0
$\begingroup$

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?

$\endgroup$
  • 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
1
$\begingroup$

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!

$\endgroup$

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.