Some Infinite Cardinality Identities

Working with infinite sets


This post aims at recapturing the main ideas of the formal proofs that I’ve read. It never tries to replace them. You may consult the references if you need any of them.

Some notations

Unless otherwise specified, all cardinalities here are infinite. Denote $\mathfrak{a} = \card{A}$, $\mathfrak{b} = \card{B}$ and $\mathfrak{i} = \card{I}$.

$\mathfrak{a} + \mathfrak{b} = \card{A \cup B}$ provided that $A \cap B =\varnothing$.
$\mathfrak{a} \, \mathfrak{b} = \card{A \times B}$
$\mathfrak{a}^\mathfrak{i} = \card{A^I}$, where $A^I = \lbrace f \mid f: I \to A \rbrace$ denotes the set of functions from $I$ to $A$.

I've chosen $I$ instead of $B$ to express the index set because this reminds me of an array of $(a_i)_i$ indexed by $I$.

Using these notations, it’s easy to recover usual power laws, distributive laws and inequalities. I’ll take them for granted.

An alternate definition of infinite set

A set which admits a proper subset bijective to itself.

To see this from the usual definition (cardinality $\ge \aleph_0$), “take $\aleph_0$ elements from the set” and build a bijection that verifies the above statement (The easiest one is “to discard one element”.) Extend this bijection the easiest way: “leave outsiders untouched”.

To see the converse, instead of reasoning directly from the above statement, we prove the contrapositive form. The reason is evident: finite sets are much easier to deal with.

Sum, product and power of infinite sets

  1. Sum

    $$ \mathfrak{a} + \mathfrak{b} = \max(\mathfrak{a},\mathfrak{b}) \tag{$\Sigma$} $$
  2. Product

    $$ \mathfrak{a} \, \mathfrak{b} = \max(\mathfrak{a}, \mathfrak{b}) \tag{$\Pi$} $$
  3. Power: Given $2 \le \mathfrak{a} \le 2^\mathfrak{b}$.

    $$ \mathfrak{a}^\mathfrak{b} = 2^\mathfrak{b} \tag{$\land$} $$

In fact, WLOG, it suffices to show

$$ \mathfrak{a} + \mathfrak{a} = \mathfrak{a} \, \mathfrak{a} = \mathfrak{a}. $$

To show $(\Sigma)$, we will factorize $\mathfrak{a} = \mathfrak{i} , \aleph_0$, so that the second factor $\aleph_0$ can be split into two identical halves (odd and even).

The morals of $(\Sigma)$ and $(\Pi)$ is that taking union or direct product of infinite sets of the same size won’t give a bigger set. Exponentiating an infinite set is the way to get a bigger set.

Digression on induction

Since we know almost nothing about $A$ (not even countability), we have to use Zorn’s lemma (to do “uncountable induction”).

Zorn’s lemma Mathematical induction
“universal set” $\mathcal{X}$ poset $\lbrace n \in \N \mid P(n) \rbrace$
examples collection of sets, function class (subsets of) $\N$
order partial: set inclusion, extension of function total: usual $\le$
base case $\mathcal{X} \ne \varnothing$ usually $P(1)$
“assumed chain” any totally ordered subset $\mathcal{C}: j \le \cdots \le m$ “finite chain”: $P(1) \Rightarrow \cdots \Rightarrow P(m)$
“inductive step” $\mathcal{C} \ni m \le \ell \in \mathcal{X}$ $P(m) \Rightarrow P(m+1)$
explanation of “inductive step” every chain is bounded above domino effect
result $\mathcal{X}$ has maximal element. The “universal set” is (an infinite subset of) $\N$.

This table is just a memory aid. It can never replace the statements themselves. There’s no implication between the two results. A proof by Zorn’s lemma should be routine, but I often forget the details.

Brézis’ Functional Analysis and PDE book, which starts with defining inductive sets for proving Hahn–Banach Theorem, has inspired me to think of terms like “assumed chain” and “inductive step” in order to capture the adjective inductive in chains $\mathcal{C}$.

The domino effect doesn’t resemble an upper bound of $\mathcal{C}$. I’ve scratched my head to think of this analogy: for any chain $\mathcal{C}$, “jump outside” of it to find a “new leader” $\ell$ linked to each member $m \in \mathcal{C}$ by either $m \le \ell$ or $P(m) \Rightarrow P(\ell)$. $\ell$ is often found by $\cup \mathcal{C}$. (Or take union on the chain of domains in case of function extensions.)

In usual situations, finding a maximal element $x_{\max}$ doesn't end the proof. One may need to argue that $x_{\max}$ coïncides with our desired element $x$. A proof by contradiction is often used in the case of set inclusions. For example, to prove the existence of a Hamel basis $\beta$ on any vector space, apply Zorn's lemma to get $\beta_{\max}$, and argue that any $v_0 \notin \span{\beta_{\max}}$ would contradict the maximality of $\beta_{\max}$.

Maximal isn’t synonymous to maximum: think of prime ideals in $\Z$.

Product vs power

I was stuck at this malformed question for a hour: how much is $(a_i)_i$?

$\color{red}{A} \backslash \color{blue}{I}$ $\color{blue}{1}$ $\color{blue}{2}$
$\color{red}{1}$ $\color{red}{1}$ $\color{red}{1}$
$\color{red}{2}$ $\color{red}{2}$ $\color{red}{2}$
$\color{red}{3}$ $\color{red}{3}$ $\color{red}{3}$

If we see $(\color{red}{a}_{\color{blue}{i}})_\color{blue}{i}$ as a function, then there's $|\color{red}{A}|^{|\color{blue}{I}|}$ many $(\color{red}{a}_{\color{blue}{i}})_\color{blue}{i}$'s.

What if we enlarge the index so that it becomes $(\color{red}{a}, \color{blue}{i})$?

To make sense of $|\color{red}{A} \times \color{blue}{I}|$, you may identify (identical copies of) $\color{red}{A}_{\color{blue}{i}}$ with $\color{blue}{i}$. (Say, $\color{red}{A}_{\color{blue}{1}} = \color{red}{A}_{\color{blue}{2}} = \lbrace \color{red}{1},\color{red}{2},\color{red}{3} \rbrace$, but we treat $\color{red}{2} \in \color{red}{A}_{\color{blue}{1}}$ and $\color{red}{2} \in \color{red}{A}_{\color{blue}{2}}$ differently by adjoining an index $\color{blue}{i}$, so that they become $(\color{red}{2},\color{blue}{1})$ and $(\color{red}{2},\color{blue}{2})$.)

In the first case, we write $\color{red}{A}^{\color{blue}{I}} = \prod_{\color{blue}{i} \in \color{blue}{I}} \color{red}{A}_{\color{blue}{i}}$; in the second case, we write $\color{red}{A} \times \color{blue}{I} = \sqcup_{\color{blue}{i} \in \color{blue}{I}} (\color{red}{A}_{\color{blue}{i}}, \color{blue}{i})$.

Return to the proofs

$A_i$ $\cdots$ $A_j$ $\lbrace a_1, \dots, a_n \rbrace$
Partitions of cardinality $\aleph_0$ indexed by $i$ finitely many elements left

To factorize $\mathfrak{a} = \mathfrak{i} , \aleph_0$, we need partitions $A_i$’s of “size” $\aleph_0$. After taking “as much $A_i$’s as possible”, only finitely many elements $a_1, \dots, a_n$ are left. Put them into any one partition $A_i$—this won’t affect its cardinality. Zorn’s lemma’s role in proving $(\Sigma)$ is to formalize the action of “taking as much $A_i$’s as possible” by creating a maximal collection of such sets. The “universal set” is the collection of partitions of cardinality $\aleph_0$, and the partial order is the usual set inclusion between the collections rather than the partitions.

Identity $(\Pi)$ is much more difficult to prove. My attempts to decompose $\mathfrak{a}$ into powers involving $\aleph_0$ have failed. Zorn’s lemma is used in this way:

  1. $\mathcal{X} = \lbrace f: C \to C^2 \mid f \mathinner{\text{bijective}}, C \subseteq A \rbrace$
  2. $\le$: function extension in $\mathcal{X}$
  3. base case: $C = \N$, $f$ is a counting funciton for $\Q$
  4. chain: $\lbrace (D_i, f_i) \rbrace_i$
  5. upper bound: $(D, f)$, where $D = \cup_i D_i$ and $f: D \to D^2$ defined by $f(d_i) = f_i(d_i)$. Easy to verify that it’s bijective.
  6. maximal element: by abuse of notation, we keep using $(D, f)$.
  7. Claim $A = D$: the most difficult part

Let $E = A \setminus D$. Write $\mathfrak{a} = \mathfrak{d} + \mathfrak{e}$.

$D^2$ $D \times E$
$E \times D$ $E^2$

From $(\Sigma)$, assume $\mathfrak{a} = \mathfrak{e} > \mathfrak{d}$. This allows us to extract $F$ of cardinality $\mathfrak{d}$ from $E$. We’re going to extend $(D, f)$ to get a contradiction.

$D^2$ $D \times F$
$F \times D$ $F^2$

The top-left grid represents $(D, f)$ and $\mathfrak{d}^2 = \mathfrak{d}$. In words, the square can be “flattened into a line segment”. Note that “all of these four grids are of the same size. From $(\Sigma)$, the remaining three grids together are as big as one grid, which can be “flattened into $F$”. This extends $(D, f)$, contradicting its maximality in $\mathcal{X}$.

As a result, the assumption $\mathfrak{a} > \mathfrak{d}$ is wrong. The equality $\mathfrak{a} = \mathfrak{d} + \mathfrak{e}$ gives $\mathfrak{a} = \mathfrak{d}$. This proves $(\Pi)$.

With the usual power laws ported onto cardinal numbers, it’s easy to show

$$ \begin{aligned} & 2 &\le \mathfrak{a} &\le 2^{\mathfrak{b}} \\ \Rightarrow \quad & 2^{\mathfrak{b}} &\le \mathfrak{a}^{\mathfrak{b}} &\le (2^{\mathfrak{b}})^{\mathfrak{b}} = 2^{\mathfrak{b} \, \mathfrak{b}} \overset{(\Pi)}{=} 2^{\mathfrak{b}}. \end{aligned} $$

Cardinals under total order

This has motivated me to find out more about cardinal numbers. The next result shows that the sizes of two sets can always be compared.

To see this, we have to apply Zorn’s lemma again.

  1. $\mathcal{X} = \lbrace f: X \to Y \mathinner{\text{bijective}} \mid X \subseteq A \land Y \subseteq B \rbrace$

  2. $\prec$: usual function extension (extend domain and codomain, keep original values unchanged).

  3. Start from any chain $\mathcal{C}$. Take $\cup \mathcal{C}$ and show that it’s in $\mathcal{X}$.

  4. By Zorn’s lemma, there exists a maximal element in $\mathcal{X}$. Denote this again by $f: C \to D$.

  5. Define $E$ and $F$ to be

    $A = C \sqcup E$ $B = D \sqcup F$
    max $C \xleftarrow[\text{bijective with}]{}$ $\xrightarrow[\text{each other}]{}D$
    rest $E := A \setminus C$ $F:= B \setminus D$
  6. If $E = \varnothing$ or $F = \varnothing$, we’re done. Otherwise, maximality of $f: C \to D$ would be contradicted.

Bijection from $\mathcal{P}(\N^*)$ to real line

Denote $\mathfrak{c} = \card{[0,1]}$. Let’s compare $\aleph_0$ and $\mathfrak{c}$ before proving a cardinality equality.

Digression: Cantor’s diagonal extraction

The aim of this subsection is to show that $\mathfrak{c} > \aleph_0$.

To show that $[0,1]$ is uncountable, a classic counter-argument proceeds as follows:

Suppose $[0,1]$ is countable, and denote it by $(r_n)_n$. Extract a number not in the list $(r_n)_n$ by a function

$$ \phi(r_n) = \begin{cases} \color{blue} 6 \color{black} &\colorbox{yellow}{\color{red}{\text{if $n$-th decimal place of $r_n$}}} \mathinner{\text{equals}} \colorbox{yellow}{7} \\ \color{blue} 7 \color{black} &\colorbox{yellow}{\text{\color{red}{otherwise}}} \end{cases} $$
$$ \begin{aligned} \mathinner{\text{``list''}} & \mathinner{\text{diagonal extraction}} & \color{blue} \phi(r_n) & \mathinner{\text{explanation}} \\ r_1 &= 0.\colorbox{yellow}{\color{red}{7}}\color{black}{56734}\cdots &\to \color{blue}{6} &\quad \color{black}{(\because} \colorbox{yellow}{\color{red}{7}} \color{black}{=} \colorbox{yellow}{7}\color{black}{)} \\ r_2 &= 0.5\colorbox{yellow}{\color{red}{6}}\color{black}{8784}\cdots &\to \color{blue}{7} &\quad \color{black}{(\because} \colorbox{yellow}{\color{red}{6}} \color{black}{\ne} \colorbox{yellow}{7}\color{black}{)} \\ r_3 &= 0.16\colorbox{yellow}{\color{red}{8}}\color{black}{567}\cdots &\to \color{blue}{7} &\quad \color{black}{(\because} \colorbox{yellow}{\color{red}{8}} \color{black}{\ne} \colorbox{yellow}{7}\color{black}{)} \\ r_4 &= 0.864\colorbox{yellow}{\color{red}{7}}\color{black}{56}\cdots &\to \color{blue}{6} &\quad \color{black}{(\because} \colorbox{yellow}{\color{red}{7}} \color{black}{=} \colorbox{yellow}{7}\color{black}{)} \\ r_5 &= 0.5565\colorbox{yellow}{\color{red}{4}}\color{black}{8}\cdots &\to \color{blue}{7} &\quad \color{black}{(\because} \colorbox{yellow}{\color{red}{4}} \color{black}{\ne} \colorbox{yellow}{7}\color{black}{)} \\ r_6 &= 0.17826\colorbox{yellow}{\color{red}{1}}\color{black}{\cdots} &\to \color{blue}{7} &\quad \color{black}{(\because} \colorbox{yellow}{\color{red}{1}} \color{black}{\ne} \colorbox{yellow}{7}\color{black}{)} \\ \vdots &= \cdots \cdots \colorbox{yellow}{\color{red} \dots} & \quad \color{blue}{\vdots} \\ & \text{new decimal not ``listed''} & d = & \color{blue} 0.677677\cdots \end{aligned} $$

The number $$ d = \sum_{n \in \N^*} \frac{\color{blue}\phi(r_n)}{10^n} $$ converges, so it’s well-defined. However, $d$ doesn’t belong to list $(r_n)$ since the $n$-th decimal place of $d$ ($\phi(r_n)$) never matches $\colorbox{yellow}{\color{red}{\text{the $n$-th decimal place of $r_n$}}}$.

Diagram illustrating the bijections

$$ T \triangleq \lbrace 0,1 \rbrace^{\N^*} \xrightleftharpoons{\displaystyle\footnotesize (a_n)_n \overset{\phi}{\mapsto} \sum_{n \in \N^*} a_n/2^n} [0,1] \xleftrightarrow{(\Sigma)} (0,1) \xleftrightarrow[\text{and shifting}]{\text{scaling}} \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \xrightleftharpoons[\tan^{-1}]{\tan} \R \tag{B} $$

The first bijection isn’t evident as $\frac12 = \sum_{n\ge2} \frac{1}{2^n}$. Therefore, a direct proof won’t work. We have to calculate cardinalities and establish injections between some infinite subset and use CSB theorem. We have to separate two cases

  • $T_1 = \lbrace (a_n)_n \mid a_n = 1 \mathinner{\text{i.o.}} \rbrace$ (“i.o.” means “infinitely often”. This is used in Durrett’s Probability Theory book.)
  • $T \setminus T_1 = \lbrace (a_n)_n \mid a_n = 0 \mathinner{\text{eventually}} \rbrace \subseteq \Q$

We have $\card{T \setminus T_1} \le \aleph_0$. To see that's indeed an equality, take $\lbrace (\delta_{ij}) \rbrace_j$. $(\Sigma)$ and Cantor's Theorem forces $\card{T_1} = 2^{\aleph_0}$. To show a cardinality inequality, we need an injection. Since $\mathfrak{c} > \aleph_0$, we can't inject $[0,1]$ into $T \setminus T_1$.

Surjectivity of $\phi$

To show $\phi$ is surjective, construct $(a_n)_n$ from any $r \in [0,1]$ recursively as follows:

  1. $i = 1$: If $r \ge \frac12$, $a_1 = 1$. Otherwise, $a_1 = 0$. The error for the first iteration is $\epsilon_1 \leftarrow r - \frac{a_1}{2}$.

  2. $i > 1$: If $\epsilon_{i-1} \ge \frac{1}{2^i}$, $a_i = 1$. Otherwise, $a_i = 0$. The error for the $i$-th iteration is $\epsilon_i \leftarrow r_{i-1} - \frac{a_i}{2^i}$ It’s clear that

    $$ \begin{aligned} \mathinner{\text{error}} \epsilon_n &= \mathinner{\text{real value}} r - \mathinner{\text{lower approximation}} \sum_{i = 1}^n \frac{a_i}{2^i}. \end{aligned} $$

Show that $(a_n)_n$ constructed from $r$ can be mapped back to $r$ by $\phi$. (i.e. $\phi((a_n)_n) = r$) This can be proved directly and inductive from the construction of $(a_n)_n$. For each iteration, the increasing partial sum is near enough to $r$.

$$ \forall n \in \N^*, \epsilon_n \le \frac{1}{2^n} \tag{P}. $$
  • base case: direct calculations

  • induction hypothesis: assume $\mathrm{P}(m)$

  • inductive step $i = m + 1$:

    $$ \begin{aligned} \epsilon_{m+1} &= \epsilon_m -\frac{a_{m+1}}{2^{m+1}} \\ &\le \begin{cases} \frac{1}{2^{m+1}} - 0 &\text{if } \epsilon_m < \frac{1}{2^{m+1}} \\ \frac{1}{2^m} - \frac{1}{2^{m+1}} &\text{if } \epsilon_m \ge \frac{1}{2^{m+1}} \end{cases} \\ &= \frac{1}{2^{m+1}} \end{aligned} \tag{$\star$} $$

    For the second case in $(\star)$, $\mathrm{P}(m)$ is applied.

As a result, error $\epsilon_n \to 0$ as $n \to \infty$.

Translate the surjectivity of $\phi$ into the cardinality inequality

$$ \mathfrak{c} \overset{\text{(B)}}{=} \card{[0,1]} \overset{\phi \text{ surj.}}{=} \card{\phi[T]} \overset{\text{(AC)}}{\le} \card{T} \\ = \card{T_1} + \card{T \setminus T_1} = 2^{\aleph_0} + \aleph_0 \overset{(\Sigma)}{=} 2^{\aleph_0} \tag{$\supseteq$} $$

$(\text{AC})$ means the Axiom of Choice.

Injectivity of $\phi|_{T_1}$

Suppose $(a_n)_n, (b_n)_n \in T_1$ such that $\phi((a_n)_n) = \phi((b_n)_n) = r$. If $a_n = b_n \forall n \in \N^*$, then $(a_n)_n = (b_n)_n$ and we’re done. Otherwise, they differ starting from the $n_0$-th term.

$$ \mathinner{\text{i.e.}} n_0 \triangleq \min\{n \in \N^* \mid a_n \ne b_n\} $$

WLOG, assume $a_{n_0} = 1$ and $b_{n_0} = 0$. We have

Two seq in $T_1$ head: $n < n_0$ they differ from here tail: $n > n_0$
$(a_n)_n$ same $a_{n_0} = 1$ infinitely many $1$’s
$(b_n)_n$ terms $b_{n_0} = 0$ dominated by $a_{n_0} = 1$
$$ \begin{aligned} \phi((a_n)_n) &= \sum_{n \in \N^*} \frac{a_n}{2^n} \quad& \\ &= \left( \sum_{1 \le n < n_0} \frac{a_n}{2^n} \right) +\frac{1}{2^{n_0}}+ \left( \sum_{n > n_0} \frac{a_n}{2^n} \right) & (a_{n_0} = 1) \\ &> \left( \sum_{1 \le n < n_0} \frac{a_n}{2^n} \right) +\frac{1}{2^{n_0}} & ((a_n)_n \in T_1, \mathinner{\text{so}} a_n = 1 \mathinner{\text{i.o.}}) \\ &= \left( \sum_{1 \le n < n_0} \frac{b_n}{2^n} \right) + \frac{b_{n_0}}{2^{n_0}} + \sum_{n > n_0} \frac{1}{2^n} & (a_n = b_n \forall n < n_0, b_{n_0} = 0) \\ &\ge \left( \sum_{1 \le n < n_0} \frac{b_n}{2^n} \right)+\frac{b_{n_0}}{2^{n_0}} + \sum_{n > n_0} \frac{b_n}{2^n} & (b_n \le 1 \forall n \in \N^*) \\ &= \phi((b_n)_n) & (\mathinner{\text{contradiction}}) \end{aligned} \tag{I} $$

The key step in $\text{(I)}$ is the use of $a_n = 1$ i.o. to get a strict inequality.

The injectivity of $\phi|_{T_1}$ gives

$$ 2^{\aleph_0} = \card{T_1} \overset{\phi|_{T_1} \text{inj.}}{=} \card{\phi[T_1]} \le \card{[0,1]} \overset{\text{(B)}}{=} \mathfrak{c} \tag{$\subseteq$} $$

Combine $(\subseteq)$ and $(\supseteq)$ and apply CSB theorem to get $\mathfrak{c} = 2^{\aleph_0}$.


Gabriel Nagy’s Real Analysis (version 2.0) Appendix B

I’ve modified his proof for $\mathfrak{c} = 2^{\aleph_0}$ by changing $T_0$ into $T_1$. This leads automatically to $\card{T \subseteq T_1} = \aleph_0$. To avoid going too deep, I simply take $r = 2$.

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