It is well known that $\overline {A \cap B} \neq \overline A \cap \overline B$

I wish to show that $\overline A \cap \overline B \not\subseteq \overline {A \cap B}$ by using the definition (instead of proof by counter example)

Note 1. Reverse case $\overline {A \cap B} \subseteq \overline A \cap \overline B$ is trivial

Note 2. the definition of closure I am using is one in Munkres:

$x \in \overline A \iff \text{ for every open set } U \text{ containing } x, U \cap A \neq \varnothing$

Wrong proof:

  • Let $x \in \overline A \cap \overline B$, then $x \in \overline A$ and $x \in \overline B$

    • Therefore $\forall U \in \tau, x \in U \implies U \cap A \neq \varnothing \text{ and } U \cap B \neq \varnothing$

      • So $\forall U \in \tau, x \in U \implies U \cap A \cap U \cap B \neq \varnothing \implies U \cap (A\cap B) \neq \varnothing$

        • Therefore $x \in \overline {A\cap B}$

Where did I go wrong in the above?

  • 3
    $\begingroup$ $\overline{A} \cap \overline{B} \subseteq \overline{A \cap B}$ holds sometimes, so it's going to be hard to prove what you want. $\endgroup$ – Hoot May 29 '16 at 1:03
  • $\begingroup$ @Hoot I see...so counter example is the only way $\endgroup$ – Carlos - the Mongoose - Danger May 29 '16 at 1:04
  • $\begingroup$ The first statement is wrong, if you have $(1,3)$ and $(2,4)$ then $\overline{(1,3)\cap(2,4)}=\overline{(1,3)}\cap\overline{(2,4)}$ $\endgroup$ – Masacroso May 29 '16 at 1:42

It looks like you're managing to confuse yourself slightly by not representing the quantifiers on $A$ and $B$ explicitly.

Apparently you know that it is not the case that $$ \tag{1} \forall A,B : \overline A\cap \overline B \subseteq \overline{A\cap B} $$ However, you seem to be confusing that fact with $$ \tag{2} \forall A,B : \neg(\overline A\cap \overline B \subseteq \overline{A\cap B})$$ but the actual negation of (1) is $$ \tag{3} \neg \forall A,B : \overline A\cap \overline B \subseteq \overline{A\cap B}$$ which is the same as $$ \tag{3'} \exists A,B : \neg(\overline A\cap \overline B \subseteq \overline{A\cap B})$$

Of the above claims, (2) is false (just consider the case where $A=B$; then both sides reduce to $\overline A$.

(3), on the other hand, is true, and in the form (3') we can see that one example is indeed all you need to prove it.

Something like $A=\mathbb Q$, $B=\mathbb R\setminus\mathbb Q$ should do.

The error in your wrong proof is that you can't reason from $U\cap A\ne\varnothing$ and $U\cap B\ne \varnothing$ to $(U\cap A)\cap(U\cap B)\ne\varnothing$. Just because each of the two sets is nonempty doesn't mean they have any elements in common!


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