Verification of proof, interior is open. Let $(\mathbb{X},d_{\mathbb{X}})$ be a metric space and $\emptyset \neq A \subseteq \mathbb{X}$ its subset. Prove that the interior $A^{\circ} =\{ a \in A | \exists\epsilon(a) > 0, B_{\epsilon}^{d_{\mathbb{X}}}(a) \subseteq A\}$ is open in $\mathbb{X}$
My answer:
$A^{\circ}$ is open in $\mathbb{X}\quad \iff \quad\forall x\in A^{\circ}\quad \exists \epsilon>0 \quad st \quad B_{\epsilon}(x)\subset A^{\circ}$
By definition of the interior we have: $\exists\epsilon(a) > 0, B_{\epsilon}^{d_{\mathbb{X}}}(a) \subseteq A$.
Since $B_{\epsilon}^{d_{\mathbb{X}}}(a)$ is open by definition, $\forall a_0 \in B_{\epsilon}^{d_{\mathbb{X}}}(a)$ we have that $\exists \epsilon(a) > n(a_0) > 0$ such that $B_{n(a_0)}^{d_{\mathbb{X}}}(a_0) \subset B_{\epsilon}^{d_{\mathbb{X}}}(a) \subseteq A$ which implies that $B_{n(a_0)}^{d_{\mathbb{X}}}(a_0) \subset A^{\circ}$ so also $B_{n(a)}^{d_{\mathbb{X}}}(a) \subset A^{\circ}$.
Can somebody please verify this proof?
 A: The proof is essentially correct (though I'm not a fan of the notation with superscripted metric), and I don't know what $n(a_0)$ is, but that seems irrelevant to the idea of the proof. But you say that by definition open bals $B_{\varepsilon}^{d_{\mathbb{X}}}(a)$ are open, and from this you get the $\varepsilon(a) > 0$. But this is a provable fact, which follows from the triangle inequality. 
You seem to have a definition of "interior of a set" and a definition of "open". The latter means (from your proof) in your text that every point of $A$ is an interior point. Interior of a set is defined in terms of metric balls. But you haven't shown in the proof that the metric balls are themselves open in the definitional sense (namely all their points are interior points). They are called "open balls", but that doesn't make them open; they actually are, but it needs a small proof:
Suppose $x \in B_r(a)$. Then $d(x,a) < r$ by definition. Define $s = r - d(x,a) > 0$ and we claim that $B_s(x) \subseteq B_r(a)$: pick $y \in B_s(x)$, so that $d(x,y) < s$. Then $d(y,a) \le d(y,x) + d(x,a) < s + d(x,a) =  (r - d(x,a)) + d(x,a) = r$, so $y \in B_r(a)$ as required. So $x$ is an interior point of $B_r(a)$, so the interior of $B_r(a)$ is all of $B_r(a)$. Or otherwise put, $B_r(a)$ is open, for all $r>0$, $a \in X$.
If this proof was covered before, then you can assume it in the proof, and you don't need to expand it. Otherwise, you need it as a lemma.
