2
$\begingroup$

Given any sets $X$ and $Y$ with $(X\times X)\cap Y\subseteq X$ must it be true that $(X\times X)\cap Y=\emptyset$?

I believe it is, I mean $(X\times X)\cap Y\subseteq X\times X$ so if its not empty $(X\times X)\cap Y$ then its a set of ordered pairs of the elements of $X$ contained in $X$ which doesn't seem right to me.

However I've made dumb/trivial mistakes with the axiom of regularity before and other times when I'm dealing with multiple sets operating on each other while also being defined in terms of each other and with my naive intuition often causing me to think of elements as objects in their own right rather then just sets themselves belonging inside other sets this sometimes can cause me to make errors.

Anyway to iterate if this is trivial/obvious I am sorry. I thought I had an okay grasp on set theory but as I've been playing around with various definitions it seems I know far, far less then I thought I did. So I'm often doubting myself when manipulating objects containing sets within sets etc. or where I have a set that is sort of related to another set it contains in such a manner that it seems like something is self referencing from within itself to another set lower down contained in it.

$\endgroup$
  • $\begingroup$ @saulspatz what? $\endgroup$ – user3865391 Mar 20 '18 at 18:06
  • $\begingroup$ @user3865391 Sorry, I've got my head on backwards. I somehow read it as union, not intersection. $\endgroup$ – saulspatz Mar 20 '18 at 18:08
7
$\begingroup$

Counterexample:

$$X=\{(1,1), 1\}$$ $$Y=\{(1,1)\}$$

Then

$$(X\times X)\cap Y=\{(1,1)\}\subseteq X$$

$\endgroup$
  • $\begingroup$ I'd accept your answer but its making me wait $4$ minutes. $\endgroup$ – user3865391 Mar 20 '18 at 18:13

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.