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For any set $A$, is $A \in P(A)$, where $P(A)$ is the power set of $A$?

I am having trouble figuring this out. The power set of a set is a set of all possible subsets of that set, so $P(A)$ contains all possible subsets of $A$, and $\in$ symbol means that $P(A)$ is an element of $A$. If $P(A)$ is all possible subsets then that means it contains potentially multiple subsets with same values? It cant possibly be an element of $A$ can it? Or am i thinking about this wrong?

Kevin Long
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GarudaAiacos
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  • Yes; the power set of a set $A$ is the set (collection) of all its subsets and $A \subseteq A$, i.e. very set is a subset of itself. – Mauro ALLEGRANZA Mar 07 '17 at 19:19
  • yes, Every subset of A is a member of P(A). And A is a subset of itself. – Doug M Mar 07 '17 at 19:19
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    $A ∈ \mathcal P(A)$ does not mean "that $\mathcal P(A)$ is an element of $A$". – Mauro ALLEGRANZA Mar 07 '17 at 19:20
  • Perhaps an example will help. With $A={1,2}$ you have $\mathcal{P}(A)={\emptyset,{1},{2},{1,2}}$. Note that in this example $A\in \mathcal{P}(A)$ and that $\mathcal{P}(A)\not\in A$. There are also elements of $\mathcal{P}(A)$ which themselves contain the element $1$. (In general if $x\in A$ half of the elements of $\mathcal{P}(A)$ will contain $x$). – JMoravitz Mar 07 '17 at 19:23
  • The only other point of potential confusion I think I can see you having is the symbol $\in$. Very specifically $x\in y$ says the element on the left, $x$, is an element of the set on the right, $y$. In the same way that $x<y$ does not imply $y<x$ one has $x\in y$ does not imply $y\in x$. These are two completely different statements, so in the initial sentence you write $A\in P(A)$ but in later sentences you seem to have reversed which set was where talking about whether or not $P(A)$ is an element of $A$ where you should have been talking about whether or not $P(A)$ contains $A$ – JMoravitz Mar 07 '17 at 19:28
  • @JMoravitz: And note that "contains" and "includes" are sometimes ambiguous. Often we say "the set S contains the zero-vector" or "sets ordered under inclusion". One could be pedantically precise by saying "$P(A)$ has $A$ as a subset". Haha.. – user21820 Mar 08 '17 at 06:01
  • Change that to "as an element" and i would agree – JMoravitz Mar 08 '17 at 06:03

2 Answers2

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There's no such thing as "multiple subsets with the same values" since a set is determined solely by what's in it.

In answer to your particular question, $A$ is a subset of itself, so it's an element of $P(A)$.

Ethan Bolker
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A subset of $A$ can also be a member of $A$. For example if $\phi \in A$ . For a different example if $A=\{\phi,\; \{\phi\},\; \{\phi, \{\phi\}\}\;\}$ then $A$ has $3$ members and each of the $3$ members of $A$ is also a subset of $A$.... In set theory a set whose members are all subsets is called a transitive set.

DanielWainfleet
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