TITLE: If $A$ and $B$ are sets of real numbers, then $(A \cup B)^{\circ} \supseteq A^ {\circ}\cup B^{\circ}$
QUESTION [5 upvotes]: I have a proof for this question, but I want to check if I'm right and if I'm wrong, what I am missing.
Definitions you need to know to answers this question: $\epsilon$-neighborhood, interior points and interiors.  Notation: $J_{\epsilon}(a)$ means a neighborhood formed around a (i.e. $(a-\epsilon, a+\epsilon)$.  An interior point in some set $A$ is a point where an $\epsilon$-neighborhood can be formed within the set.  The set of all interior points in $A$ is denoted as $A^0$ and is called the interior.
My proof for the question:  I'm proving based on most subset proofs where you prove that if an element is in one set, then it must be in the other.  Say $x \in A^0 \cup B^0$.  Then there is a $\epsilon > 0$, where $J_{\epsilon}(x) \subseteq A$ or $J_{\epsilon}(x) \subseteq B$.  This implies that $J_{\epsilon}(x) \subset A \cup B$ which then implies $x \in (A \cup B)^0$.  We can conclude from here that $(A \cup B)^0 \supseteq A^0 \cup B^0$.

REPLY [8 votes]: Here is a slightly shorter proof:
We have $A^\circ \subset A \cup B$ and $B^\circ \subset A \cup B$, so we must have
$A^\circ \cup B^\circ \subset A \cup B$.
Since $A^\circ \cup B^\circ$ is open and the interior is the largest open set contained in a set, we must have
$A^\circ \cup B^\circ \subset (A \cup B)^\circ$.