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My doubt is relative to the power set and the notation.

If we have a set $A = \{a,b,c\}$, its power set is $$\mathcal{P}(A) = \{\emptyset,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},A\}$$

So we get rightly these notations

$a \in A$, $\{a\} \subset \mathcal{P}(A)$, $\{a\} \in \mathcal{P}(A)$?

Thanks in advance!

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    $\begingroup$ Why $\{a\}\subset P(A)$? Is $a$ a member of $P(A)$? $\endgroup$ – almagest Sep 27 '19 at 10:33
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You already wrote down correctly what $\mathcal{P}(A)$ is. So you can just check manually.

  • Definitely $a \in A$, because that is how you defined $A$.
  • For $\{a\} \subset \mathcal{P}(A)$ to hold, we must have that every element of $\{a\}$ is an element of $\mathcal{P}(A)$. So that would mean that $a \in \mathcal{P}(A)$, but as you can see $a$ never appears as an element of $\mathcal{P}(A)$. So $\{a\} \not \subset \mathcal{P}(A)$.
  • Finally, $\{a\}$ does appear as an element in $\mathcal{P}(A)$, so indeed $\{a\} \in \mathcal{P}(A)$.

When getting confused about notation, it is often best to just write out exactly what the notation means according to its definition.

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  • $\begingroup$ Therefore is it right $\{\{a\},\{b\}\} \subset \mathcal{P}(A)$? $\endgroup$ – LH8 Sep 27 '19 at 10:41
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    $\begingroup$ @LH8 Yes! I think you got it now. $\endgroup$ – Mark Kamsma Sep 27 '19 at 10:41
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    $\begingroup$ However, $\left{\left{a\right}right} \subset \mathcal{A}$ $\endgroup$ – Samuel Bodansky Sep 27 '19 at 10:46
  • $\begingroup$ And also it is true $\{\{a\}\} \subset \mathcal{P}(A)$? $\endgroup$ – LH8 Sep 27 '19 at 10:51
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    $\begingroup$ @LH8 Yes, that is actually what Samuel tried to write in the comment above (something went wrong with formatting). $\endgroup$ – Mark Kamsma Sep 27 '19 at 10:53
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We have $a\in A$ and also $\{a\}\in\mathcal P(A)$.

Further we will have $\{a\}\subseteq\mathcal P(A)$ if and only if $a\in\mathcal P(A)$.

Looking at $\mathcal P(A)$ we conclude that this occurs if and only if at least one of the following conditions is satisfied:

  • $a=\varnothing$
  • $a=\{b\}$
  • $a=\{c\}$
  • $a=\{b,c\}$

Accepting the axiom of regularity we excluded on forehand that the conditions $a=\{a\},a=\{a,b\},a=\{a,c\}$ and $a=A=\{a,b,c\}$ are satisfied.

This because this axiom "forbids" that $a\in a$.

Note that this concerns special cases. In general if $A=\{a,b,c\}$ then it is not true that $\{a\}\subseteq\mathcal P(A)$. The answers of José and Mark are in that context.

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Yes, $a\in A$, and, yes, $\{a\}\in\mathcal P(A)$. But it is not true that $\{a\}\subset\mathcal P(A)$.

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