# Interior topology

Let $$(X,\tau)$$ be a topological space. Show that $$int(A) \cap int(B)$$ $$\subset$$ $$int(A \cap B)$$

Proof: x $$\in$$ $$int(A)$$ and $$x\in int(B)$$ means that $$\exists$$ $$U_1 (open)$$ containing x so that $$U_1 \subset A$$ and $$\exists U_2$$ open containing x so that $$U_2 \subset B$$ hence $$U_1 \cap U_2$$ is open and a subset of $$A \cap B$$. Since x is contained in $$U_1$$ and $$U_2$$ it follows that x is in the interior of $$A\cap B$$.

Is this proof correct?

• Yes, it is correct. – Kavi Rama Murthy Jul 17 at 10:22
• @KaviRamaMurthy thank you very much sir. – topologicalmagician Jul 17 at 10:31
• Since $C\subset D\implies int (C)\subset int (D)$ we also have $int (A\cap B)\subset int (A)$ and $int (A\cap B)\subset int (B) .$ So $int (A\cap B)\subset int(A)\cap int(B).$ Combined with the Q we have $int(A)\cap int(B)=int (A\cap B).$....Your proof is correct. – DanielWainfleet Jul 18 at 9:15

Yes, correct.

Alternative way that makes use of the following characterization of "interior":

The interior of set $$C$$ is the union of all open subsets of $$C$$.

(If this characterization is not yet familiar to you then I advice you to make it familiar to you)

You could also say: $$\mathsf{int}(C)$$ is the largest open subset of $$C$$. This in the sense that $$\mathsf{int}(C)$$ is an open subset of $$C$$, and secondly that every open subset of $$C$$ is a subset of $$\mathsf{int}(C)$$.

Proof:

We have $$\mathsf{int}(A)\subseteq A$$ and $$\mathsf{int}(B)\subseteq B$$ by definition and consequently $$\mathsf{int}(A)\cap\mathsf{int}(B)\subseteq A\cap B$$.

As a finite intersection of open sets the set $$\mathsf{int}(A)\cap\mathsf{int}(B)$$ is an open set, hence must be a subset of $$\mathsf{int}(A\cap B)$$.