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Could anybody check my solution to this question please?

Question: Prove $2^{X \cap Y \cap Z} = 2^X \cap 2^Y \cap 2^Z$ for any three sets $X, Y, Z.$

My solution:

If $a \in 2^X \cap 2^Y \cap 2^Z$, then $a \in 2^X, a \in 2^Y, a \in 2^Z$.

So $a \subset X, a \subset Y, a \subset Z$. So a $\subset X \cap Y \cap Z$.

Then $a \in 2^{X \cap Y \cap Z}$

So $2^{X \cap Y \cap Z} \subset 2^X \cap 2^Y \cap 2^Z$.

Conversely if $a \in 2^{X \cap Y \cap Z}$ then $a \subset X \cap Y \cap Z$ and so $a \subset X, a \subset Y$ and $a \subset Z$

Then $a \in 2^X, a\in 2^Y, a \in 2^Z.$ Clearly $a \in 2^X \cap 2^Y \cap 2^Z$

So $2^X \cap 2^Y \cap 2^Z \subset 2^{X \cap Y \cap Z}$ and $2^X \cap 2^Y \cap 2^Z = 2^{X \cap Y \cap Z}$

CrossGuard
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1 Answers1

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You've done the proof in one direction. You need to show two things:

  • If $a\in 2^X \cap 2^Y \cap 2^Z$ then $a\in 2^{X\cap Y\cap Z}$;
  • If $a \in 2^{X\cap Y\cap Z}$ then $a\in 2^X \cap 2^Y \cap 2^Z$.

You've done one of those correctly, but you haven't yet done the other. So you've shown that $2^X \cap 2^Y \cap 2^Z \subseteq a\in 2^{X\cap Y\cap Z}\vphantom{\dfrac 1 1}$, but not that $2^{X\cap Y\cap Z} \subseteq 2^X \cap 2^Y \cap 2^Z$.

Some people may say $2^X$ means the set of all functions from a the set $X$ into the set $\{0,1\}$. In a sense that's the same as the set of all subsets of $X$. But if you want to draw a distinction between those two things, the latter might be denoted as $\mathcal P(X)$. The letter $\mathcal P$ stands for "power"; the set $\mathcal P(X)$ is called the "power set" of $X$, and probably that's because $2^X$ is $2$ "raised to the power" $X$.

Michael Hardy
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