Let $X$ be a topological space and let $A,B\subseteq X.$ We know that in general, we only have $$int(A)\cup int(B)\subseteq int(A\cup B). $$

My question is: When do we say that equality holds?

I came with this idea because if there was such condition for which the equality holds, then I can manage to prove that $$\partial (A\cup B)=\partial A\cup \partial B$$ whenever $\overline{A}\cap \overline{B}=\varnothing.$

Any tips?:)

  • 2
    $\begingroup$ For sure, it holds if $A,B$ are open sets. $\endgroup$
    – Sigur
    Jan 26, 2013 at 15:05
  • 1
    $\begingroup$ @Sigur Ahh...thanks If $A,B$ are open then we get $$int(A\cup B)=A\cup B=int(A)\cup int(B).$$ $\endgroup$
    – Juniven
    Jan 26, 2013 at 15:14
  • $\begingroup$ What is $b$? .. $\endgroup$
    – M.B.
    Jan 26, 2013 at 15:15
  • 3
    $\begingroup$ I think that if $\partial A\cap\partial B=\emptyset$ then the equality holds. $\endgroup$
    – Tomás
    Jan 26, 2013 at 15:31
  • 1
    $\begingroup$ @Tomas. Thanks for the suggestion. I have to verify it on my own(If I can do it.:)) $\endgroup$
    – Juniven
    Jan 26, 2013 at 15:35

1 Answer 1



$ int(A\cup B) = int(A) \cup int(B) \iff \partial A \cap \partial B \subset \partial (A \cup B) $


Since you are more intrested in the ($\Leftarrow$) part of the claim I will first prove that. For both parts of the proof, remeber that interior, exterior and boundary of a set is a partitioning of the whole space i.e. $int(A) \cup \partial A \cup ext(A) = X$ where all three sets are disjoint. Also, interior and exterior are open sets.

($\Leftarrow$) As you have said $int(A) \cup int(B) \subset int(A\cup B)$. So, I will only focus on proving $int(A\cup B) \subset int(A) \cup int(B)$.

Let $x \in int (A \cup B)$. If $x \in int(A)$ or $x \in int(B)$ then we are done. So, assume that $x \notin int(A)$ and $x \notin int(B)$.

By hypothesis, if $x \in \partial A \cap \partial B$ then $x \in \partial(A\cup B)$ but interior and boundary of $(A \cup B)$ are disjoint which is a contradiction. So, $x \notin \partial A \cap \partial B$.

Without loss of generality, assume that $x \notin \partial B$. Then, $x \in ext(B)$ and either $x \in \partial A$ or $x \in ext(A)$.

Case 1: $x \in \partial A$,

Since $x \in ext(B)$, there is a neighborhood $V \subset B^C$ of x. Since $x \in int(A \cup B)$ there is a neighborhood $U \subset (A\cup B)$ of x. Then, $U \cap V$ is a neighborhood of $x$ in $X$. Note that $(U \cap V) \subset (A\cup B) \cap B^C \subset A$. But since $x \in \partial A$, any neighborhood of $x$ must intersect non-trivially with $A^C$. Hence, it is a contradiction. So, $x \notin \partial A$.

Case 2: $x \in ext(A)$,

$x \in int(A \cup B) \subset A \cup B \subset \bar A \cup \bar B = (int(A) \cup \partial A) \cup (int(B) \cup \partial B)$. However $x \in ext(A)$ and $x \in ext(B)$ implying that $x \notin int(A) \cup \partial A$ and $x \notin int(B) \cup \partial B$ which is a contradiction.

Therefore, $x \in int(A)$ or $x \in int(B)$ i.e. $x \in int(A) \cup int (B)$.

$ (\Rightarrow) $ Let $x \in \partial A \cap \partial B$.

So, $x \notin int(A) $ and $ x \notin int(B)$ implying $x \notin int(A) \cup int(B) = int(A\cup B)$.

Thus, $x \in \partial (A \cup B) \cup ext(A\cup B)$.

Assume $x \in ext(A \cup B)$ then there is a neighborhood $P \subset (A \cup B)^C $ of $x$ (where $A^C$ means the complement for set $A$).

But $(A \cup B)^C \subset A^C$. That is, we have a neighborhood of $x$ which is contained in $A^C$ meaning $ x \notin \partial A$ which is a contradiction.

Hence, $x \in \partial(A\cup B)$.

  • 2
    $\begingroup$ Thanks. You have done too much to this problem. $\endgroup$
    – Juniven
    Jan 26, 2013 at 20:48
  • 4
    $\begingroup$ I enjoyed it and it was a nice exercise to find out such a relation. $\endgroup$
    – user21965
    Jan 26, 2013 at 21:39
  • $\begingroup$ This result eventually got published: semanticscholar.org/paper/… $\endgroup$
    – bubba
    Jan 30 at 11:20

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