a
fairly easy one,

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and we'll solve it as a warm.

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So as an example,
just to illustrate

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what we're talking
about, if we look at all

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sets containing the element 1.

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Well, this collection of sets
is intersecting because they all

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have the element 1.

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So every pair intersects in some
set that contains the element

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1 in particular non-empty.

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This collection of sets has
size 2 to the n minus 1,

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because once you
fix the element 1,

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the other elements can
be in or not in the sets,

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and there are 2 to the n
minus 1 such possibilities.

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So a binary choice for each
of the elements outside 1.

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OK.

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Now this is just an example.

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But it turns out this is the
best that we can do, because--

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so the claim's that
we cannot do better,

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or we cannot get
an even larger set.

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And this is because for every
subset A of 1 through n,

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and most 1 of A and
the complement of A,

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so the elements 1
through n subtract A,

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so the complement of A.
And most one of these

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can be in the
intersecting family.

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So what we're doing here
is pairing up the sets

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with their complements.

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And because it's an
intersecting family,

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you cannot simultaneously have
a set and its complement in this

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intersecting family.

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And so this eliminates--

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so this means that you can
have at most half of all the

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2 to the n sets, thereby showing
that this 2 to the n minus 1

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is indeed the best
that you can do.

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OK.

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So this finishes part
one, question one.

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And now let's move on to the
more interesting and harder

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question two.

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Again, let's start
with some examples.

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First, here's an easy case.

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If n is less than 2K, then--

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so this is the easy setting.

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If n is less than 2K then
by pigeonhole principle,

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any pair of K element subsets
of 1 through n intersect.

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So we can take all
and choose K sets,

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and they form an
intersecting family.

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So there's nothing, really, to
do here, because we can just

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take all the K element sets.

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So the harder part
of the question,

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the more interesting part, is
what happens when n is at least

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2K?

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Here, as an example, we can
take as earlier all sets

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containing the element 1.

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And this gives us n minus
1 choose K minus 1 sets.

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And that's before
the sets, they all--

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this set form-- this
collection of sets

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form an intersecting
family because any pair

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their intersection contains the
element 1, and is thus nonempty.

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So this is the construction
and gives you n minus 1

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choose K minus 1 sets.

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But is this the best you can do?

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Perhaps through some other
methods of construction,

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through other
examples, maybe there

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are even larger intersecting
families of K element subsets.

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That turns out not
to be the case.

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And that is the main theorem
that we'll prove in this video.

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And this result is known as
the Erdos-Ko-Rado theorem.

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So this is a seminal
and beautiful result

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in extremal set theory.

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And The Erdos-Ko-Rado theorem
says that if n is at least 2K,

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then any intersecting family,
and we'll give this family

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a name, Curly F, of
subsets of 1 through n.

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OK, subsets and any intersecting
family of K element subsets

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of one through n has size at
most n minus 1 choose k minus 1.

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In other words, the construction
that we gave just now

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is indeed optimal.

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You cannot get a bigger
intersecting family of k element

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subsets of 1 through n.

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Let us now prove the
Erdos-Ko-Rado theorem.

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And we'll introduce
randomness into the problem

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and use some beautiful ideas
from the probabilistic method.

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So here's how we
start the proof.

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Let us order the numbers 1,
2, 3, and so on through n

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randomly around a circle.

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So what do I mean by this?

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Take a circle and
then basically take

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a uniform random
permutation of 1 through n

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and then place the numbers
around the circle in the order

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of that permutation.

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So for example, I'm
putting down nine numbers,

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here, n equals to nine,
in some circular order

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chosen uniformly at random.

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I'm missing a number here.

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Six.

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So here, nine numbers.

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Uniformly around the circle.

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And let us call a subset
of 1 through n contiguous.

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Here, I'm defining
the term "contiguous."

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I'll call the subset contiguous
if its elements, the way they're

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ordered around the circle,
form an arc in this ordering.

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So for example, the set
of numbers 4, 3, 9, 5,

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this four-element
set is contiguous.

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So 3, 4, 5, 9 is contiguous.

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Because the order's there--

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according to the
circular ordering,

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these four numbers form
a contiguous block.

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On the other hand, the
set 1, 3 is not contiguous

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because they do not
form a contiguous block.

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So that's just a definition.

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And now for a given set,
sub A, a subset of 1

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through n with k elements.

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So for a given k
element subset of 1

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through n, what is
the probability?

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So here, A is fixed.

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A is not random.

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So we're given this A.
What is the probability

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that A is contiguous?

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So the ordering is random.

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So given a set, what
is the probability

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that under this random
ordering is contiguous?

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Well, let's think about that.

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So the ordering is random,
but here, if A has k elements,

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you can think about how
many different ways these k

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elements can be placed
around the circle.

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There are n different positions
for where geometrically,

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this length k arc lies.

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And once you fix that
position, the probability

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that the elements,
the k elements of A

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actually falls into the
desired positions is 1 over k--

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1 over n choose k.

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So this is the probability
that A is a contiguous

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set under this random ordering.

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And now by the linearity
of expectations,

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the expected number
of contiguous sets

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in this collection F
is equal to the size

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of F times the probability
that each individual set

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is contiguous.

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And that was calculated
earlier to be this quantity.

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So n divided by n choose k.

00:12:50.260 --> 00:12:53.250 align:middle line:90%


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On the other hand, the property
of F being intersecting,

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I claim, implies that in
any given circular ordering

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there are at most k
contiguous sets for--

00:13:37.370 --> 00:13:42.020 align:middle line:90%
so in a given circular ordering.

00:13:42.020 --> 00:13:45.380 align:middle line:84%
So let's pause and think
about why this claim is true.

00:13:45.380 --> 00:13:49.460 align:middle line:84%
If I give you an
ordering and ask, well,

00:13:49.460 --> 00:13:52.140 align:middle line:84%
how many contiguous
sets can there be?

00:13:52.140 --> 00:13:56.990 align:middle line:84%
And now these contiguous sets,
they have to be intersecting.

00:13:56.990 --> 00:14:02.450 align:middle line:84%
So if we have one set-- so here,
let me denote the elements,

00:14:02.450 --> 00:14:04.400 align:middle line:90%
these circles as--

00:14:04.400 --> 00:14:07.880 align:middle line:84%
think about them as
positions on the circle.

00:14:07.880 --> 00:14:09.950 align:middle line:84%
So these dots as
positions on the circle.

00:14:09.950 --> 00:14:13.730 align:middle line:84%
So if that's one set, well,
where can the other sets be?

00:14:13.730 --> 00:14:16.230 align:middle line:84%
Well, they have to
intersect this block.

00:14:16.230 --> 00:14:21.420 align:middle line:84%
So maybe another set
is like this one.

00:14:21.420 --> 00:14:25.560 align:middle line:84%
But if that set is
in, then it excludes--

00:14:25.560 --> 00:14:28.980 align:middle line:84%
if this blue set is in,l then
it excludes this green set from

00:14:28.980 --> 00:14:32.150 align:middle line:90%
being a possibility and so on.

00:14:32.150 --> 00:14:50.720 align:middle line:90%


00:14:50.720 --> 00:14:51.220 align:middle line:90%
OK.

00:14:51.220 --> 00:14:58.600 align:middle line:84%
So you see that given that this
red set is in, well, at most,

00:14:58.600 --> 00:15:01.780 align:middle line:84%
one of the blue and green on
the second line can be in.

00:15:01.780 --> 00:15:04.210 align:middle line:84%
And at most, one of the blue
and green on the third line

00:15:04.210 --> 00:15:04.900 align:middle line:90%
can be in.

00:15:04.900 --> 00:15:07.240 align:middle line:84%
And at most, one of blue
and green on the fourth line

00:15:07.240 --> 00:15:08.000 align:middle line:90%
can be in.

00:15:08.000 --> 00:15:10.150 align:middle line:84%
And these are the
all the possibilities

00:15:10.150 --> 00:15:15.670 align:middle line:84%
of how another contiguous set
can intersect the red set.

00:15:15.670 --> 00:15:19.990 align:middle line:84%
So that's an argument that
can translate into this claim,

00:15:19.990 --> 00:15:22.210 align:middle line:84%
that F being
intersecting implies

00:15:22.210 --> 00:15:27.100 align:middle line:84%
at most k contiguous
sets in any given order.

00:15:27.100 --> 00:15:30.052 align:middle line:84%
So this argument is very
analogous to the argument

00:15:30.052 --> 00:15:31.510 align:middle line:84%
that we did at the
beginning, where

00:15:31.510 --> 00:15:33.693 align:middle line:90%
we pair up complementary sets.

00:15:33.693 --> 00:15:35.860 align:middle line:84%
But this time, we're only
looking at contiguous sets

00:15:35.860 --> 00:15:38.810 align:middle line:90%
that intersect a given set.

00:15:38.810 --> 00:15:40.220 align:middle line:90%
OK.

00:15:40.220 --> 00:15:44.600 align:middle line:84%
Well, if you can only get
at most k contiguous sets

00:15:44.600 --> 00:15:48.440 align:middle line:84%
in any given order, then
this quantity, which

00:15:48.440 --> 00:15:52.420 align:middle line:84%
is a random variable,
it is always at most k.

00:15:52.420 --> 00:15:55.640 align:middle line:90%


00:15:55.640 --> 00:16:00.560 align:middle line:84%
So in expectation, then, this
quantity that we got at the end

00:16:00.560 --> 00:16:04.030 align:middle line:90%
is also at most k.

00:16:04.030 --> 00:16:06.830 align:middle line:90%


00:16:06.830 --> 00:16:15.840 align:middle line:84%
Thus the size of F is
at most k over n times

00:16:15.840 --> 00:16:22.370 align:middle line:84%
n choose k, which one can then
expand the binomial coefficient

00:16:22.370 --> 00:16:27.990 align:middle line:84%
to see that the final
quantity is equal to n minus 1

00:16:27.990 --> 00:16:29.900 align:middle line:90%
choose k minus 1.

00:16:29.900 --> 00:16:31.780 align:middle line:84%
And that's exactly
what we claimed.

00:16:31.780 --> 00:16:36.370 align:middle line:90%


00:16:36.370 --> 00:16:36.870 align:middle line:90%
All right.

00:16:36.870 --> 00:16:40.190 align:middle line:84%
So this finishes the proof of
the Erdos-Ko-Rado theorem, which

00:16:40.190 --> 00:16:43.200 align:middle line:84%
is a foundational result
in extremal set theory.

00:16:43.200 --> 00:16:46.400 align:middle line:84%
And it's a beautiful application
of the probabilistic method

00:16:46.400 --> 00:16:49.800 align:middle line:84%
to extremal set theory and
to combinatorics in general,

00:16:49.800 --> 00:16:54.570 align:middle line:84%
where we start with a claim,
a theorem, that is not random

00:16:54.570 --> 00:16:55.070 align:middle line:90%
at all.

00:16:55.070 --> 00:16:56.690 align:middle line:90%
It is completely deterministic.

00:16:56.690 --> 00:16:59.450 align:middle line:84%
It's about finding the
maximum possible set

00:16:59.450 --> 00:17:03.050 align:middle line:84%
size, the size of something
that satisfies some constraints.

00:17:03.050 --> 00:17:07.220 align:middle line:84%
And yet the proof goes by
introducing new randomness

00:17:07.220 --> 00:17:08.599 align:middle line:90%
into the setup.

00:17:08.599 --> 00:17:13.570 align:middle line:84%
That allows us, then, to derive
this beautiful conclusion.

00:17:13.570 --> 00:17:20.000 align:middle line:90%