MHB Are All Countable Sets Closed?

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Hello everyone!

I want to show that all countable sets are closed. I can show that finite sets are closed, and the set of all natural numbers is closed by showing its complement to be a union of open sets. Now, can I start like this:

A is a countable set. Every element in A can be "mapped" to an element in N by the property of countability (I presume). N is finite, so A is finite too.

Is there proof correct, if it is but technically incorrect, could you suggest a better proof.

Thanks! :o
 
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$\Bbb N$ is not finite!

And not all countable sets are closed: take the real line with usual topology, and $S:=\{n^{-1},n\in\Bbb N\}$ is countable, but not closed (as $0$ is in the closure but not in the set).
 
Another example: the set of all rational numbers is countable but not closed- its closure is the set of all real numbers.
 
I apologize about saying N is finite, I forgot to edit that out. I believe I must review what countability strictly means.
 
A set is countable if it is finite or there is a bijection with $\mathbb{N}$. :D
 
If you consider the naturals (any subset) or rationals or something with the discrete metric then these are open, so you have (at least) countably many countable sets that are open :)
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...
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