A question about interior.

Question

Let $X$ a topological space. Let $A,B$ and $C$ subsetes of $X$.

If $$A\subseteq\text{Int}(B\cup C)$$ Under what conditions can we say that $\text{Int}(A)\cap\text{Int}(B\cup C)\ne\emptyset$?

Could you give me some examples and / or counterexamples?

Thanks!

Answer 1

Since you want $\mathrm{Int}(A) \cap \mathrm{Int}(B \cup C) \neq \emptyset,$ the first requirement is that $A \neq \emptyset,$ this also prevents $\textrm{Int} (B \cup C)$ from being empty. The next condition needed is that $\textrm{Int} (A) \neq \emptyset.$ With these two conditions we have that $\textrm{Int} (A) \subseteq A \subseteq \textrm{Int} (B \cup C)$ where $\textrm{Int} (A)$ is non empty. From the above inclusion we have $\mathrm{Int}(A) \cap \mathrm{Int}(B \cup C) = \textrm{Int} (A) \neq \emptyset.$

Answer 2

Assuming $Int(A) \neq \emptyset$. We know that $Int(A) \subseteq A$, so we always have $$ Int(A) \cap Int(B \cup C) \neq \emptyset $$ The only way for this to fail is if $Int(A) = \emptyset$ or $Int(B \cup C) = \emptyset$