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I would like to find out if my proof is alright or if it is missing any steps...thanks.

$B$ is a non-empty open set. Show that every point in $B$ is a limit point of $B$.

Since $B$ is open, $\forall b \in B$, there exists an $\epsilon$-neighborhood $V_\epsilon (b) \subseteq B$. This means that for all $b \in B$, every single possible $\epsilon$-neighborhood of b intersects $B$ at some point other than $b$. By definition, this means that all elements of B are limit points of B.

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  • $\begingroup$ Once again, that is correct. However, note that every limit point of the open set need not be inside it. $\endgroup$ – астон вілла олоф мэллбэрг Nov 3 '16 at 1:02
  • $\begingroup$ Note that this applies to a topological space that has no isolated points , A point $x$ in a space $S$ is an isolated point of $S$ iff $\{x\}$ is open in$S.$ $\endgroup$ – DanielWainfleet Nov 3 '16 at 1:15
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That is, for your $V_{\varepsilon}(b)\subset B$, we have $(V_{\varepsilon}(b)\smallsetminus\{b\})\subset V_{\varepsilon}(b)\subset B$. Then $(V_{\varepsilon}(b)\smallsetminus\{b\})\cap B\neq\varnothing$.

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