MHB Are All Countable Sets Closed?

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Not all countable sets are closed, as demonstrated by examples like the set of natural numbers and the set of rational numbers, both of which have closures that include additional points. The initial claim that all countable sets are closed is incorrect, as shown by the set S = {n^-1 | n ∈ ℕ}, which does not include 0 in its set but has it in its closure. The discussion highlights a misunderstanding of countability, clarifying that a set is countable if it is finite or can be put in a one-to-one correspondence with the natural numbers. Additionally, subsets of countable sets can be open in certain topologies, such as the discrete metric. Overall, the conversation emphasizes the need for a clearer understanding of topological concepts related to countability.
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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 :)
 

Thread 'About the definition of topological manifold using closed sets'

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.
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