Tried to prove it. wasn't 100% sure I did everything correctly so I would appreciate a lot your advice. this is what I've done:enter image description here

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    $\begingroup$ Your proof looks good. $\endgroup$ – drhab Dec 4 '15 at 18:08
  • $\begingroup$ Your proof is correct. $\endgroup$ – Taroccoesbrocco Dec 4 '15 at 18:47

This is basically fine. But you don't even need to prove anything for part (a). If $A= B$ then you can substitute $B$ in for $A$ in any expression and the two expressions will be equal. $A$ and $B$ are exactly the same, so all you're doing is calling $A$ by a different name. Also, I would try to write in complete, grammatical sentences.

Suppose $a \in A$. Then $(a,a) \in A \times A$, so ...


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