In a metric space $(X,d)$, let $E$ be a nonempty subset of $X$.

Let $x$ be a closure point of $E$. Thus for each $n\in \Bbb N$, $B(x,\frac 1n)$ contains a point of $E$; we set it as $x_n$. We thus constructed a sequence $(x_n)$ from elements of $E$ that converges to $x$. Thus the closure point is the limit of a sequence in $E$.

Let $(x_n)$ be a sequence in $E$ converging to $x$. So to any positive $\epsilon$, there corresponds an $N\in \Bbb N$ such that $\forall n≥N$, $x_n\in B(x,\epsilon)$. Thus every neighborhood of $x$ contains a point of $E$. So the limit of the sequence is a closure point of $E$.

Is the claim and proof correct? Also, is the result true in any topological space? If not, why? What property of metric spaces (or the topological spaces where this is true) causes this to happen?


It is correct for metric spaces, but it is not true in general. What's spectial for metric spaces is that, for every point $p$, there is a sequence $(V_n)_{n\in\mathbb N}$ of neighbourhoods of $p$ such that, for every neighbourhood $V$ of $p$, there is some $n\in\mathbb N$ such that $V\supset V_n$ (take $V_n=B_{1/n}(p)$, for instance). This is not true in general for topological spaces. Those for which this is true are called first countable spaces.


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