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Chapter 2: Problem 29

Prove that \(\mathcal{P}(X \cap Y)=\mathcal{P}(X) \cap \mathcal{P}(Y)\) for all sets \(X\) and \(Y\).

Short Answer

Expert verified
\(\mathcal{P}(X \cap Y)=\mathcal{P}(X) \cap \mathcal{P}(Y)\) is proven to be true.

Step by step solution

01

Recall the Definition of Power Set

First, we recall the definition of a power set. The power set of a set \(S\), denoted \(\mathcal{P}(S)\), is the set of all subsets of \(S\). That is, for any set \(A\), \(A \in \mathcal{P}(S)\) if and only if \(A \subseteq S\).
02

Prove that \(\mathcal{P}(X \cap Y) \subseteq \mathcal{P}(X) \cap \mathcal{P}(Y)\)

Let \(A\) be an arbitrary set in \(\mathcal{P}(X \cap Y)\), which means that \(A\) is a subset of \(X \cap Y\). Because every element of \(A\) is in both \(X\) and \(Y\), \(A\) is a subset of both \(X\) and \(Y\). Thus, \(A\) is in both \(\mathcal{P}(X)\) and \(\mathcal{P}(Y)\). Hence, \(A \in \mathcal{P}(X) \cap \mathcal{P}(Y)\). This proves that \(\mathcal{P}(X \cap Y) \subseteq \mathcal{P}(X) \cap \mathcal{P}(Y)\).
03

Prove that \(\mathcal{P}(X \cap Y) \supseteq \mathcal{P}(X) \cap \mathcal{P}(Y)\)

Let \(A\) be an arbitrary set in \(\mathcal{P}(X) \cap \mathcal{P}(Y)\), which means \(A \in \mathcal{P}(X)\) and \(A \in \mathcal{P}(Y)\). Thus, \(A \subseteq X\) and \(A \subseteq Y\) implying that \(A \subseteq (X \cap Y)\). Hence, \(A \in \mathcal{P}(X \cap Y)\). This proves that \(\mathcal{P}(X \cap Y) \supseteq \mathcal{P}(X) \cap \mathcal{P}(Y)\).
04

Conclusion

Because we have shown both \(\mathcal{P}(X \cap Y) \subseteq \mathcal{P}(X) \cap \mathcal{P}(Y)\) and \(\mathcal{P}(X \cap Y) \supseteq \mathcal{P}(X) \cap \mathcal{P}(Y)\), we can conclude that \(\mathcal{P}(X \cap Y) = \mathcal{P}(X) \cap \mathcal{P}(Y)\).

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