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PROFESSOR: Cardinality
is the word

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that's used to refer to
the size of infinite sets.

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And before we go further,
let's take a quick look

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at why we're interested
in infinite sets

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in a course that's Mathematics
for Computer Scientists.

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Why does computer science
care about infinite sets?

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Well, like every data
structure that you'd

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examine in computer
memory is finite

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and the integers
individually are finite,

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you only calculate
with finite things.

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But, the infinite abstraction
happens right at the beginning.

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Although any given integer is
finite, the set of all integers

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is infinite.

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And although any given
matrix is finite,

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the set of all the
matrices that might

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be represented in a computation
are an infinite set.

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So, we take infinite sets
for granted and reason

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about them all the time.

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The second, from a
pedagogical point of view,

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introducing the concept
of infinite sets

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and reasoning about
them carefully

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forces you to go
beyond your intuition

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and really follow
the rules and reason

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in a careful mathematical way.

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Because although some properties
that you're familiar with from

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finite sets carry over to
infinite sets, others don't.

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And in order to
know which is which,

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you have to be thinking
carefully about the rules

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and properties that
they have, as opposed

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to just going by intuition
and familiar properties.

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And finally, the
reasoning that goes

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into comparing the sizes
of infinite sets, which

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is the topic of today's video,
has profound implications

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in computer science
because it leads

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to the insight about the
logical limits of computation

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and examples of
specific problems

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that computers can't
solve, which we'll

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be taking up in a later video.

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But for now, let's go back
to the topic of cardinality.

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So, there was this mathematician
in the late 19th century

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named Cantor who was actually
working on Fourier's series.

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And he discovered that
the kind of series

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that he was working
with diverged

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in infinitely many places,
which sounds kind of bad.

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But he wanted to
get across the idea

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that it didn't diverge in
very many infinite places.

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And that led him to
this idea of comparing

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the sizes of infinite sets.

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So, this is Cantor's idea.

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We know from the mapping
dilemma that if you're

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looking at finite sets A
and B, then the size of A

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is greater than or
equal to the size of B

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if and only if A
surj B, were surj

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is this technical relation which
means there exists a surjection

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function from A to B that
is a function with greater

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than or equal to one arrow
into every element of B.

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And Cantor's idea
was saying, well,

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it works fine for finite sets.

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Why don't we take this as the
definition of what we mean by A

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is at least the size
of B for infinite sets?

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So, we're going to
think of A surj B

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now as saying, A is as big
as B. And for finite sets,

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it's literally true that A surj
B if and only if the size of A

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is greater or equal
to the size of B.

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Now, let me take a moment to
say that this notion of size

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or cardinality, when you're
talking about infinite sets,

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it's kind of a no-no.

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There's an abstract concept of
what cardinal numbers are, what

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these infinite numbers are.

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But the truth is, they're
technical and not of very much

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use.

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So, we will never actually be
talking about the cardinality

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or size of an infinite set.

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But what we will
do is compare them.

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We're going to have a nice
elementary theory of the idea

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that the cardinality
of one set is

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greater than or equal to the
cardinality of another set.

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And the basic definition is
going to be based on surj.

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Similarly, bijection
is even easier.

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A bij B means that there's
a bijection from A to B.

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And we're going
to interpret that

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as saying that A and
B are the same size.

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That is, for finite sets,
it literally means A

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and B have the same
number of elements.

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We're going to adopt the
notion of a bijective relation

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for infinite sets as meaning, I
don't know what their size is,

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but I know it's the
same because there's

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a bijection between them.

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There's a perfect one to one
correspondence between As

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and Bs.

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Let's look at an example of
where bijection comes up.

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The power set of N-- if N is
the non-negative integers--

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the power set of N
is all the subsets

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of non-negative integers.

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And let me just remark that
there is an obvious bijection

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between the subsets of integers
and the infinite bit-strings,

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the infinite strings
of zeroes and ones.

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So, N is the set of
non-negative integers, 0, 1, 2.

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If you take any subset of
N, here's one with has 0,

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missing 1, has 2 and
3, missing 4, 5, has 6,

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and so on, then what I can do
is represent such a subset,

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possibly an infinite subset
now, by an infinite sequence

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of ones and zeros.

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Put in ones in the position
where elements in the subset

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occur and zeroes in positions
where elements don't occur.

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This was exactly
the same bijection

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that we had found between the
non-negative-- the bit-strings

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and the finite subsets of
the non-negative integers.

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But now, we're just extending
it to arbitrary subsets

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of the non-negative integers.

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So, this defines a
bijection between any subset

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of integers corresponds to
an infinite bit-strings.

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And conversely, from
any infinite bit-string,

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you can reconstruct what
subset it refers to.

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So, we use this notation
[? 0,1 ?] to the omega,

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meaning the infinite bit-strings
that are infinite to the right.

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They have a beginning.

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In comparison to [? 0,1 ?]
superscript star,

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which refers to the finite
sets of bit-strings.

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So now, let's examine the
standard size properties

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that you would expect if these
relationships of surj and bij

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behaved like relationships
between sizes.

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So, one basic property
that finite sizes have

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is that if A is equal to B
and B is equal to C in size,

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then the size of A and the
size of C are the same.

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That's certainly
true for finite sets.

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Does it hold for infinite
sets, where now equality is

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going to be replaced by bij?

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Well, we have to check it.

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Is it true that if A bij B and
B bij C implies A bij C, well,

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how do you prove that?

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Well, it's true, and here's how.

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By definition,
since A bij B, that

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means that you have a
bijection g from A to B.

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And since B bij C, you have a
bijection f from B to C. Now,

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I need from these two
bijections that I'm

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given, I need to find a
bijection between A and C.

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Well, that's easy.

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What you do is you use
g to go from A to B,

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and then you use f to go from
B to C. And you compose them,

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and that gives you the needed
bijection from A to C. So,

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define h to be the
composition of f and g.

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And it's easy to check that if
g and f are bijections, then

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their composition
is a bijection.

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So, that's how I find the needed
bijection from A to C. So,

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this property works
out just fine.

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The similar property applies
to at least as big as,

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greater than or equal to.

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For finite sets, if A is
greater than or equal to B

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and B is greater than
or equal to C in size,

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then A is greater
than or equal to C.

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And actually, the same
argument that worked for bij

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works for surj,
because the composition

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of surjective functions
is a surjective function.

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So, if A surj B and B
surj C implies A surj C.

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Now again remember, although
we're thinking of surj

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as meaning greater than
or equal to in size,

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you cannot take these size
properties for granted.

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They have to be proved.

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Surj has a technical
definition having

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to do with surjective functions,
functions with greater

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than or equal to one arrow in.

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That is not the same as
talking about equality

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of some kind of sizes.

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Well, let's look at an example
where the size properties hold

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but they're less obvious.

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Because here's another
familiar size property.

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If A and B are each
of size greater than

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or equal to the other one,
then they're the same size.

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So, if the size of A is greater
than or equal to the size of B

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

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then A and B are the same size.

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Now, this is certainly
true for finite sets.

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It's kind of, you don't
even think about that fact.

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And it holds for infinite sets.

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But, it's not so obvious.

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So, what we're saying is that
if I have a surjective function

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from A to B and I have another
surjective function from B

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to A, then there's a
bijection between A and B.

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And the problem here is that
this surjection from A to B

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might not be a bijection.

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And this surjection from B to A
might also not be a bijection.

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So, where's the bijection
going to come from?

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I have to build it.

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And so, this is not an
obvious property, it's true.

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It's called the
Schroeder-Bernstein Theorem.

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And the trick, basically, is you
take the bijection from A to B

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and the bijection from B to
A and you take parts of one

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and combine it with
parts of the other.

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And in a slightly
ingenious way that

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actually is contained in
a problem in the text,

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you can find the
bijection from A to B,

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but it does take a
little bit of ingenuity.

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So this is a size property
that works for surj and bij,

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but you can't say it's obvious.

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Well, let's look at an
unfamiliar size property.

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Something that's not
true of finite sets

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where we have to start being
careful and not just hand wave

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and use our intuition
about finite sets.

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Namely, for infinite sizes,
size plus 1 is equal to size.

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Now, what exactly
does that mean?

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Well, let's just illustrate
it with an example.

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In fact, in some ways,
you can the definition

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of an infinite set is
that its size plus 1

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is equal to its size.

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Let's look at a simple example.

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So, on the bottom, I have
the non-negative integers.

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And on the top, I have
the positive integers.

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So, I can get from
the positive integers

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to the non-negative integers
just by throwing in zero.

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So, that's where the
plus one comes from.

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Here's a nice infinite set.

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I add another element to it,
and I get another infinite set,

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but they are the same size.

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I have to show a
bijection between them

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to show they're the set size.

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Well, you know what
the bijection is.

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Map 0 to 1, 1 to 2, 2 to 3.

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This is a bijection which you
know as the add one function.

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The add one function maps
the non-negative integers

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to the positive integers,
and it's a perfect bijection.

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Therefore, adding one element
to the non-negative integers--

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to the positive integers
does not get me a larger set,

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it gets me another
set of the same size.

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And this argument
actually generalizes

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to any infinite set.

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If you throw in
one extra element,

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you could still find a bijection
between the original set

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and the set with
one extra element.

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So, N is the same size
as the positive integers.

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Well, in fact, let's
look at this one.

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I can enumerate on the
top all the integers,

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both positive and negative.

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0, 1, minus 1, 2,
minus 2, and so on.

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And that gives me the set
consisting of all the integers.

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And over here, I
can have 0, 1, 2,

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just the non-negative integers.

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And you can see the
orderly way in which I've

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listed the integers at the top.

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00:11:53,150 --> 00:11:54,980
That implicitly
defines a bijection.

241
00:11:54,980 --> 00:11:59,840
I'm going to map zero to the
zeroeth element of the sequence

242
00:11:59,840 --> 00:12:00,340
above.

243
00:12:00,340 --> 00:12:03,880
1 to 1, 2 to minus 1,
3 to 2, 4 to minus 2.

244
00:12:03,880 --> 00:12:07,410
And in this way, I
have actually defined

245
00:12:07,410 --> 00:12:11,370
a bijection between the
non-negative integers and all

246
00:12:11,370 --> 00:12:12,120
the integers.

247
00:12:12,120 --> 00:12:15,852
In other words, you take
half of the integers, namely

248
00:12:15,852 --> 00:12:17,560
the non-negative
integers, and it's still

249
00:12:17,560 --> 00:12:19,390
the same size as all of them.

250
00:12:19,390 --> 00:12:23,080
There's a bijection
between N and Z.

251
00:12:23,080 --> 00:12:26,000
Now, you could write
a formula, actually,

252
00:12:26,000 --> 00:12:29,580
if you were trying to figure out
what does the number N go to?

253
00:12:29,580 --> 00:12:32,130
What positive or
negative integer?

254
00:12:32,130 --> 00:12:34,770
There's some not very
hard formula involving

255
00:12:34,770 --> 00:12:36,999
dividing N by two and rounding.

256
00:12:36,999 --> 00:12:38,040
But, that doesn't matter.

257
00:12:38,040 --> 00:12:41,100
Once I've figured
out some sensible way

258
00:12:41,100 --> 00:12:44,760
to list all the elements
of the integers in a row,

259
00:12:44,760 --> 00:12:47,480
then I can wind them up against
the non-negative integers.

260
00:12:47,480 --> 00:12:51,310
And that listing, in
effect, defines the mapping

261
00:12:51,310 --> 00:12:55,370
in a perfectly clear way without
necessarily having a formula.