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WTS: The interior of a metric space A is open.

Proof:

Let a $\in$ $int(A)$, we can find a $\delta$ $>0$ : $N_{\delta}(a)$ $\subset A$, hence a is in some open neighborhood, so a $\in$ $N_{\delta}(a)$, for some $\delta$ $>0$. Hence $int(A) \subset $ $\bigcup\limits_{a \in int(A)}N_{\delta}(a)$. But since every neighborhood is contained within the interior of A, it follows that $\bigcup\limits_{a \in int(A)}N_{\delta}(a)$ $\subset int(A)$, hence the two sets are equal. May someone tell me how to improve it, and what I can do to make it better, clearer and logical?

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    $\begingroup$ I don't see anything wrong. Your proof is concise and accurate. $\endgroup$ – Don Thousand Feb 11 at 20:28
  • $\begingroup$ Your notation ($\bigcup\limits_{a \in int(A)}N_{\delta}(a)$) suggests that $\delta$ is the same for all $a$. $\endgroup$ – d.k.o. Feb 11 at 20:32
  • $\begingroup$ Thank you so much for the feedback! $\endgroup$ – topologicalmagician Feb 11 at 21:09
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    $\begingroup$ Every neighbor hood selected was selected to be a neighborhood contained in $A$; not the interior of $A$. So we know $N_{\delta_a} (a) \subset A$ but we don't know $N_{\delta_a}(a) \subset int A$ $\endgroup$ – fleablood Feb 11 at 21:22
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    $\begingroup$ @topologicalmagician An important point to note: in several of your posts you are replacing the term '$A$ is a subset of a metric space' by '$A$ is a metric space'. The interior of any metric space $X$ is $X$ itself so it is obviously open. $\endgroup$ – Kabo Murphy Feb 11 at 23:30
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Your proof is perfectly fine.

Maybe an easier approach would be just to state that the interior of A is the union of all open sets contained in A, and any union of open sets is again open.

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  • $\begingroup$ That depends on the definition of interior. $\endgroup$ – enedil Feb 11 at 20:33
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If you know already that open (in the sense of being defined by $<r$) balls $N_r(a)$ are open (in the metric topology), your argument is fine, writing the interior as a union of open balls this way.

Otherwise maybe show that first as a lemma of sorts.

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For every $a\in A$ we have $a\in N_1(a)\subset A$ so $a\in N_1(a)\subset int (A)$ so every $a\in A$ belongs to $int(A).$

So $A\subset int (A).$

And we also have $int (A)\subset A$.

So $A\subset int (A)\subset A,$ which implies $A=int(A).$

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