If $A$ is a non empty subset of the metric space ($\mathbb{R}$,d) and $A$ = $A'$, then $A$ is uncountable.

Question

I have to prove this proposition "If $A$ is a non empty subset of the metric space ($\mathbb{R}$,d) and $A$ = $A'$, then $A$ is uncountable. " but I dont know how. Can anybody help me ?

Sets with this property are known as perfect sets. The result that perfect sets are uncountable (in a complete metric space) is "famous" if you will: see here for some proofs. — , Mar 25 '21 at 14:26

score 0 · Answer 1 · answered Mar 26 '21 at 15:48

The set A cannot be finte as every element in that finte set is an isolated point.Now if A is countably infinite then A can be written as a countable union of closed sets (since every singleton set is closed).Now A being a closed set of a complete space is complete.But by Baire category th then at least one closed set must have non -empty interior but singleton set has empty interior . So A must be uncountable.

If $A$ is a non empty subset of the metric space ($\mathbb{R}$,d) and $A$ = $A'$, then $A$ is uncountable.

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