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I have a feeling that this isn't true, but I cannot for the life of me think of a counter example. I've tried thinking about the empty set, but then that leaves me with:

$ A \cup \emptyset = A \\A \cap \emptyset = \emptyset$

And so it's the case that $A \cap \emptyset \subset \ A \cup \emptyset $, which is the opposite of what I wanted. All other examples I've tried involved using sets A, B such that A=B. Any hints would be much appreciated!

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  • $\begingroup$ One might read this out as "Every element in either $A$ or $B$ is in both" which makes it a little clearer that this is true. $\endgroup$ – Milo Brandt Oct 11 '16 at 2:21
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Let $x\in B \implies x\in B \lor x\in A \implies x\in (A \cup B) \implies x\in (A \cap B) \implies x \in A \land x\in B \implies x\in A$

In the same way

Let $x\in A \implies x\in A \lor x\in B \implies x\in (A \cup B) \implies x\in (A \cap B) \implies x \in A \land x\in B \implies x\in B$

Then, $A=B$ $\blacksquare$

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Hint: Suppose $A\cup B \subset A\cap B$. Then $$A\subset A\cup B \subset A\cap B \subset A$$ so you have equality throughout. Repeat with $B$ instead of $A$.

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  • $\begingroup$ Did you mean to write $A\subset A\cup B\subset A\cap B\subset B$? $\endgroup$ – bof Oct 11 '16 at 1:56
  • $\begingroup$ @bof, No, I did not. The intention was to find that $A=A\cup B$ and also that $B=A\cup B$, hence, $A=B$, using the anti-symmetry of set inclusion. But yes, that also works out well. $\endgroup$ – Hayden Oct 11 '16 at 1:58
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Hint: $\quad A \Delta B = (A \cup B) \setminus (A \cap B) = \emptyset \quad \implies \quad A = B$

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