int $A$ is open

Question

For a topological space $(X, \mathcal{O})$, I want to show that for arbitrary $A \subset X$, int $A$ is open without using the fact that every open set contained in $A$ is contained in int$A$ and that $A$ is open iff $A =$int $A$.

int$A = \{x \in A | \exists U \in \mathcal{O} \text{ s.t. } x \in U, U \subset A\}$.

If we take arbitrary $y \in $ int $A$, then we have that there exists an open neighbourhood of $y$, $U_y \subset A$, but I need to show that $U_y \subset $ int $A$.

To do this, I take an arbitrary element, $a \in U_y$, and assume that it is not in the interior of $A$. Then, it must either be in the exterior of A or on the boundary of $A$. We can rule it being an exterior point since $U_y \subset A$, but I don't see how to rule out that it might be a boundary point, since this seems like a feasible value for $a$ (open ball that touches the boundary).

Answer 1

If $a\in U_y$, then, since:

1. $U_y$ is an open set;
2. $a\in U_y$;
3. $U_y\subset A$,

$a\in\operatorname{int}A$.

Answer 2

Instead of ruling out its being a boundary point, show directly that $a \in \operatorname{int} A$ by showing $\exists U \in \mathcal O$ such that $a \in U$ and $U \subset A$. Which $U$ can you use?

Answer 3

Let $(X,\mathcal{T})$ be a topological space. Fix $A\subseteq X$. Observe

$$\operatorname{int}A:=\bigcup\{U\in\mathcal{T}:U\subseteq A\}.$$

Because we take as an axiom that the union of open sets are open, we have that the interior is always open.