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I need to prove «If P(A) is a subset of P(B) => A is a subset of B», generally, I understand the main way I should prove it, but the problem is in the formal, pedantic language I have to use to prove such statement.

General proof is:
1. Let suppose «a» is any element of A
2. If «a» is in A, thus «a» is a subset of P(A)

The problem is in the formal and pedantic way to right it, can I just write, that element «a» is a subset of P(A) or should I define additional set, which will be a subset of P(A) and element «a» will be a part of this new set.

Please, help me to write such prove in a formal way with the correct syntax.

Thanks.

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  • $\begingroup$ Did you try to search the site? It's been asked at least twice before. $\endgroup$ – Asaf Karagila Nov 8 '13 at 20:37
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    $\begingroup$ $A\in\mathcal{P}(A)$, so $A\in\mathcal{P}(B)$ by hypothesis; therefore $A\subseteq B$. $\endgroup$ – egreg Nov 8 '13 at 20:49
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    $\begingroup$ @egreg: +1, elegant. $\endgroup$ – hmakholm left over Monica Nov 8 '13 at 20:49
  • $\begingroup$ I'm a newbie on this site and it's quite difficult to search because of all these math special symbols… $\endgroup$ – Mike B. Nov 8 '13 at 20:52
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You need to distinguzuish better between elements and sets, that is between $\in $and $\subseteq$. So assume $P(A)\subseteq P(B)$. Let $a\in A$. Then $\{a\}\subseteq A$, hence $\{a\}\in P(A)$, hence $\{a\}\in P(B)$, hence $\{a\}\subseteq B$, hence $a\in B$.

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    $\begingroup$ Why going down to elements? $\endgroup$ – egreg Nov 8 '13 at 20:49
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  1. $\mathcal{P}(A)$ is the set of subsets of $A$. In particular $A\in\mathcal{P}(A)$.

  2. Since, by hypothesis, $\mathcal{P}(A)\subseteq\mathcal{P}(B)$, we infer that $A\in\mathcal{P}(B)$.

  3. From $A\in\mathcal{P}(B)$ we get, by definition, $A\subseteq B$.

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The element $a$ is neither an element nor a subset of $\wp(A)$, but the set $\{a\}$ is an element of $\wp(A)$, and that’s what you need to use:

Let $a\in A$. Then $\{a\}\subseteq A$, so $\{a\}\in\wp(A)$. And $\wp(A)\subseteq\wp(B)$, so $\{a\}\in\wp(B)$, and therefore ...?

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