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?

  • 2
    $\begingroup$ Yes, it is correct. $\endgroup$ – Kavi Rama Murthy Jul 17 at 10:22
  • $\begingroup$ @KaviRamaMurthy thank you very much sir. $\endgroup$ – topologicalmagician Jul 17 at 10:31
  • $\begingroup$ 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. $\endgroup$ – 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 $\mathsf{int}(C)$.


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 $\mathsf{int}(A\cap B)$.


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