2
$\begingroup$

Given sets $X$ and $B$ contained in $E$, prove whether the following propositions are true, or give a counterexample if they're false:

$(X \subseteq A \land \overline{X} \subset B) \implies A \cap B = \emptyset$

I'm not sure what am I supposed to do here. I tried to prove it false with an example:

Take:

$X = \{1\}$

$A = \{1\}$

$B = \{2,3\}$

This fulfills $X \subseteq A$ because all elements in $X$ are in $A$.

And fulfills $\overline{X} \subset B$ because none of the elements in $X$ is in $B$....? (This is what I don't know - can I make this affirmation?)

So assuming that it fulfills both, then $\{1\} \cap \{2,3\}$ would be $\emptyset$

$\endgroup$
  • $\begingroup$ You should try to use $\subsetneq$ for is a proper subset and $\subset$ for is a subset of. That's the modern convention I see every where. $\endgroup$ – BananaCats Author Sep 26 '13 at 3:55
2
$\begingroup$

Since $\bar{X}$ is properly (not equal to) contained in $B$ (the '$\subset$') there exist $x\in B$ that are also in $B - \bar{X} \subset X \subset A$ so $A\cap B$ is not empty.

Since $\bar{X}$ is properly contained in $B$, by definition there must exist $x\in B$ such that $x \notin \bar{X}$. But $x\notin \bar{X} \iff x \in X$. So we have the existence of $x$ in $B\cap X$. But $B\cap X = B\cap \bar{X}^c$, where $c$ also stands for complement as well as $\bar{bar}$ but the LaTeX doesn't render right if I use two bars. $B\cap \bar{X}^c = B - \bar{X}$ by definition. I.e. the $rhs$ of that is defined to be that operation on sets in the $lhs$.

$rhs =$ right hand side of mentioned equation.

$\endgroup$
  • $\begingroup$ I didn't quite get the there exist $x\in B$ that are also in $B - \bar{X} \subset X \subset A$ part. What did you do there? $\endgroup$ – Zol Tun Kul Sep 26 '13 at 4:02
  • $\begingroup$ I edited my post for you. You can now see easily why $B - \bar{X} \subset X$. $\endgroup$ – BananaCats Author Sep 26 '13 at 5:03
  • $\begingroup$ For two sets $B,C$, $B-C$ is the set difference operation also written $B \backslash C$. Remove all the elements of $B$ that are in $C$ to get the result of the operation. Or intersect with $C$'s complement. These operations are equal. $\endgroup$ – BananaCats Author Sep 26 '13 at 5:04
  • $\begingroup$ Would it be pretty much the same thing if it happened to be an $\subseteq$ instead of $\subset$? $\endgroup$ – Zol Tun Kul Sep 26 '13 at 5:28
  • $\begingroup$ If you mean proper subset versus regular subset. If $\bar{X} = B$, then you couldn't use my argument. But if it's asking you to consider any $\bar{X}$, then you can definitely consider the case where $\bar{X}$ is proper. In fact you can choose that $\bar{X}$ be proper, and thus you've proved that in general the statement isn't true (i.e. true for any $\bar{X}$). $\endgroup$ – BananaCats Author Sep 26 '13 at 7:05

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.