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Are all countable sets closed?

OhMyMarkov

Member
Mar 5, 2012
83
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! :eek:
 

girdav

Member
Feb 1, 2012
96
$\Bbb N$ 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).
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Another example: the set of all rational numbers is countable but not closed- its closure is the set of all real numbers.
 

OhMyMarkov

Member
Mar 5, 2012
83
I apologize about saying N is finite, I forgot to edit that out. I believe I must review what countability strictly means.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
A set is countable if it is finite or there is a bijection with $\mathbb{N}$. :D
 

hmmm16

Member
Feb 25, 2012
31
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 :)