It's kind of a simple proof (I think) but I´m stuck!

I have to show that $\operatorname{int} (A \cap B)=\operatorname{int} (A) \cap \operatorname{int}(B)$.

(The interior point of the intersection is the intersection of the interior point.)

I thought like this:

Intersection: there's a point that is both in $A$ and $B$, so there is a point $x$, so $\exists ε>0$ such $(x-ε,x+ε) \subset A \cap B$.I don´t know if this is right.

Now $\operatorname{int} (A) \cap \operatorname{int}(B)$, but again with the definition ,there is a point that is in both sets,there's an interior point that is in both sets,an $x$ such $(x-ε,x+ε)\subset A \cap B$. There we have the equality.

I think it may be wrong. Please, I'm confused!

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    $\begingroup$ It´s better now,@Peter Tamaroff! $\endgroup$ – HipsterMathematician Jun 9 '12 at 21:08

If $x\in\mathrm{int}(A\cap B)$, then there exists $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq A\cap B$. And since $A\cap B\subseteq A$ and $A\cap B\subseteq B$, then...

If $x\in\mathrm{int}(A)\cap\mathrm{int}(B)$, then there exists $\epsilon_1\gt 0$ such that $(x-\epsilon_1,x+\epsilon_1)\subseteq A$, and there exists $\epsilon_2\gt 0$ such that $(x-\epsilon_2,x+\epsilon_2)\subseteq B$. Can you find a single $\epsilon$ that works for both sets? Then what can you say about $(x-\epsilon,x+\epsilon)$?


Always remember the trivial inclusion using the property $A\subset B \implies \operatorname{int}A\subset \operatorname{int}B$. Then:

$$A\cap B\subset A,\ A\cap B\subset B \implies \operatorname{int}(A\cap B)\subset \operatorname{int}A,\ \operatorname{int}(A\cap B)\subset \operatorname{int}B$$

therefore $\operatorname{int}(A\cap B)\subset \operatorname{int}A\cap\operatorname{int}B$. The other inclusion is in Arturo Magidin answer.

In same form we can prove the trivial inclusion $\operatorname{int}A\cup\operatorname{int}B\subset\operatorname{int}(A\cup B)$. Using only the fact that $A\subset A\cup B$ and $B\subset A\cup B$.

If you known what is the closure of a set you can prove that if $A\subset B$ then $\overline{A}\subset \overline{B}$. Then the following facts are inmediate:

$$\overline{A\cap B}\subset\overline{A}\cap\overline{B},$$ $$\overline{A}\cup\overline{B}\subset \overline{A\cup B}.$$

Please don't forget it. This observation is crucial and is always used. The other inclusion sometimes is false or sometimes is true, generally you must to use definition indeed this basic properties.

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    $\begingroup$ So,I could have used only the properties,with no definitions? $\endgroup$ – HipsterMathematician Jun 9 '12 at 21:43
  • $\begingroup$ Only in the trivial inclusion. $\endgroup$ – Gaston Burrull Jun 9 '12 at 21:48
  • $\begingroup$ Try to prove the statments that I wrote but I didn't prove. Using set properties :) $\endgroup$ – Gaston Burrull Jun 9 '12 at 21:51
  • $\begingroup$ Ok,Thanks!I think i understood.sometimes i have problems is proofs ,using epsilons...it´s a very stupid mistake,I know,I just can´t make things right ,it stays a little confused.... $\endgroup$ – HipsterMathematician Jun 9 '12 at 21:56
  • $\begingroup$ with this property you probably will make faster some of your proofs $\endgroup$ – Gaston Burrull Jun 9 '12 at 22:32

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