0:00
Hi.
00:01
In this question, we have this collection, which is just the first n integers, positive integers.
00:11
And so if we have a collection of subsets of c, which has this property that any two pairs has at least one element in common, we want to prove that two to the end, there's at most two to the n minus one subsets in this collection.
00:28
So let's say s is our collection of subsets with this property.
00:38
And we want to show that s is less than or equal to 2 to the n minus 1.
00:49
So the way that i want to prove this is by saying, well, how many, what is the total number of subsets of c? so sometimes that's denoted by p of c, the power set.
01:01
And that's the set of all subsets of c.
01:05
And you might know this fact that the number of elements is 2 to the n, 2 to the number of elements of c, which in our case is n.
01:15
So if we want to show that s has less than 2 to the n minus 1 elements, we have to show that half of the elements can't be in here.
01:28
And the logic we can use to see why this is true is we can just, let's call a, we can define a bijection.
01:48
So here let's say this.
01:52
Let's let d, or not d, what did i call it? so s is the collection of subsets.
02:00
So let t be equal to p of c set minus s.
02:07
So in other words, the subsets of c that are not in s.
02:19
And so it's clear that the number of elements of t plus the number of elements of s is equal to the number of elements in the power set, which is 2 to the n.
02:32
Now what we want to do is we want to define a bijection between, so i'll call it f from, well, it doesn't really matter, we'll go from s to t.
02:53
And what we do is we just say, well, f of little s, so s is going to be some subset.
03:01
Of c and what do we know then if s is in if little s is in s then that means its complement can't be in it can't be in s right because its complement won't share any elements with it so f of s we can just define to be c set minus s maybe let me use a different variable f of a c minus a right and this is well defined by what i just said so c set minus s so c set minus a, it must be in t since, i'll just say by property, right, by that they, any two sets in s must share, must share an element.
03:58
So the complements, they don't share any elements, so they, so c set minus a must be in t.
04:06
Now we have to show it's an actual bijection.
04:08
We showed it's well defined...