Is $2^{\aleph_0}$ well-defined?

Question

Ok so I was thinking about power sets and mathematics that utilize infinity, and I ended up thinking about power sets for $\aleph_0$. Knowing that to get a power set you take $2$ and raise it to the $n$ power ($n$ being the number of values in the set). So I am wondering about the possibility of taking an actual integer and raising to infinity (an amount of things in a set that spans on forever/not technically a number).

If anyone has info on this idea please answer this: Can an integer be raised to an idea such as infinity and be written in terms of infinity.

Answer 1

There are a few facts that may help you, but I'm still unsure what you're question is.

It is an axiom of $\text{ZFC}$ that for any set $X$, the collection of all subsets exists and is a set. We denote that collection by $\mathscr{P}(X)$ and call it the powerset.

When $A$ and $B$ are sets, the notation $A^B$ denotes the set of functions with domain $B$ and codomain $A$.

It can be easily proven that $\mathscr{P}(X)$ and $2^X$ (here we are taking $2 = \{0,1\}$) have the same cardinality. In particular, for finite sets $X$, this means that $|\mathscr{P}(X)| = |2^X| = 2^{|X|} = 2^n$ where $|X| = n$.

Nothing is stopping you from looking at the set $n^{\aleph_0}$ where $n$ is some finite integer; it just means the set of all functions with domain $\aleph_0$ to the set $ n = \{n-1, n-2, \ldots, 1, 0\}$.

Answer 2

Yes, you can. For any infinite cardinal number $\aleph_n$, you can think of $2^{\aleph_n}$ as the set of functions from a set of size $\aleph_n$ to the set $\{0,1\}$. Namely, if $A$ is an infinite set with size $\aleph_n$, and $S$ is an element of the powerset of $A$, then the corresponding function is $f(a)=1$ when $a\in S$ and $f(a)=0$ otherwise. This is known as the "indicator function" for $S$.

Analgoously, $k^{\aleph_n}$ can be thought of as the set of functions from a set of size $\aleph_n$ to the set $\{0,1,\dots,k-1\}$. Instead of $2$ choices for each element ($0$ or $1$, in or out), there are $k$ choices for each element.

However, this concept is not especially interesting, since it turns out that $$ |k^{\aleph_n}|=|2^{\aleph_n}|\qquad \text{for all $k\ge 2$, for all $n$} $$ That is, there are the same number of functions from $A\to \{0,\dots,k-1\}$ as there are from $A\to \{0,1\}$, when $A$ is an infinite set with cardinality $\aleph_n$. Therefore, the two exponents with different bases represent the same infinite quantity.

Certainly $k^{\aleph_n}\ge 2^{\aleph_n}$, since there are more options for each element of $A$. Proving the reverse inequality is a good exercise in elementary set theory.

Answer 3

One of the common misunderstanding when first approaching infinity, particularly in the context of set theory, is that Infinity is not defined as a limit. It is a concrete and well-defined object, in a mathematical universe that is fixed and concrete as well.

We are not computing and "trying to figure out" $\Bbb N$ or its power set. These exist. Period. Full stop. End of story. Moving on.

The exponentiation in cardinal arithmetic is not somehow a limit of a process, or a continuation of exponentiation from real analysis. Cardinals are not real numbers, as I so often remarked on this website.

The exponentiation is defined in terms of the cardinality of a set of functions between two particular sets. We can easily show that this set exists, and that its cardinality does not depend on the choice of our two particular sets. (As long as we assume some basic and common set theory, e.g. that $\Bbb N$ is a set, and that every set has a power set.)