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Let $X$ be the set of all finite subsets of $\mathbb{N}$. Thus some elements of $X$ include $\emptyset$ and $\{1, 5, 9\}$ and $\{3, 346\}$ and $\{1\}$; however, the set of even natural numbers is not an element of $X$. Prove that $X$ is countable.

I think I might need to prove that all sets of $X$ are countable and there's a countable number of sets, but I'm not sure. I also don't know what " the set of even natural numbers is not an element of $X$. Prove that $X$ is countable" means?

Anton Vrdoljak
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Q G
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    Welcome to Mathematics Stack Exchange. The set of even natural numbers is not an element of $X$, because that set is not a finite subset of $\mathbb N$ – J. W. Tanner Aug 30 '20 at 13:34
  • "The set of even natural numbers is not an element of $X$" means that the set ${0,2,4,6,\dots} \notin X$. This is because it's not a finite set. – Dark Malthorp Aug 30 '20 at 13:35
  • "Prove that $X$ is countable" means that you need to prove that $X$ is a countable set, it does not mean you need to prove every element of $X$ is countable. You're trying to prove that there are only countably many finite subsets of $\mathbb{N}$. – Dark Malthorp Aug 30 '20 at 13:36
  • Do you know that a (finite) product of countable sets is countable, and the union of countably many countable sets is countable? – J. W. Tanner Aug 30 '20 at 13:37
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    Thank you guys so much! – Q G Aug 30 '20 at 13:41
  • Also, cf. this question and this question – J. W. Tanner Aug 30 '20 at 13:43
  • No, I don't really know that a (finite) product of countable sets is countable, or that the union of countable sets is countable. We didn't go over it in class yet. Is there a way to systematically enumerate X? – Q G Aug 30 '20 at 13:43

3 Answers3

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You have to find a map $\phi: \mathbb N \to X$ which is surjective. Then you are done. The Set of uneven numbers is not in $X$ because that set would be infinite.

You can first look at the case of a fixed $n\in \mathbb N$ and then realise that there are countable infinitely many subsets of $\mathbb N$ that have $n$ elements by finding a surcjective map $\phi_n : \mathbb N \to X_n$. Where $X_n$ are those subsets of length $n$. ($\mathbb N^n \cong \mathbb N$)

If you have that you can define $\Phi: \mathbb N^2 \to X; (n,m) \mapsto \phi_n(m)$ and note that $\mathbb N^2$ is itself countably finite.

crush3dice
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It is already answered, but you can also write a 1-1 map from $X$ to $\mathbb{N}$ to prove it. Let $p_k$ be the $k$-th prime number, so $p_1 = 2$, $p_2 = 3$, ... Define $f: X\rightarrow \mathbb{N}$ as follows. Any element of $X$ is a finite subset of the set of natural numbers of the form $A=\{n_1, n_2, ..., n_k\}$ where $n_1 < n_2 < ... < n_k$. Define $$f(A) = {p_1}^{n_1} ... {p_k}^{n_k}$$ Then $f$ is one-to-one.

user114285
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If $P(A)$ denote the power set of $A$, you can write $X$ as follows:

$$X = \cup_{k=1}^{\infty} \{ Y : Y\in P(\{1,2,\dots ,k\}) \}$$

So X is countable union of countable sets, hence countable.

Paulo
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