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Let $(X, d)$ be a metric space.

We suppose that $A\subseteq X $ is open in $X$ and $B\subseteq X $ is a not open set in $X$, where $A,B \neq \emptyset $. Is $A\cap B$ a not open set in $X$?

Ι have tried to prove it by considering that $A\cap B$ is open set in $X$ but I can't find the contradiction.

We say that a subset $A$ is open in $X$ if, whenever $a \in A$, we can find a $δ_1 > 0$ such that $ B(a,\delta_1)\subseteq A $.

On the other hand, $B$ is a not open set in $X$ if, whenever $\delta_2>0$, we can find a $b \in B $ such that $ B(b,\delta_2)\nsubseteq B$.

I suppose that $A\cap B$ is open set in $X$. So, whenever $z \in A\cap B $, we can find a $δ_3 > 0$ such that $ B(z,\delta_3)\subseteq A\cap B $.

If anybody could help me, I would be grateful.

Spy93
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1 Answers1

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That's wrong. Consider the reals with the usual metric, $A=(0,4)$ open and $B=(1,5]$ not open. Then $A \cap B=(1,4)$ is open.

balddraz
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  • @Spy93 Here's a more interesting question for you: given a metric space, suppose $A$ is open, $B$ is closed, $A \cap B \neq \emptyset$ and neither $A$ nor $B$ contain one another. Is $A \cap B$ open, closed or neither? – balddraz Jun 27 '17 at 14:21
  • Supposing that we work we the usual metric the intersection of $A $ and $B$ is neither open or closed set. If $A= (3,6)$ and $B=[5,7]$ , then $A\cap B=[5,6)$ , which is neither open or closed set. – Spy93 Jun 27 '17 at 14:39
  • @Spy93 Yes, for that particular example the intersection is neither open nor closed. But does the pattern hold for any metric space and any $A$ and $B$ as defined? I think that is a harder question to answer worth investigating. – balddraz Jun 27 '17 at 14:54