Let $(X,T)$ be a topological space and $A \subset X$. Show that $int(A)$ is the union of all open subsets of $A$.
We have defined the intererior of $A$, $int(A)$ to be the set of interior points. Furthermore $x \in X$ is an interior point if there is $U \in U_x$ with $U \subset A$. Where $U_x$ is the neighbourhood system of $x$.
For every point $x_i \in int(A) \ \exists U \in U_x \ : \ U \subset A$. Since $U \in U_x \ \exists V \in T (open) \ : \ x_i \in V_i$ and $V_i \subset U \subset A$. Therefor we can conclude $\cup_i V_i \subset A$.
Now I need help with how I can proceed.
Edit from comment.
$U_x = \{U \in X : U \text{ neighbourhood of } x\in X \}$ A set $U \subset X$ is called a neighbourhood if there is $V \in T$ with $x \in V$ and $V \subset U$.
Best Answer
Let $Y$ be the union of all open subsets of $A$.
Therefore, $Y\subseteq A$ and $Y$ is open (since the union of arbitrary open sets is open). Therefore, if $y\in Y$, then $y\in \operatorname{int}(A)$ as we can use $Y$ for $U_y$.
Suppose $x\in\operatorname{int}(A)$, then $U_x$ is an open subset of $A$ containing $x$, so $U_x$ is in the union defining $Y$. Therefore, $U_x\subseteq Y$ and so $x\in Y$.