November 17-21

**Definition:** For \(n\in\mathbb{N}\) (where \(\mathbb{N}\) are the natural numbers), we let \([n]\) denote the set \(\{1,2,\ldots, n\}\).

Two sets \(A\) and \(B\)

*have the same cardinality*if there is a bijective map \(f:A\to B\).A set \(A\) is finite if there exists an \(n\) so that \(A\) has the same cardinality as \([n]\). Otherwise, \(A\) is infinite.

A set \(A\) is countable if it is finite or if it has the same cardinality as \(\mathbb{N}\). In the latter case it is called “countably infinite.”

An infinite set that is not countably infinite is called

*uncountable.*

**Proposition:** The relation “having the same cardinality” is an equivalence relation.

**Proof:** First, \(A\) has the same cardinality as \(A\) because the identity map (which sends \(a\in A\) to itself) is a bijection. Second, if \(A\) and \(B\) have the same cardinality, that means there is a bijection \(f:A\to B\). But then the inverse function \(f^{-1}:B\to A\) is a bijection in the other direction, so the relation is symmetric. Finally, suppose \(A\) has the same cardinality as \(B\), and \(B\) has the same cardinality as \(C\). Then there is a bijection \(f:A\to B\) and \(g:B\to C\). The composition \(g\circ f:A\to C\) is again a bijection, so \(A\) and \(C\) have the same cardinality.

**Proposition:** An infinite subset \(T\) of \(\mathbb{N}\) is countably infinite.

Proof: We construct a map \(\mathbb{N}\to T\). Let \(f(1)\) be the smallest element of \(T\). For \(n>1\), let \(T(n)=\{x\in T: x>f(n-1)\}\). If \(T(n)\) is empty, then \(T\) is finite, which isn’t true; so by the well-ordering principle \(T(n)\) has a smallest element. Let \(f(n)\) be the smallest element of \(T(n)\) for \(n\ge 2\).

\(f\) is injective because \(f(1)<f(2)<\cdots\). We will show that \(f\) is surjective. Suppose not. Then there is a smallest element x of \(T\) for which there is no \(n\) with \(f(n)=x\). Choose the largest element of \(T\) which is smaller than \(x\) and then there is an \(m\) with \(f(m)=x\). But then \(x\) is the smallest element of \(T(m+1)\) so \(f(m+1)=x\), contradicting our choice of \(x\). Thus \(f\) is surjective and therefore bijective.

**Proposition:** An infinite subset \(T\) of a countably infinite set \(S\) is countably infinite.

**Proof:** There is a bijection \(f:\mathbb{N}\to S\) since \(S\) is countable. Let \(U\) be the subset of \(\mathbf{N}\) consisting of \(\{x\in \mathbb{N}: f(x)\in U\}.\) This is an infinite subset of \(\mathbb{N}\) which is therefore countably infinite and in bijection with \(U\), so \(U\) is countably infinite.

**Proposition:** If \(A\) and \(B\) are countable, then so is \(A\times B\).

**Proof:** this is the “digaonals” argument.

**Propositon:** If \(A_1,\ldots, A_n\) are countable then so is \(A_1\times A_2\times\cdots\times A_n\).

**Proof:** By induction. We know \(A_1\times A_2\) is countable. Assume \(A_1\times\cdots\times A_n\) is countable. Then \(A_1\times\cdots\times A_{n+1}\) is countable because it is \((A_1\times\cdots\times A_n)\times A_{n+1}\).

**Proposition:** If \(A\) and \(B\) are countable, so is \(A\cup B\).

**Proof:** The union of \(A\) and \(B\) is a subject of the disjoint union of \(A\) and \(B\). Construct bijections \(f\) and \(g\) of \(\mathbb{N}\) to each of \(A\) and \(B\). Define a new function \(a\) with \(a(2n)=f(n)\) and \(a(2n-1)=g(n)\) for \(n=1,\ldots\). Then \(a\) is a bijection of \(\mathbb{N}\) with the disjoint union of \(A\) and \(B\), which is therefore countable, so its subset \(A\cup B\) is also countable.

**Proposition:** The integers are countable.

**Proof:** \(\mathbb{Z}\) is the union of \(0,1,\ldots\) and \(-1,-2,\ldots\) each of which is countable.

**Proposition:** The rational numbers are countable..

**Proof:** The set of ordered pairs of integers \((a,b)\) is countable, and the rational numbers is in bijection witha subset of this set.

The set of infinite sequences \([a_1,a_2,a_3,\ldots]\) with \(a_i\in\mathbb{N}\) is uncountable. (this is the diagonal argument)

The real numbers are uncountable. (diagonal argument)

The open open infinite interval \((0,\infty)\) has the same cardinality as \(\mathbb{R}\). Use the \(\log(x)\) and \(\exp(x)\) as bijections.

If a subset \(U\) of \(A\) is uncountable, so is \(A\). Proof: Suppose not. Then \(A\) is countable, so \(U\) is a subset of a countable set and therefore countable.

**Theorem:** (Cantor) The power set of a set \(A\) has a “larger” cardinality than \(A\).

**Proof:** The map \(a\to\{a\}\) puts \(A\) inside \(\mathcal{P}(A)\) so \(\mathcal{P}(A)\) is bigger than or equal to \(A\) in cardinality. Suppose there is a bijection \(f:A\to \mathcal{P}(A)\). Let \(U=\{a\in A: a\not\in f(a)\}\). This is a subset of \(A\) so an element of \(\mathcal{P}(A)\). Suppose \(f(x)=U\). Now if \(x\in U\), then \(x\in f(x)\), so \(x\not\in U\). But if \(x\not\in U\), then \(x\not\in f(x)\) so \(x\in U\). Neither is possible, so \(U\) cannot be in the range of \(f\), so \(f\) is not bijective.