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For my point-set topology class, I'm working on proving the theorem: The set of all finite subsets of a countable set is countable.

Please don't post the proof of the theorem. The proof was easy for the case of the set being finite, so now I'm working on the case that the set is infinite. I'm trying to use a result that we already proved: If $S$ and $T$ are countable sets, then $S \times T = \{ (s,t) | s \in S, T \in T\}$ is also countable. Then I'm saying, if $S$ is the set under consideration, and $n \in \mathbb{Z},n \geq 0$, then define $T_n = \{ U \subset S |\ |U| = n\}$. Then if $T$ represents the set of all finite subsets of $S$, it is clear that $T = T_0 \cup T_1 \cup T_2 \cup \cdots$. Also, since there are countably many different $T_i$, then from another theorem we already proved, as long as every $T_i$ is countable, then their union will also be countable. So, now I'm trying to just demonstrate that every $T_i$ is countable.

Can I do it the following way? I think I'm wrong, because I think I might be confusing a finite set of cardinality $n$ with an ordered $n-$tuple. \begin{align*} T_0 & = \{ \{ \} \} \text{ is countable.} \\ \Rightarrow T_1 & = T_0 \times S \text{ is countable.} \\ \Rightarrow T_2 & = T_1 \times S \text{ is countable.} \\ \Rightarrow T_3 & = T_2 \times S \text{ is countable.} \\ \vdots \end{align*} Hence all of the $T_i$ are countable.

I'm fairly certain this is wrong, so if you could explain to me what I'm doing wrong (and maybe point me in the right direction) without telling me the full answer, I would be very grateful!

Update - I just got the idea, could I argue that $T_2 \subset T_1 \times S$, and since $T_1$ and $S$ are each countable, $T_1 \times S$ is countable, so $T_2$ is also countable? And then continue this argument for $T_3, T_4, \ldots$? I'm still a little worried that it might not be valid to claim that $T_2 \subset T_1 \times S$.

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To the question in the title, a finite set is just a set, which is finite. An ordered $n$-tuple is an ordered $n$-tuple, which means that it allows repetition of the same element, and the order matters.

For example, $\{1,2,2\}$ is a set with two elements, whereas $\langle 2,1,2\rangle$ and $\langle 1,2,2\rangle$ are both $3$-tuples which are distinct.

Another important difference is that an ordered tuple is ordered. That means that it carries a structure; whereas a set is by definition a structureless object. It might be that a set is [linearly] ordered by $\in$ or by $\subseteq$, but that is usually "a happy accident". If $A$ is an arbitrary finite set, then it is not often that $A$ has a canonical ordering, whereas an $n$-tuple (even if we require that no two coordinates are equal) is always ordered.

Finally, for the proof, you can do that in several ways.

  1. You can show that the set of all tuples (every $n$), is countable; and that there is a surjection from the set of tuples onto the set of finite subsets. Then use the theorem that the image of a countable set is countable.

  2. You can show that it suffices to solve this for the case of the set being $\Bbb N$, and then use the additional structure to construct a bijection.

  3. Or you can show directly that for every $n$, the set $[S]^n=\{A\subseteq S\mid |A|=n\}$ is countable, and conclude that the set of all finite subsets if the countable union of countable sets.

Your current proof is closer in trajectory to the first case, but it tries to be the third case. You need to adjust it along either hints, and you'll be fine.

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$T_n$ is a set containing subsets of $S$.

$T_n \times S$ is a set of ordered pairs where the first element is a subset of S of cardinality $n$, and the second element is in $S$.

So you cannot have for example $T_2 \subset T_1 \times S$, but it is true that $|T_2| \leq |T_1 \times S|$.

What I think you meant to say is that

$\displaystyle T_{n+1} = \{U \cup \{s\} : U \in T_n \land s \in S \land s \notin U\}$

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