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I have a problem with the following exercise:

Let be $A,B$ subsets of a topological space $X$. Prove that

$$\overline{A\cap B}\subset \bar{A}\cap\bar{B}.$$

I only know is that $\bar{A}=A\cup\partial A$.

Emanuele Paolini
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MathCracky
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    It's also quite instructive to consider whether $\bar A \cap\bar B = \overline {A \cap B}$ is true. – BigbearZzz Oct 18 '15 at 14:27
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    Well, the equality can't be true. Let me give you an example:

    Let be $X=\mathbb{R}$, then we have $A=(0,1)$ and $B=(1,2)$. Then is $A\cap B=\emptyset$, so $\overline{A\cap B}=\emptyset$. But: $\bar{A}=[0,1]$ and $\bar{B}=[1,2]$. So, $\bar{A}\cap\bar{B}={1}$. That means $\overline{A\cap B}\neq \bar{A}\cap\bar{B}$.

     – MathCracky Oct 18 '15 at 14:36
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    @Alex $Q$ ana $Q^c$ – R.N Oct 18 '15 at 14:39
  • Consider that $Cl (A\cap B)\subset Cl(A)$. – DanielWainfleet Oct 18 '15 at 15:15

3 Answers3

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Show that $\overline A\cap\overline B$ is a closed set and contains $A\cap B$. Then it must also contain the smallest closed set containing $A\cap B$.

Hagen von Eitzen
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${A\cap B}\subseteq {A}\Rightarrow \overline{{A\cap B}}\subseteq\bar {A}$ similarly $${A\cap B}\subseteq {B}\Rightarrow \overline{{A\cap B}}\subseteq\bar {B}$$ so $$ \overline{{A\cap B}}\subseteq\bar {A}\cap \bar{B}$$

R.N
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$\newcommand{\op}[1]{{#1^*}}$Indeed, if you can show that some operation $^*$ on sets obeys $A \subset B \Rightarrow \op{A} \subset \op{B}$ then $$A \cap B \subset A,B \Rightarrow \op{(A \cap B)} \subset \op{A}, \op{B} \Rightarrow \op{(A \cap B)} \subset \op{A} \cap \op{B}$$ and similarly for unions, so that $$\op{(A \cap B)} \subset \op{A} \cap\op{B} \subset \op{A}, \op{B} \subset \op{A} \cup\op{B} \subset \op{(A \cup B)}$$ Closure is one such set operation.

Chris Kerridge
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