# Is $2^\infty$ uncountable and is cardinality a continuous function?

I apologize if the title seems too vague, but this is how I was asked the question. So one of my friends intended to write an infinite sum like $\displaystyle \sum_{i=1}^{\infty} a_{2^i}$ .

However, he wrote $\displaystyle \sum_{i=2^0}^{2^\infty} a_{i}$. Then he was worried whether $2^\infty$ was uncountable [since we had proved $|2^{\mathbb{N}}| = c$ in class and $2^X$ stands for the power set of $X$] and the sum did not make sense since he believed he is adding uncountable number of elements. I resolved his difficulties by explaining that $2^n \to n$ is an injection and thus the set of powers of two can have a cardinality of at most that of naturals and thus the set is countable. He was not very convinced :( Is there a more convincing argument?

I guess he was blinded by some sort of a notion that cardinality is a continuous function of sets. To make it precise, lets define:

${\cal F}_n := 2^{\{1,2,3,\cdots, n\}}$. $\mathbb{N}$ is the set of naturals.

We note that ${\cal F}_n \subset {\cal F}_{n+1}$ and $\displaystyle \bigcup_{n \geq 1} {\cal F}_n = 2^{\mathbb{N}}$ (Edit: $\displaystyle \bigcup_{n \geq 1} {\cal F}_n = 2^{\mathbb{N}}$ is wrong, see comments below).

Now $|{\cal F}_n| = 2^n$ and thus we must have ${\cal F}_n \to 2^{\mathbb{N}}$ and thus $|{\cal F}_n| \to c$.

So I believe cardinality is not a continuous function, but then one needs a natural topology on the cardinals to say such a thing. So my question is: Do we have a natural topology on the cardinals where one can show cardinality is not a continuous function of sets?

Thank you,

Iso

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When you write a sum like $\displaystyle \sum_{i=1}^{\infty} a_i$, you aren't really including $\infty$ in the sum; you're only summing over the natural numbers, and thinking of the natural numbers as a subset of $\mathbb{N} \cup \{ \infty \}$ the upper limit there is never actually reached. The confusion your friend is afflicted with seems to be ultimately due to this unfortunate choice of notation. –  Qiaochu Yuan Jan 30 '12 at 5:19
You are mixing the notion of cardinality with the notion of infinity in calculus. They are very different things. –  azarel Jan 30 '12 at 5:19
To extend @Qiaochu’s comment, the notation $\sum\limits_{i\ge 1}a_i$ means exactly the same thing and may be less confusing. –  Brian M. Scott Jan 30 '12 at 5:27
You may want to take a look: math.stackexchange.com/questions/90758/… –  Asaf Karagila Jan 30 '12 at 7:38
Just a short comment about something that seemed strange. You define $\mathcal{F}_n = 2^{\{1,2,\ldots ,n\}}$, and then write that $\mathcal{F}_n \subseteq \mathcal{F}_{n+1}$ (which there is no problem with), but also that $\bigcup_{n \geq 1} \mathcal{F}_n = 2^{\mathbb{N}}$, which there is a problem with. You can show (quite easily) that $\bigcup_{n \geq 1} \mathcal{F}_n$ consists of all (and only) the finite subsets of $\mathbb{N}$. It does not, for instance, contain $\mathbb{N}$ itself. (You can also show that $\bigcup_{n \geq 1} \mathcal{F}_n$ is countable, but $2^{\mathbb{N}}$ isn't.) –  Arthur Fischer Jan 30 '12 at 9:55

Cardinals form an ordered class, that is we have a good notion of "larger cardinality". Assuming the axiom of choice this class is totally ordered and in fact well ordered, namely every nonempty collection of cardinals has a miniumum element.

The topology on cardinals is indeed the natural topology on orders: open sets are unions of intervals. This means that there are many isolated points, all the finite cardinals for example. However there is no way to write $\aleph_0$ in an open interval without infinitely many finite cardinals.

This means that $\aleph_0$ is a limit cardinal we can then continue to higher inifinities, the idea is the same, you have a cardinal $\mu$ then there is a next one: $\mu^+$. After infinitely many steps you have a limit cardinal.

Now, what is a continuous function in this topology? It is a function which respect limits, that is: $$\sup\{f(\kappa_i):i\in I\}=f(\sup\{\kappa_i:i\in I\}$$

With respect to this topology, if so, to say that the function $\mu\mapsto2^\mu$ is to say that for a limit cardinal $\lambda$ we have $\sup\{2^\kappa:\kappa<\lambda\}=2^\lambda$.

This is not true, even in $\aleph_0$ since we have:

1. $\sup\{n:n\in\mathbb N\}=\aleph_0$.
2. If $n$ is finite then $2^n$ is finite, therefore $\sup\{2^n:n\in\mathbb N\}=\aleph_0$.
3. As you know $2^{\aleph_0}>\aleph_0$.

Therefore the continuum function has a discontinuity at $\aleph_0$.

On the other hand it is possible to have continuity in limit cardinals for trivial reasons. For example if all the values below the limit are constant and are above the limit we can have continuity points as well.

I should probably answer the question in the title: Is $2^\infty$ uncountable?

Well, $\infty$ is not an exact mathematical object, it's a notion of "larger than all real numbers", however this is not a cardinal notion. Real numbers do not signify cardinality but rather length, so in a sense $\infty$ means "longer than all finite lengths".

If you mean $\lim_{n\to\infty} 2^n$ one can replace $n$ by the set definitions of finite numbers: $$0=\varnothing; n+1=\{0,\ldots,n-1\}$$

It follows that $n<m\iff n\in m$, and $n\le m\iff n\subseteq m$. Then we have that $\lim_{n\to\infty} 2^n = |\bigcup_{n\in\mathbb N}\mathcal P(n)|=\aleph_0$. This is indeed countable, and it represents all the finite subsets of $\mathbb N$.

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I wrote all that on my iPhone and now my wrists hurt! :-) –  Asaf Karagila Jan 30 '12 at 17:37
Now I edited and added stuff on a real keyboard. But still the first part... it looks even more impressive when I look at it on a real computer monitor! –  Asaf Karagila Jan 30 '12 at 21:10
I appreciate your answer. I will need to work through it though. The idea of a limit cardinal is not very clear to me though. So I will read it and get back to you if I have any doubts. Thank you very much :) –  Isomorphism Jan 31 '12 at 4:18
@Isomorphism: Are you slightly familiar with the idea of ordinals? –  Asaf Karagila Jan 31 '12 at 7:49
:Not rigorously, so I am studying Halmos Naive set theory. –  Isomorphism Feb 1 '12 at 3:32