Set Theory/Cardinals

Cantor–Bernstein (Schröder–Bernstein). If ${\displaystyle |A|\leq |B|}$ and ${\displaystyle |B|\leq |A|}$, then ${\displaystyle |A|=|B|}$.

Assume there are injections ${\displaystyle f_{1}:A\to B}$ and ${\displaystyle f_{2}:B\to A}$.

Step 1 (Reduce to the nested case ${\displaystyle B\subseteq A}$ via equipollence). Let ${\displaystyle B'=f_{2}[B]\subseteq A}$. Since ${\displaystyle f_{2}}$ is injective, it restricts to a bijection ${\displaystyle f_{2}:B\to B'}$ with inverse defined by ${\displaystyle f_{2}^{-1}(b')=b}$ whenever ${\displaystyle f_{2}(b)=b'}$. Hence ${\displaystyle |B|=|B'|}$. Because equipollence is an equivalence relation, it suffices to prove ${\displaystyle |A|=|B'|}$; then transitivity gives ${\displaystyle |A|=|B|}$. Therefore, replacing ${\displaystyle B}$ by ${\displaystyle B'}$ we may assume from now on that ${\displaystyle B\subseteq A}$.

Define

${\displaystyle f=f_{2}\circ f_{1}:A\to A}$ (injective) and ${\displaystyle A_{1}=f[A]\subseteq f_{2}[B]=B\subseteq A}$.

Thus ${\displaystyle A_{1}\subseteq B\subseteq A}$ and ${\displaystyle f:A\to A_{1}}$ is a bijection of ${\displaystyle A}$ onto its image ${\displaystyle A_{1}}$.

Step 2 (Iterate ${\displaystyle f}$ on ${\displaystyle A}$ and ${\displaystyle B}$). Define descending sequences

${\displaystyle A_{0}=A,\quad A_{n+1}=f[A_{n}]\quad (n\in \omega )}$,

${\displaystyle B_{0}=B,\quad B_{n+1}=f[B_{n}]\quad (n\in \omega ).}$

Then ${\displaystyle A_{n+1}\subseteq A_{n}}$ and ${\displaystyle B_{n+1}\subseteq A_{n+1}\subseteq B_{n}}$ for all ${\displaystyle n}$.

Let

${\displaystyle S=\bigcup _{n<\omega }(A_{n}\setminus B_{n}),\qquad T=A\setminus S}$.

Then ${\displaystyle S\cup T=A}$ and ${\displaystyle S\cap T=\varnothing }$.

Step 3 (Define the candidate bijection ${\displaystyle g:A\to B}$). Set

${\displaystyle \displaystyle g(x)={\begin{cases}f(x)&{\text{if }}x\in S\\x&{\text{if }}x\in T.\end{cases}}}$

We verify that ${\displaystyle g}$ is bijective ${\displaystyle A\to B}$.

Claim 1 (Points in ${\displaystyle T}$ already lie in ${\displaystyle B}$ and are fixed). Since ${\displaystyle S=\bigcup _{n}(A_{n}\setminus B_{n})}$, if ${\displaystyle x\in T}$ then ${\displaystyle x\notin A_{0}\setminus B_{0}=A\setminus B}$, hence ${\displaystyle x\in B}$. By definition ${\displaystyle g|_{T}=\mathrm {id} _{T}}$, so ${\displaystyle g[T]=T\subseteq B}$.

Claim 2 (Layers map forward: ${\displaystyle f[A_{n}\setminus B_{n}]=A_{n+1}\setminus B_{n+1}}$). Because ${\displaystyle B_{n+1}=f[B_{n}]}$, for ${\displaystyle x\in A_{n}}$ we have

${\displaystyle f(x)\in B_{n+1}\iff x\in B_{n}.}$

Thus ${\displaystyle x\in A_{n}\setminus B_{n}}$ iff ${\displaystyle f(x)\in A_{n+1}\setminus B_{n+1}}$, proving the equality.

Taking unions over ${\displaystyle n}$,

${\displaystyle \displaystyle f[S]=\bigcup _{n<\omega }{\big (}A_{n+1}\setminus B_{n+1}{\big )}=\bigcup _{n\geq 1}(A_{n}\setminus B_{n}).}$

For ${\displaystyle n\geq 1}$, ${\displaystyle A_{n}\subseteq A_{1}\subseteq B}$, hence ${\displaystyle A_{n}\setminus B_{n}\subseteq B}$. Also ${\displaystyle A_{n}\setminus B_{n}\subseteq S}$ by definition. Therefore

${\displaystyle f[S]={\Big (}\bigcup _{n\geq 1}(A_{n}\setminus B_{n}){\Big )}=B\cap S=B\setminus T.}$

By the definition of ${\displaystyle g}$ on ${\displaystyle S}$, we conclude

${\displaystyle g[S]=f[S]=B\setminus T.}$

Surjectivity. Using Claims 1–2,

${\displaystyle g[A]=g[S]\cup g[T]=(B\setminus T)\cup T=B.}$

Injectivity. Let ${\displaystyle x,y\in A}$.

If ${\displaystyle x,y\in T}$ and ${\displaystyle g(x)=g(y)}$, then ${\displaystyle x=y}$ since ${\displaystyle g|_{T}}$ is the identity.

If ${\displaystyle x,y\in S}$ and ${\displaystyle g(x)=g(y)}$, then ${\displaystyle f(x)=f(y)}$, and ${\displaystyle f}$ injective gives ${\displaystyle x=y}$.

If ${\displaystyle x\in S}$, ${\displaystyle y\in T}$, then ${\displaystyle g(x)=f(x)\in B\setminus T}$ by Claim 2, while ${\displaystyle g(y)=y\in T}$ by Claim 1; these are disjoint, so ${\displaystyle g(x)\neq g(y)}$. (Symmetrically for ${\displaystyle x\in T}$, ${\displaystyle y\in S}$.)

Hence ${\displaystyle g}$ is injective. Together with surjectivity, ${\displaystyle g}$ is a bijection ${\displaystyle A\to B}$, and therefore ${\displaystyle |A|=|B|}$.

(Cantor.) For every set ${\displaystyle X}$, ${\displaystyle |X|<|P(X)|}$.

Let ${\displaystyle f:X\to P(X)}$ be any function. Define the “diagonal” subset

${\displaystyle Y={x\in X:\ x\notin f(x)}.}$

We show that ${\displaystyle Y}$ is not in the range of ${\displaystyle f}$. Suppose, for a contradiction, that there exists ${\displaystyle z\in X}$ with ${\displaystyle f(z)=Y}$. Then:

By definition of ${\displaystyle Y}$, we have

${\displaystyle z\in Y\iff z\notin f(z).}$

But if ${\displaystyle f(z)=Y}$, this becomes

${\displaystyle z\in Y\iff z\notin Y,}$

a contradiction. Hence no such ${\displaystyle z}$ exists, so ${\displaystyle Y\notin \operatorname {ran} (f)}$.

Therefore no function ${\displaystyle X\to P(X)}$ is surjective, so ${\displaystyle |P(X)|\neq |X|}$. The map ${\displaystyle x\mapsto {x}}$ is an injection ${\displaystyle X\hookrightarrow P(X)}$, hence ${\displaystyle |X|\leq |P(X)|}$. Combining these, we conclude ${\displaystyle |X|<|P(X)|}$.

If ${\displaystyle |A|=\kappa }$, then ${\displaystyle |P(A)|=2^{\kappa }}$.

${\displaystyle X\subseteq A\mapsto \chi _{X}\in \{0,1\}^{A}}$ is a bijection.

For every ordinal ${\displaystyle \alpha }$ there exists a cardinal strictly greater than ${\displaystyle \alpha }$. More generally, for every set ${\displaystyle X}$ there exists an ordinal ${\displaystyle h(X)}$ such that no injection ${\displaystyle h(X)\to X}$ exists; moreover, ${\displaystyle h(X)}$ is a cardinal (an initial ordinal).

Step 1 (Collect all well-orderings of subsets of ${\displaystyle X}$ as a set). Consider the class of all binary relations on ${\displaystyle X}$, which is the set ${\displaystyle P(X\times X)}$ by the Axiom of Power Set. Using Separation, form the subset

${\displaystyle \mathrm {WO} (X)=\{R\subseteq X\times X:\ R\ {\text{is a well-ordering of some }}D\subseteq X\}.}$

Thus ${\displaystyle \mathrm {WO} (X)}$ is a set.

Step 2 (Take order-types; Replacement). For each ${\displaystyle R\in \mathrm {WO} (X)}$, let ${\displaystyle \operatorname {otp} (R)}$ be the unique ordinal isomorphic to ${\displaystyle (D,R)}$ (where ${\displaystyle D=\mathrm {dom} (R)}$). The map ${\displaystyle R\mapsto \operatorname {otp} (R)}$ is functional; by the Axiom Schema of Replacement, its range

${\displaystyle S=\{\operatorname {otp} (R):R\in \mathrm {WO} (X)\}}$

is a set of ordinals. Intuitively, ${\displaystyle S}$ is the set of all ordinals that inject into ${\displaystyle X}$ (via the isomorphisms to well-ordered subsets of ${\displaystyle X}$).

Step 3 (Define ${\displaystyle h(X)}$ and verify its key property). Let ${\displaystyle \displaystyle \theta =\bigcup S=\sup S}$ (existence by Union; ordinals are transitive so union is supremum), and let the successor

${\displaystyle h(X)=\theta \cup \{\theta \}=\theta +1}$

(existence by Pairing and Union). Then every ordinal ${\displaystyle \beta <h(X)}$ satisfies ${\displaystyle \beta \leq \theta }$, hence ${\displaystyle \beta \in S}$, so there is an injection ${\displaystyle \beta \to X}$. On the other hand ${\displaystyle h(X)\notin S}$, so there is no injection ${\displaystyle h(X)\to X}$. Thus ${\displaystyle h(X)}$ is the least ordinal with no injection into ${\displaystyle X}$.

Step 4 (${\displaystyle h(X)}$ is an initial ordinal, hence a cardinal). Suppose, towards a contradiction, that ${\displaystyle |h(X)|=|\gamma |}$ for some ${\displaystyle \gamma <h(X)}$. Since ${\displaystyle \gamma <h(X)}$, there is an injection ${\displaystyle i:\gamma \to X}$ (by Step 3). Let ${\displaystyle b:\gamma \to h(X)}$ be a bijection. Then ${\displaystyle i\circ b^{-1}:h(X)\to X}$ is injective—contradicting the defining property of ${\displaystyle h(X)}$. Hence no ordinal ${\displaystyle \gamma <h(X)}$ is equipollent with ${\displaystyle h(X)}$; i.e. ${\displaystyle h(X)}$ is an initial ordinal (a cardinal).

Step 5 (Specialize to ${\displaystyle X=\alpha }$). For an ordinal ${\displaystyle \alpha }$, every ${\displaystyle \beta \leq \alpha }$ injects into ${\displaystyle \alpha }$ (the inclusion map), so ${\displaystyle \alpha \leq \theta }$ and therefore ${\displaystyle \alpha <h(\alpha )=\theta +1}$. Thus there exists a cardinal ${\displaystyle h(\alpha )}$ strictly greater than ${\displaystyle \alpha }$.

If ${\displaystyle X}$ is a set of cardinals and ${\displaystyle \alpha =\sup X}$, then ${\displaystyle \alpha }$ is a cardinal.

Assume for contradiction that ${\displaystyle \alpha }$ is equipollent to some ${\displaystyle \beta <\alpha }$ (i.e. there is a bijection ${\displaystyle b:\beta \leftrightarrow \alpha }$). By the definition of ${\displaystyle \sup X}$, for this ${\displaystyle \beta }$ there exists ${\displaystyle \kappa \in X}$ with ${\displaystyle \beta <\kappa \leq \alpha }$.

Now compare sizes:

Because ${\displaystyle \kappa \leq \alpha }$ (as ordinals), there is an injection ${\displaystyle \kappa \hookrightarrow \alpha }$, so ${\displaystyle |\kappa |\leq |\alpha |}$.

Because ${\displaystyle \beta <\kappa }$, there is an injection ${\displaystyle \beta \hookrightarrow \kappa }$, so ${\displaystyle |\beta |\leq |\kappa |}$.

Because ${\displaystyle |\alpha |=|\beta |}$ (via ${\displaystyle b}$), we get

${\displaystyle |\kappa |\ \leq \ |\alpha |\ =\ |\beta |\ \leq \ |\kappa |,,}$

and by Cantor–Bernstein, ${\displaystyle |\kappa |=|\beta |}$.

But ${\displaystyle \kappa \in X}$ is a cardinal (an initial ordinal), so it cannot be equipollent to the strictly smaller ordinal ${\displaystyle \beta <\kappa }$—a contradiction. Therefore no such ${\displaystyle \beta }$ exists, and ${\displaystyle \alpha }$ is an initial ordinal, i.e. a cardinal.

The relation ${\displaystyle <}$ above is a well-ordering of the class ${\displaystyle \mathrm {Ord} ^{2}}$. Moreover, for each ordinal ${\displaystyle \alpha }$, the set ${\displaystyle \alpha \times \alpha }$ is exactly the initial segment below ${\displaystyle (0,\alpha )}$.

Linearity. For any two pairs, either their maxima differ (deciding ${\displaystyle <}$), or the maxima agree and then the first coordinates decide unless they agree again, in which case the second coordinates decide. Hence any two pairs are comparable.

Well-foundedness. Let ${\displaystyle X\subseteq \mathrm {Ord} ^{2}}$ be nonempty. Consider the set of maxima ${\displaystyle M={\max {\xi ,\eta }:(\xi ,\eta )\in X}}$, which has a least element ${\displaystyle m}$ since ordinals are well-ordered. Restrict to ${\displaystyle Y={(\xi ,\eta )\in X:\max {\xi ,\eta }=m}}$. Among first coordinates in ${\displaystyle Y}$ choose the least ${\displaystyle a}$; among pairs in ${\displaystyle Y}$ with first coordinate ${\displaystyle a}$ choose the least second coordinate ${\displaystyle b}$. Then ${\displaystyle (a,b)}$ is the least element of ${\displaystyle X}$.

Initial segment at ${\displaystyle (0,\alpha )}$. A pair ${\displaystyle (\xi ,\eta )}$ satisfies ${\displaystyle (\xi ,\eta )<(0,\alpha )}$ iff ${\displaystyle \max {\xi ,\eta }<\alpha }$, i.e. iff ${\displaystyle \xi <\alpha }$ and ${\displaystyle \eta <\alpha }$. Thus the initial segment below ${\displaystyle (0,\alpha )}$ is exactly ${\displaystyle \alpha \times \alpha }$.

For every ordinal ${\displaystyle \alpha }$, ${\displaystyle \operatorname {otp} (\alpha \times \alpha )=\Gamma (0,\alpha )}$.

By the initial-segment identification, ${\displaystyle \alpha \times \alpha =I(0,\alpha )}$, hence ${\displaystyle \operatorname {otp} (\alpha \times \alpha )=\operatorname {otp} {\big (}I(0,\alpha ){\big )}=\Gamma (0,\alpha )}$.

For every (well-orderable) infinite cardinal ${\displaystyle \aleph _{\alpha }}$, ${\displaystyle \aleph _{\alpha }\cdot \aleph _{\alpha }=\aleph _{\alpha }}$. Equivalently, ${\displaystyle \operatorname {otp} (\omega _{\alpha }\times \omega _{\alpha })=\omega _{\alpha }}$.

Assume toward a contradiction that there is a least ${\displaystyle \alpha }$ with ${\displaystyle \gamma (\alpha )>\omega _{\alpha }}$.

Since ${\displaystyle \gamma (\alpha )}$ is an ordinal strictly greater than ${\displaystyle \omega _{\alpha }}$. By definition of ordering of ordinals, we have ${\displaystyle \omega _{\alpha }\in \gamma (\alpha )}$. By the definition of order-type, there exists an isomorphism ${\displaystyle \Gamma '}$ between ${\displaystyle \gamma (\alpha )}$ and ${\displaystyle \omega _{\alpha }\times \omega _{\alpha }}$. Let ${\displaystyle (\beta ,\gamma )=\Gamma '(\omega _{\alpha })}$. Using "initial segments respects isomorphisms," there is an isomorphism between the initial segment by ${\displaystyle (\beta ,\gamma )}$ in ${\displaystyle \mathrm {Ord} ^{2}}$ and the initial segment by ${\displaystyle \omega _{\alpha }}$ in ${\displaystyle \mathrm {Ord} }$, meaning

${\displaystyle \Gamma (\beta ,\gamma )=\omega _{\alpha }.}$

Choose ${\displaystyle \delta <\omega _{\alpha }}$ with ${\displaystyle \delta >\beta }$ and ${\displaystyle \delta >\gamma }$. Then ${\displaystyle (\beta ,\gamma )\in \delta \times \delta }$, so the initial segment ${\displaystyle \delta \times \delta }$ contains a point of rank ${\displaystyle \omega _{\alpha }}$, and hence

${\displaystyle \operatorname {otp} (\delta \times \delta )\ \geq \ \Gamma (\beta ,\gamma )\ =\ \omega _{\alpha }.}$

Therefore ${\displaystyle |\delta \times \delta |\geq |\omega _{\alpha }|=\aleph _{\alpha }}$. But by minimality of ${\displaystyle \alpha }$, for every ${\displaystyle \delta <\omega _{\alpha }}$ we must have ${\displaystyle \operatorname {otp} (\delta \times \delta )=\delta }$, hence ${\displaystyle |\delta \times \delta |=|\delta |<\aleph _{\alpha }}$. Contradiction.

Thus no such ${\displaystyle \alpha }$ exists, and ${\displaystyle \gamma (\alpha )=\omega _{\alpha }}$ for all ${\displaystyle \alpha }$, i.e. ${\displaystyle |\omega _{\alpha }\times \omega _{\alpha }|=|\omega _{\alpha }|}$.