1
$\begingroup$

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?

$\endgroup$
  • 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
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.