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I have to show that

$$ \bigcup_{n=1}^{\infty}A_{n} := (a: \exists n \in \mathbb{N}: a \in A_{n}) $$

EDIT: So beneath this edit were just my thoughts. Above is the task. Is there any elementary way to prove that this Union of countable sets is countable? (besides counting like $\mathbb Q$)

As far as I understand countability, I have to show that the union is a bijection to $\mathbb{N}$.

$$ f: \bigcup_{n=1}^{\infty}A_{n} \to \mathbb{N} $$

Well, surjection is given by definition above, isn't it?

Now I only have to show injection. So if $f(a)=f(b)$ then $a=b$. How do I show that for the union?

Seen
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  • Surjection isn’t “given by definition.” You have to show that there exists a bijection (a function) from the union to $\mathbb{N}$. It doesn’t make sense to say “the union is a bijection.” What is $f$ in your question? – Steve Kass Nov 02 '14 at 23:38
  • f is the function from the union to $\mathbb{N}$. I thought the definition $(a: \exists n \in \mathbb{N}: a \in A)$, would already state surjection – Seen Nov 02 '14 at 23:41
  • $f$ is what function from the union to $\mathbb{N}$? (I only noticed your definition of the union just now. I’m not sure what it means. The notation $(a:\exists n\in\mathbb{N}:a\in A)$ doesn’t mean much. What is $A$, for example?) Your question says “I have to show defined to be .” – Steve Kass Nov 02 '14 at 23:45
  • A is one of the countable sets of the union. A_{1}, A_{2} … all countable – Seen Nov 02 '14 at 23:48
  • Maybe there is an error in the task. It is really that definition for the union. Maybe there is an quantifier missing… Ok corrected it. You were right. The n was missing – Seen Nov 02 '14 at 23:50
  • Ok, your union should be defined as ${a:\exists n\in\mathbb{N}:a\in A_n}$ The subscript you left out is important to include. Still, this definition does not define a function $f$, and you can’t let $f(a)$ be the subscript of the particular set $a$ is in, because $a$ could be in none, one, or many of the sets. – Steve Kass Nov 02 '14 at 23:50

2 Answers2

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One way of looking at this is that a countable union of countable sets is countable for the same reason that $\mathbb Q$ is countable.

Write $$A_1 = \{ a_{11}, a_{12}, a_{13}, \dots \}$$ $$A_2 = \{ a_{21}, a_{22}, a_{23}, \dots \}$$ $$etc\dots$$

and then count their union using the same zig-zag pattern we use to count $\mathbb Q$ in the standard proof of its countability.

gamma
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  • This only works if the union is disjoint. – user2345215 Nov 02 '14 at 23:45
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    @user2345215 It works for $ \mathbb Q $ with redundant entries such as $ \frac{1}{1}, \frac{2}{2}, \dots $ where we are supposed to "skip" redundancies. – gamma Nov 02 '14 at 23:46
  • @user2345215 even if it isn't disjoint it is a surjective function from $\mathbb{N}$ onto the union. It is obvious how to define a surjective function from the union onto $\mathbb{N}$ so this should still prove they are both "at least as big" as the other. –  Nov 02 '14 at 23:48
  • @Gage So we are back to Cantor-Bernstein? I doubt he knows that theorem. – user2345215 Nov 02 '14 at 23:49
  • @user2345215 I don't know that we need to invoke Cantor-Bernstein in an elementary setting. We learn to count $ \mathbb Q $ without reference to Cantor-Bernstein, and the OP is posing the question in an elementary setting. – gamma Nov 02 '14 at 23:52
  • Next week is my third week at university, never heard of it. But we had to count $\mathbb Q$ like that – Seen Nov 02 '14 at 23:53
  • @NickR Of course not. I was reacting to Gage's post. – user2345215 Nov 02 '14 at 23:55
  • @Seen Third week. I'm surprised. I didn't encounter this material until my second year (except through private reading). Do you know the standard argument for the countability of the rationals? – gamma Nov 02 '14 at 23:56
  • I looked at my papers and it was just really informal. Same for whole numbers. $\mathbb Q$ it was just with a grid and arrows and for $\mathbb Z$ just 0,1,-1,2,-2 and so on.. Btw. I study computer science – Seen Nov 03 '14 at 00:02
  • @Seen The same grid and arrows works fine here. We have the grid, just draw the arrows. Informal arguments can be very convincing. – gamma Nov 03 '14 at 00:06
  • ok cool, at least I got something. And its just an excersice task. Thanks for the help! :) – Seen Nov 03 '14 at 00:07
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$$\aleph_0=|A_1|\le\Big|\bigcup_{n\in\mathbb N} A_n\Big|\le\aleph_0\cdot\sup_{n\in\mathbb N}|A_n|=\aleph_0\cdot\aleph_0=\aleph_0$$

user2345215
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  • This is basically a circular argument, I think, because it depends on the fact that $\aleph_0\cdot\aleph_0=\aleph_0$. I doubt that a student learning about countability already has that fact available to use. – Steve Kass Nov 02 '14 at 23:39
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    @SteveKass Even if he knew that, I also need Cantor-Bernstein, so I didn't want to elaborate. And I disagree it's circular, the original union is not required to be disjoint. – user2345215 Nov 02 '14 at 23:43
  • we didn't have that yet, so no. Just basic countabilty. I dont really know what this means.. – Seen Nov 02 '14 at 23:44