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PROFESSOR: So we're
finally ready to tie up

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these ideas about
mapping properties

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to counting properties.

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And let's begin with
the remark we already

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made, that if there's a
bijection from one set

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to another, if there's a
bijection from a finite set

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A to a finite set B, then
A and B have the same size.

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By the way, those vertical
bars, absolute value

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of A, when it's a set
refers us to the size

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of the set for finite sets.

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So it means if A has n elements,
then the absolute value of A

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is n.

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OK.

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Let's use this bijection
idea immediately.

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Here's a nice example.

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Suppose you want to figure
out how many subsets

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are there of a finite set A?

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Suppose you didn't know yet.

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All right.

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So what we're asking
about is, what's

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the size of the power set of A?

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Remember the power
set of A is all

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of the possible subsets
of the set A. Well suppose

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A has three elements.

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What is the power
set of A look like?

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If capital A has
elements little a,b,c.

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It's a set of size three.

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Then the power set
of A is going to have

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one subset with no elements,
three subsets with one element,

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they're listed there.

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It's going to have three more
subsets with two elements,

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and one subset with
three elements.

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For a total of eight subsets
in the power set of A.

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So let's see what the counting
argument in general would be.

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So suppose that
A has n elements.

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And we'll number them from
a 0 up through a n minus 1,

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because computer scientists
usually zero origin index.

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So A is a set of n
elements, indicated there.

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Suppose I have some arbitrary
subset of A. Let's suppose

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that it has a 0,
then there's no a 1,

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that's what this space is for.

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It has a 2.

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It has a 3.

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Doesn't have a 4.

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And it goes on for awhile
in some uncertain way.

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And it ends it actually
has the last element a

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and minus 1 in it.

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Well I can take the
subset like this

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and have it correspond
to a bit-string,

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where I put a 1 where the
element is there in the subset,

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and a 0 where the element
is not there in the subset.

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So there's a 1 because a 0 is
in the subset followed by a 0,

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because a 1 is not in the
subset, followed by two 1's

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because a 2 and a 3 are in
the subset, followed by a 0

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because a 4 is not, and
so on, ending with a 1,

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because a n minus
1 is in the subset.

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So in short, the bit-string the
[? case ?] bit in the string

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is 1, if and only if,
a sub k is in the set.

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Now this clearly
defines a bijection

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between subsets and strings.

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Because given a subset, I can
uniquely determine the string,

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and given the string, I can
uniquely determine the subset.

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So there's one arrow
in and one arrow out

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of each of those things.

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So I could immediately conclude
then by this bijection theorem

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that the number of
n bit-strings is

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equal to the size of the
power set of A. Well,

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now every computer
scientist knows how many

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n bit-strings there are, right?

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There are 2 to the
n n-bit bit-strings.

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And therefore, the power set
of A has 2 to the n elements.

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We can get rid of the n
because it was the size of A.

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We have this nice formula.

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The power set-- the size of
the power set of a finite set A

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is 2 to the size of A. Well
there are some more counting

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rules like the
bijection rule that

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relates sizes of
sets by inequalities

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according to whether
the mappings are

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surjective injective
functions and so on.

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So let's take a
quick look at those.

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Suppose that I have a
surjective function from A

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to B. Well that means there's
less than or equal to 1

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arrow out of every element of
A, that's what function means.

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And there is greater
than or equal to 1 arrow

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in to every element of B.
That's what surjective means.

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So here's this function from
A to B. That means less than

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or equal to 1 arrow out.

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And that means that there
have to be more elements in A

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than there are
arrows since there's

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at most, one arrow out
of every element of A

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and that accounts
for all the arrows.

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So the size of A has
to be greater than

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or equal to the
number of arrows when

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you have a partial function.

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When it's a surjection,
that means that there's

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an arrow coming into
every element of B.

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That means that there have
to be at least as many arrows

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as there are in B.
Because everything in B

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has at least one
arrow coming in.

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So if you look at these
two pieces, the size of A

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has to be greater than or
equal to the number of arrows.

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And the size of B
has to be less than

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or equal to the
number of arrows.

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Putting them together,
we have the mapping rule

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for surjections.

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If I have a surjective
function from A to B

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it means the size
of A is greater than

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or equal to the size of
B for finite A and B.

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The same argument
goes for injections.

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With an injection, I have less
than or equal to 1 arrow in.

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And what does that tell me?

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Well if it's total, then
there's at least 1 arrow out

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of everything.

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And that means that the size
of A has to be less than

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or equal to the
number of arrows.

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Because every element
in A contributes

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an arrow, maybe more than one.

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OK.

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If it's an injection,
there's less

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than or equal to 1 arrow
coming into each element of B.

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That means that B had
better be big enough

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to catch all the arrows.

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So the number of
arrows has to be

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less than or equal
to the size of B.

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Put those two
inequalities together

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and you find that if you have a
total injective relation from A

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to B, that implies
that the size of A

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is less than or equal to the
size of B for finite A and B.

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The statement here that it's
a totally injective relation

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is for generality.

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But the truth is,
whenever there's

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a total injective
relation there's also

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a totally injective function.

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I'll leave that to a class
exercise to work out.

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So to summarize,
what we can do is

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characterize these kinds of
jection relations between sets.

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So let's define A
bij B. So this is

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going to be a binary
relation between two

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sets A and B. And its meaning is
that there's a bijection from A

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to B. So the definition of
bij is a binary relation where

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the domain and the co domain
are the class of finite sets

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or all sets for that matter.

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surj B, A surj B
means that there

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is a surjective function
from A to B. Again,

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surj is a binary
relation between sets.

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And finally, A inj
B says, there's

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a total injection
relation from A to B.

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So we have those three
relations between sets.

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And what we've just
shown is that if there's

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a bijection between
A and B, that's true,

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implies that the size of A
is equal to the size of B.

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And in fact, it's not hard
to prove the converse.

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If the size of A and the
size of B are the same,

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then it's easy enough to find
a bijection between them.

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Similarly, if there's a
surjection from A to B,

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that's true, if and
only if, the size of A

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is greater than or
equal to the size of B.

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And if there's an
injection from A to B,

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that's true, if and
only if, the size of A

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is less than or equal
to the size of B.

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So this-- the existence of these
kinds of relations between sets

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is a handle on their
relative sizes.

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And this applies for
finite sets A and B.

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Because we really know what
the size of infinite sets are.

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So it wouldn't make sense
to talk about the mapping

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Lemma for infinite sets.

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However, we can ask some
questions about infinite sets.

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So one natural question
to ask is, let's look

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at some finite properties.

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If the size of A and B are the
same and the size of B and C

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are the same, obviously, the
size of A and C are the same.

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And that corresponds
directly to an insertion

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about bij because the size
of A equals the size of B,

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for finite sets, is
the same as A bij B.

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So what this says is
that if A bij B and B

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bij C, then A bij C. If
there's a bijection from A to B

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and a bijection from
B to C, then there's

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a bijection from A to C.
Corresponding property

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for finite sets, again, is
greater than or equal to-- if A

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is greater than or equal to
B greater than or equal to C,

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then A is-- the
size of A is greater

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than or equal to the size of
C. And that would correspond

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to a similar statement about
the surjection relation,

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A surj B, B surj C,
implies A surj C.

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And finally, more
interesting one,

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is that if A and B
are each greater than

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or equal to each other, if A
is greater than or equal to

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B and B-- the size of B is
greater than or equal to A,

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then in fact, they're
the same size.

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A is equal to B.
That would correspond

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to the following statement
in terms of jections,

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if A surj B and B surj A, then
A bij B. Now all of these facts

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follow immediately for
finite sets A and B.

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But we're going to be
interested in whether they

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hold for infinite sets.

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So they follow immediately for
finite sets from the Mapping

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Lemma because the Mapping Lemma
says that these bij and surj

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relations are the same
is equal and greater

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than or equal to [INAUDIBLE].

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So again, this is immediate
from the Mapping Lemma.

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But now I can ask whether
or not these same properties

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hold for infinite sets.

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00:10:02,610 --> 00:10:05,550
It's an exercise in
reasoning about properties

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of mappings and relations.

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And the answer is that the first
two claims go through easily

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if you're trying to prove
them for finite sets,

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00:10:16,170 --> 00:10:17,800
for infinite sets,
the basic ideas.

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Let's look at the first one.

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00:10:18,966 --> 00:10:21,900
It says that if A-- If there's
a bijection from A to B

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and there's a
bijection from B to C,

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00:10:23,830 --> 00:10:27,110
then there ought to be one from
A to C. How do you find it?

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00:10:27,110 --> 00:10:28,110
Well it's actually easy.

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00:10:28,110 --> 00:10:30,220
You just compose
the bijection from A

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00:10:30,220 --> 00:10:33,170
to B with the
bijection from B to C.

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00:10:33,170 --> 00:10:36,620
And it's a very easy
exercise that the composition

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of bijection is a bijection.

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Likewise, the composition of
surjections is a surjection.

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And that proves the
second claim easily.

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00:10:44,680 --> 00:10:49,957
But the third claim is much more
interesting and is not obvious.

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00:10:49,957 --> 00:10:51,790
It's called the
Schroeder-Bernstein theorem.

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00:10:51,790 --> 00:10:56,290
And it will come up a little
bit in a few more lectures.