# Proving that the interior of a metric space is open.

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?

• I don't see anything wrong. Your proof is concise and accurate. – Don Thousand Feb 11 at 20:28
• Your notation ($\bigcup\limits_{a \in int(A)}N_{\delta}(a)$) suggests that $\delta$ is the same for all $a$. – d.k.o. Feb 11 at 20:32
• Thank you so much for the feedback! – topologicalmagician Feb 11 at 21:09
• 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 $N_{\delta_a}(a) \subset int A$ – fleablood Feb 11 at 21:22
• @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. – Kabo Murphy Feb 11 at 23:30

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.

• That depends on the definition of interior. – enedil Feb 11 at 20:33

If you know already that open (in the sense of being defined by $$) balls $$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.

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 $$A=int(A).$$