Proving that the interior of a metric space is open.

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $A=int(A).$