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Let $A$ be a subset of the set of real numbers containing none of its limit points i.e. $A \cap A'=\phi$ where $A'$ is the set of all limit points of $A$ ; then must $A$ be countable ?

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  • $\begingroup$ Yes. (for any points in $x$, try to find an open interval $U$ so that $U \cap A = \{x\}$) $\endgroup$ – user99914 Dec 8 '14 at 4:38
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Here's a proof:

For $x \in A$, we have that $x$ is not a limit point of $A$. Thus (unless $A$ is just a single point, which is trivial) there's some nonzero infimum of the set of distances from $x$ to other points in $A$. Let $U_x$ be an open ball around $x$ of half this in radius.

Then the $U_x$ are disjoint.

There are countably many $U_x$ with radius larger than $1/n$. (Since only finitely many such can be a subset of any finite interval $[-N,N]$.)

Thus there are countably many in total (as a countable union of countable sets is countable).

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