# Does it make sense to define $\aleph_{\infty}=\lim\limits_{n\to\infty}\aleph_n$? Is its cardinality “infinitely infinite”?

I recently read a book about infinity, which introduced the basic notions of different kinds of infinity. I'm a total layman concerning this topic, and one question fascinated me:

Can we, in some sense, define: $$\aleph_{\infty}=\lim_{n\to\infty}\aleph_n$$ Such that there exists a set whose cardinal is $\aleph_{\infty}$, i.e. whose cardinality is infinitely infinite?

• If there is such a set, its powerset is even bigger. – Stefan Perko Oct 9 '15 at 19:01
• This is part of the question; sometimes we define $\infty+1=\infty$ and here, we would have to have $2^{\aleph_{\infty}}=\aleph_{\infty}$ – Redundant Aunt Oct 9 '15 at 19:04
• See Cantor's paradox. – pregunton Oct 9 '15 at 19:04
• Check my answer, cantor's theorem says that $2^\kappa>\kappa$ for any cardinal $\kappa$. I recommend strongly that you define your term 'infinitely infinite'. – YoTengoUnLCD Oct 9 '15 at 19:04
• @user109899 You have to be careful. In some situations it may make sense to say $\infty + 1 = \infty$. In this situation here, where we're dealing with ordinal numbers and would call "the thing that comes after all the natural numbers" $\omega$, $\omega + 1$ is a new infinite (ordinal) number, larger even than $\omega$. – Mike Haskel Oct 9 '15 at 19:19

There are two senses of "infinite number" in play here: ordinal and cardinal. Roughly, cardinal numbers count "how many," and ordinals count "which step in a progression." The $\aleph$-numbers are cardinals. By counting $\aleph_0$, $\aleph_1$, $\aleph_2$, etc., we can see that the subscripts are ordinals, however. Just like the $\aleph$ numbers give us our infinite cardinals, we have infinite ordinals, also. If we count $0,1,2,\ldots$, there is an infinite ordinal that comes "next" after all those natural numbers; we call it $\omega$. You can keep going, and get $\omega + 1,\omega + 2,\omega+3,\ldots,\omega+\omega$, etc. The $\aleph$ numbers keep going in this same sense: after $\aleph_0,\aleph_1,\aleph_2,\ldots$, we get $\aleph_\omega,\aleph_{\omega+1},\aleph_{\omega+2},\ldots,\aleph_{\omega+\omega}$, and on and on.

$\aleph_0$ is the cardinality of the set of finite ordinal numbers $0,1,2,3,4,\ldots$,

$\aleph_1$ is the cardinality of the set of all ordinal numbers of cardinality $\le\aleph_0$.

$\aleph_2$ is the cardinality of the set of all ordinal numbers of cardinality $\le\aleph_1$.

$\aleph_3$ is the cardinality of the set of all ordinal numbers of cardinality $\le\aleph_2$.

and so on.

$\aleph_\omega$ is the cardinality of the set of all ordinals of cardinality $\aleph_n$ for some $n$. This is the smallest cardinal number $\ge\aleph_n$ for every finite ordinal number $n$. $\omega$ is the smallest infinite ordinal number.

Short Answer: It does make sense, but I will adjust your notation (notice $\aleph_0\subseteq\aleph_1\cdots\subseteq\aleph_n$) $$\aleph_\infty\overset{\mathrm{def}}{=}\lim_{n\to \infty}\aleph_n=\bigcup_{n=0}^\infty\aleph_n\overset{?}{=}\aleph_\omega$$ The ? will be addressed in the long answer. The cardinality is $\aleph_\omega$.

Prerequisite(s): ordinal numbers, well-formed formulas

Theorem (Transfinite Recursion): For each formula $\phi(x,y)$, if for each $x$, there exists a unique $y$ such that $\phi(x,y)$ holds, then there exists a formula $\Phi(\alpha,z)$ such that for each ordinal $\alpha$, there exists a unique $z$ such that $\Phi(\alpha,z)$ holds, and for each ordinal $\alpha$ and for each function $f$ such that $$\mathrm{domain}(f)=\alpha\qquad\text{and}\qquad \Phi(\beta,f(\beta))$$ for every $\beta\in\alpha$ and for each $z$, $$\phi(f,z)$$ if and only if $\Phi(\alpha,z)$ holds.

Definition Let $\Phi(\alpha,z)$ be a formula such that for each ordinal $\alpha$, there exists a unique $z$ such that $\Phi(\alpha,z)$ holds, and for each ordinal $\alpha$ and for each function $f$ such that $$\mathrm{domain}(f)=\alpha\qquad\text{and}\qquad\Phi(\beta,f(\beta))$$ for every $\beta\in\alpha$ and for each $z$, $$\text{z is the least infinite cardinal strictly greater than every element of \mathrm{range}(f)}$$ if and only if $\Phi(\alpha,z)$ holds. Let $\alpha$ be an ordinal. Aleph number-$\alpha$ (or aleph-$\alpha$) is $z$ such that $\Phi(\alpha,z)$ holds.

Notation Let $\alpha$ be an ordinal. "$\aleph_\alpha$" is notation for "aleph-$\alpha$."

Remark

• For all ordinals $\alpha$ and $\beta$, if $\alpha<\beta$, then $\aleph_\alpha<\aleph_\beta$
• Aleph-$0$ is the least infinite cardinal
• Aleph-$1$ is the least uncountable cardinal
• For each ordinal $\alpha$, $\aleph_{\alpha+1}=\aleph_\alpha^+$ (the least cardinal greater than $\aleph_\alpha$)
• For each limit ordinal $\gamma$, $\aleph_\gamma=\bigcup_{\beta\in\gamma}\aleph_\beta$ (notice $\omega=\{0,1,2,\ldots\}$ is a limit ordinal which addresses ? in the short answer)
• For each ordinal $\alpha$, $\alpha\le\aleph_\alpha$
• For each infinite cardinal $\kappa$, there exists an ordinal $\eta$ such that $\kappa=\aleph_\eta$.

"Infinitely infinite" is a fairly vague/ill-defined term.

Consider the set $\Bbb N$, whose cardinality is $\aleph_0$. Now, by Cantor's theorem $|\Bbb N|=\aleph_0<2^{\aleph_0}= P(\Bbb N)$.
Now consider the sequence $$A_0=\Bbb N \\ A_n=P(A_{n-1})$$
• I don't think this answer addresses the (admittedly vague) question. The OP didn't ask for a "largest cardinal" they asked for an "infinitely infinite" cardinal, and suggested an imprecise "limit of all the $\aleph_i$." Such cardinals do exist, but this answer doesn't build any. – Mike Haskel Oct 9 '15 at 19:09