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Set theory proofs

Could someone help me out with these set theory proofs? \(A\), \(B\) and \(C\) are sets, and the questions are independent of each other.

Q1. Show that if \(A\subset B\), then \(C-B\subset C-A\).

Q2. If \(P(A)=P(B)\), show that \(A=B\).

Note: \(P(A)\) denotes the power set of \(A\).


Also, could you provide me an faster solution to this question? The question is 'Show that \(A=(A\cup B)\cap(A-B)\)'. My method is to assume that \(a\in A\), and prove that \(a\in (A\cup B)\cap(A-B)\), and hence say that \(A\subseteq(A\cup B)\cap(A-B)\), and prove vice versa. I find this method extremely lengthy. Is there a faster method?

Note by Omkar Kulkarni
1 year, 11 months ago

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Q3: Let's suppose that \(\mathscr{P}(A)=\mathscr{P}(B)\).

Let \( x \in A\). Then \( \exists C \in \mathscr{P}(A) : x \in C \). Let \( C \in \mathscr{P}(A) \) be in that conditions. By hypothesis, we have that \( C \in \mathscr{P}(B) \), this means that \( x \in C \subseteq B \), so \( x \in B\).

Now, \( y \in B\). Then \( \exists D \in \mathscr{P}(B) : x \in D \). Let \( D \in \mathscr{P}(B) \) be in that conditions. By hypothesis, we have that \( D \in \mathscr{P}(A) \), this means that \( y \in D \subseteq A \), so \( y \in A\).

In conclusion, \(A=B \). Paulo Guilherme Santos · 1 year, 8 months ago

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Q1: Show that if \(x \in C \setminus B\), then \(x \in C \setminus A\).

Q2: This basically says that if you are given \(P(A)\), then you can uniquely determine the set \(A\). For example, if \[P(A) = \{\emptyset, \{1\}, \{4\}, \{5\}, \{1,4\}, \{1,5\}, \{4,5\}, \{1,4,5\}\},\] then what is \(A\)? You just need to generalize this idea.

As for your last question, I don't think there's a faster way. That's the traditional approach for showing that two sets are equal. Sometimes, you may be able to use some identity that you have proven before, but I doubt that's the cas here. Jon Haussmann · 1 year, 11 months ago

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@Jon Haussmann Oh okay. Makes sense. Thank you! I'll come back to you with the second one if I don't get it. Omkar Kulkarni · 1 year, 11 months ago

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