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I have an given set $S$ where $S=$ set of finite subsets of $\mathbb{N}.$ We need to prove $S$ is countably infinite.

My approach: I need to prove there is one-to-one correspondence between $S$ and ${\mathbb{N}}.$

Suppose $S = \{\{1,2,3\}, \{1,3,4,5\},\{4,5\}, \{\emptyset\},.............\} =\{f_1,f_2,f_3,f_4..............\}$where $f_i\subseteq\mathbb{N}.$

$\mathbb{N}=\{1,2,3,4...... ..............\}$

Now $1$ map to $f_1,$$2$ map to $f_2,$ $3$ map to $f_3..........$ and so on. Now apply Cantor diagonalisation theorem $f'=\{2,3.............\}$ where $2$ comes from $f_2$ because $2\notin f_2,$$3$ comes from $f_3$ because $3\notin f_3...........$ and so on, $f'$ is different from $f_i.$ $f'$ isn't covered in bijection. $f:\mathbb{N}\to S$ isn't bijection.

So $S$ is uncountable. Where did I wrong don't understand?

Mark Saving
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    $f'$ isn't finite. So $f' \notin S$. – Mark Saving Oct 06 '21 at 21:02
  • Hint for actual proof: Use the fact that the countable union of countable sets is countable. Can you characterize each "type" of finite subset in such a schema? – Alan Oct 06 '21 at 21:04
  • @Mark $f'$ making from $f_i$, so why $f_i$ is infinite? –  Oct 06 '21 at 21:05
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    @User4567 I should be more precise. You do not know that $f'$ is finite. When you eventually do come up with a surjection $\mathbb{N} \to S$, the fact that it is a surjection will show that $f'$ is not finite since it is not in the range of $f$. Your argument only works to prove that $f$ isn't a surjection if you assume that $f'$ is finite, which you cannot do. – Mark Saving Oct 06 '21 at 21:07
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    @Mark if you insert your comment in answer, I will accept it. –  Oct 06 '21 at 21:10
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    Related: what is the scheme for choosing the elements of $f'$? The point is that it should be finite but not be any of the elements of the set, so it should differ from $f_i$ in the membership of some element, but this choice could be anything. One concrete construction would be to choose $f'$ to contain $i$ if and only if $i \not \in f_i$. This would produce a subset of $\mathbb{N}$ not in the list, but it's "probably" infinite. – Ian Oct 06 '21 at 21:11
  • @User4567 Here is how f' could be infinite: maybe 1 is not in f_1, and so you put 1 in f'. Likewise, if 2 is not in f_2, you could pick 2 to put into f'. Etc. So, in the end, f' becomes the set of all natural numbers. So: it is possible for f' to end up as infinite. You would have not proved that f' is finite. In fact, your definition of f' isn't even well-defined: you say that for each f_i you pick some number that is not in f_i, but there are infinitely such numbers. – Bram28 Oct 06 '21 at 21:11

1 Answers1

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Your proof that $S$ is not countable goes as follows:

Consider any $f : \mathbb{N} \to S$. Define $f' = \{n \in \mathbb{N} \mid n \notin f_n\}$.

Then we see that $f'$ is not in the range of $f$. Therefore, $f$ cannot be surjective. Thus, $f$ can't be a bijection.

However, there is a flaw in this reasoning. It assumes that $f' \in S$. In other words, it assumes that $f'$ is finite. If $f'$ is not finite, then there is no problem at all with the fact that $f'$ is not in the range of $f$.

In fact, it is indeed possible to construct a bijection $f : \mathbb{N} \to S$. The resulting $f'$ will be an infinite set.

For how to prove that $S$ is countable, see this answer.

Mark Saving
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