Lecture 6 - Cardinality

A lot of this may be review for you.

Recall

Let $f : A \to B$ be a map (a function) from set $A$ to set $B$ . Then:

The domain of $f$ is $A$ . The codomain of $f$ is $A$ .
The range of $f$ is ${f (a) | a \in A}$
$f$ is one-to-one when $\forall a_{1}, a_{2} \in A (a_{1} \neq a_{2} \Rightarrow f (a_{1}) \neq f (a_{2})$ . This is called being injective, or being a monomorphism. Alternatively it's $f (a_{1}) = f (a_{2}) \Rightarrow a_{1} = a_{2}$ .
$f$ is onto if $\forall b \in B \exists a \in A$ where $f (a) = b$ . This is called being surjective, or being an epimorphism.
$f$ is bijective or an isomorphism if it's both injective and surjective. We call this a 1-to-1 correspondance.

See Review of Set Theory and Functions for a more thorough review of these topics.

See the drawing for an example of a valid bijection:

Finite/Cardinality

$A$ is finite if $A \neq \emptyset$ or $\exists$ a one-to-one cover:

f : A \to {1, 2, . . ., n}

for some $n \in N$

$A$ is infinite if it is not finite.

Infinite Cardinality Equivalence

$A, B$ have the same cardinality if $\exists$ a bijection $f : A \to B$ .
We write $A \sim B$ if they have the same cardinality.

Countability

$A$ is countable iff $A \sim N$ . Then we can order the elements $a_{n}$ for each $n \in N$ .

$A$ is at most countable if it is either finite or countable.

We talked about Hilbert's Hotel. The idea is that $N \sim Z$ since we have the bijection:

f (n) = {\begin{cases} \frac{n}{2} & n is even \\ 0 & n = 1 \\ \frac{n - 1}{2} & otherwise \end{cases}

Now you can check that $f$ is both injective and surjective.

(0, 1) \sim R

Proof
The idea is that we want a function that has the following shape:

Thus for simplicity we can have $f (\frac{1}{2}) = 0$ for our case. We also want undefined at $x = 0, 1$ thus have:

f (x) = \frac{x - \frac{1}{2}}{x (1 - x)}

Will work (you can verify yourself).
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More Facts from Set Theory

We have:

If $A \subseteq B$ and $B$ is countable, then $A$ is at most countable.
If $A_{1}, . . ., A_{k}$ are each countable, then $A_{1} \cup \dots \cup A_{k}$ is countable.
If $A_{1}, A_{2}, A_{3}, . . .$ are each countable, then $⋃_{i = 1}^{\infty} A_{i}$ is still countable.
If $A_{1}, A_{2}, A_{3}, . . .$ are each at most countable, then $⋃_{i = 1}^{\infty} A_{i}$ is at most countable.

We also have the following theorem:

Schroder-Bernstein Theorem

If $f : A \to B$ is 1-to-1 and $g : B \to A$ is 1-to-1, then $A \sim B$ .

Let's prove (3):

Proof
For each $A_{i}$ for $i = 1, . . .$ they can be counted. For instance:

A_{1} = {a_{1}, a_{2}, . . .}

Then denote $a_{11}, a_{12}, . . .$ to be counting these elements. Namely each $a_{i j}$ is elements $a_{j} \in A_{i}$ .

To say the union is countable, we should be able to make a 1-to-1 correspondance to our $a_{11}, a_{12}, . . .$ list. We can do this by using a diagonal counting of the elements:

☐ # Get to the Point!

Okay okay here's the big theorem:

Q

is countable.

Proof
For $n \in N$ , let:

A_{n} = {\frac{a}{b} | a, b \in Z, - n \leq a \leq n, 1 \leq b \leq n}

Notice that $Q = ⋃_{i = 1}^{\infty} A_{i}$ . Each $A_{n}$ is countable so using (3) then $Q$ is countable.
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R

is uncountable.

Proof
Let $A = {x_{1}, x_{2}, x_{3}, . . .}$ where $x_{n} \in R$ for each $n \in N$ . We'll show $A \neq R$ .

Have each $I_{i}$ be a closed interval where $x_{i} \notin I_{i}$ and each $I_{i} \supseteq I_{i + 1}$ . We can use the Nested Interval Property to say that $\exists x \in ⋂ I_{i}$ .

But for any $n \in N$ if $x = x_{n}$ then $x_{n} \notin I_{n}$ so clearly $x \notin I_{n}$ which is a contradiction. Thus then there is no $n$ to make this possible, so our list is false.
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