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:


$X = \{1\}$

$A = \{1\}$

$B = \{2,3\}$

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 $\emptyset$

  • $\begingroup$ 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. $\endgroup$ Commented Sep 26, 2013 at 3:55

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – Saturn
    Commented Sep 26, 2013 at 4:02
  • $\begingroup$ I edited my post for you. You can now see easily why $B - \bar{X} \subset X$. $\endgroup$ Commented Sep 26, 2013 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. $\endgroup$ Commented Sep 26, 2013 at 5:04
  • $\begingroup$ Would it be pretty much the same thing if it happened to be an $\subseteq$ instead of $\subset$? $\endgroup$
    – Saturn
    Commented Sep 26, 2013 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 $\bar{X}$). $\endgroup$ Commented Sep 26, 2013 at 7:05

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