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If $x$ is a limit point of a non empty subset $A$ of a metric space $X$, must it also be a limit point of $Y$ where $A\subset Y\subset X$?

This seems to be trivially true to me. If $x$ is a limit point of $A$ then for any $r > 0$ there exists $B_r(x)$ such that this open ball will contain at least one point of $A$ other than $x$ itself.

As $A \subset Y$ this open ball will clearly contain at least one point in $Y$ other than $x$ itself as all points in $A$ are in $Y$.

Is that correct or am I missing something?

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It’s fine; the result is, as you say, trivially true. – Brian M. Scott Oct 16 '12 at 20:40
@jim_CS You have a type in your title, should be $A\subset Y\subset X$ – Jean-Sébastien Oct 16 '12 at 20:45
up vote 1 down vote accepted

Yes, you are right.

Another (maybe more natural) way of thinking to limit points in metric spaces is that

$x$ is a limit point of $A$ iff $\exists a_n\in A\setminus\{x\}$: $\lim a_n = x $

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