Equivalent definitions for Interior of sets in Topological Space In my Topology class the professor gave the first definition (*) as the definition of the Interior of a set, however, I have seen, in many cases, where the alternate definition (**) is given instead. I am wondering if someone can validate my proof, and if I am not correct give a correct one. Thanks!
Let $(\mathbb{X},\tau)$ be a Topological Space and $A \subseteq \mathbb{X}$. Then the following are equivalent definitions of $A^o$.
(*) $\bigcup \{\mathcal{O}: \mathcal{O} \subseteq A \text{ and }\mathcal{O} \text{ is open}\}$
(**) $\{x : \exists \mathcal{O} \text{ open}: x \in \mathcal{O} \subseteq A\}$
Proof:
[$(*) \implies (**)$]
Let $x \in \bigcup \{\mathcal{O}: \mathcal{O} \subseteq A \text{ and }\mathcal{O} \text{ is open}\} \implies
\exists \mathcal{O} \subseteq A \text{ open}: x \in \mathcal{O} \subseteq A$. So, we have the following $x \in \{x : \exists \mathcal{O} \text{ open}: x \in \mathcal{O} \subseteq A\}$. Therefore, the folloing inclusion holds: $$\bigcup \{\mathcal{O}: \mathcal{O} \subseteq A \text{ and }\mathcal{O} \text{ is open}\} \subseteq \{x : \exists \mathcal{O} \text{ open}: x \in \mathcal{O} \subseteq A\}. \quad (1)$$
[$(**) \implies (*)$]
Let $x \in \{x : \exists \mathcal{O} \text{ open}: x \in \mathcal{O} \subseteq A\} \implies \exists \mathcal{O} \text{ open}: x \in \mathcal{O} \subseteq A \implies x \in \bigcup \{\mathcal{O}: \mathcal{O} \subseteq A \text{ and }\mathcal{O} \text{ is open}\}$. Therefore, the folloing inclusion holds: $$\{x : \exists \mathcal{O} \text{ open}: x \in \mathcal{O} \subseteq A\} \subseteq \bigcup \{\mathcal{O}: \mathcal{O} \subseteq A \text{ and }\mathcal{O} \text{ is open}\}. \quad (2)$$
By (1) and (2) we have that the following holds $$\bigcup \{\mathcal{O}: \mathcal{O} \subseteq A \text{ and }\mathcal{O} \text{ is open}\} = \{x : \exists \mathcal{O} \text{ open}: x \in \mathcal{O} \subseteq A\}.$$
 A: Your proof is correct (well done!), but I have a slight nitpick about your organizational notation:
Saying $(**) \implies (*)$ doesn't really make sense, because neither $(**)$ nor $(*)$ is a proposition. It would be much clearer if you said $(**) \subseteq (*)$. Likewise for $(*) \implies (**)$.
It's likely you framed this in your head as "proving two things are equivalent", and thus expected there to be two parts to the proof: a forwards implication and a backwards implication. But this isn't a proof of equivalence of propositions, it's a proof of equivalence of definitions! In other words, you're really trying to show that two mathematical objects are equal, not that two propositions are logically equivalent.
This isn't a big deal, but it's not just a matter of taste – it really is not correct to say $(*) \implies (**)$ here.
Edit: It may be of interest to you to prove that $A^o$ is also the largest open subset of $A$ (make this precise)
A: It's better to show that both definitions
$$A(1):= \bigcup \{O \subseteq X : O \in \mathcal{T} \land O \subseteq A\}\tag{1}$$
$$A(2):= \{ x \in X \mid \exists O_x \in \mathcal{T}: x \in O_x \subseteq A\}\tag{2}$$
are the same set for any topological space $(X, \mathcal{T})$ and $A \subseteq X$. This comes down to tow inclusions.
$A(1) \subseteq A(2)$: If $x \in A(1)$, then by definition $x$ lies in some $O$ from that union, so $O$ open and $O \subseteq A$. So that $O$ can serve as the $O_x$ that shows that $x \in A(2)$ as well.
$A(2) \subseteq A(1)$: It\s clear that the set $A(2)$ is open (as a union of open sets, from the axioms) and is a subset of $A$ (because all $O_x$ are too). So $A(2)$ is just one of the subsets we take a union of in the definition of $A(1)$ and the inclusion is trivial. QED.
Note that the $A(2)$ ("pointwise") approach is the focus in metric spaces (it's the set of interior points, that contain balls that stay inside $A$ etc.), while $A(1)$ makes more sense from a general topology point of view (a straightforward way to define a maximal open subset of $A$ directly from the axioms). But they indeed come down to the same thing.
