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I originally believed that the answer is yes, as all elements of A are subsets of A that would be included in its power set. And, the empty set would be in the power set but not in A. However, the textbook states that this is sometimes but more often not the case. Would someone mind explaining?

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    $\begingroup$ What is a proper set? $\endgroup$ – Jacob Wakem Mar 22 '17 at 0:03
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No. There is a world of difference between a thing, and a set containing that thing. It is the same as the difference between a cat and the word cat.

Take a set of children: A = {Sam, Claire}. Its power set is {{Sam, Claire}, {Sam}, {Claire}, {}}. Note that every element of the power set of A is a set. Every element of A is a child. Therefore A is not a subset of its power set, and not being a subset is not a proper set either.

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    $\begingroup$ Unfortunately for your explanation, set theory deals with pure sets that don't involve cats or children: if $\mathtt{Sam} = \emptyset$ and $\mathtt{Claire} = \{\emptyset\}$ and $A = \{\mathtt{Sam}, \mathtt{Claire}\} = \{\emptyset, \{\emptyset\}\}$ then $A \subseteq \Bbb{P}(A)$ $\endgroup$ – Rob Arthan Mar 21 '17 at 23:50
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    $\begingroup$ @RobArthan The standard axioms of set theory do not construct or deny the possibility of constructing sets out of things that have nothing to do with sets. (The existence or not of such sets is independent of ZFC.) But the normal usage of the word set in mathematics supports it, and it leads to clear intuition about the result. $\endgroup$ – btilly Mar 22 '17 at 3:26
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    $\begingroup$ well actually, the axiom of foundation effectively does deny that possibility. However, I did not write that set theory deals exclusively with pure sets. $\endgroup$ – Rob Arthan Mar 22 '17 at 6:24
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    $\begingroup$ @RobArthan Upon a close read you are right and I was wrong. I still think that my answer was more likely to be helpful, but yours is clearly more accurate in every key respect. $\endgroup$ – btilly Mar 23 '17 at 5:25
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Sets that are subsets of their power sets are quite special. They are called transitive sets. If you work through the definitions you will find that $A \subseteq \Bbb{P}(A)$ means that for any $x \in A$ and for any $y \in x$, $y \in A$, i.e., it says that members of members of $A$ are members of $A$ (hence the name transitive sets, because the membership relation on the members of $A$ and their members is transitive). If (as is usual in set theory) you use the von Neumann representation of the natural numbers: $$ \begin{align*} 0 &= \emptyset \\ 1 &= \{0\} \\ 2 &= \{0, 1\}\\ &\ldots \end{align*} $$ then the natural numbers are all represented by transitive sets while $\{0, 2\}$ gives you the simplest example of a set that is not transitive (because $2 \in \{0, 2\}$ but $2 = \{0, 1\} \not\subseteq \{0, 2\}$).

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    $\begingroup$ This is the correct answer. $\endgroup$ – ChocolateAndCheese Mar 22 '17 at 0:03
  • $\begingroup$ @ChocolateAndCheese This is the answer that can be used to formally prove the existence of non-transitive sets from your favorite axiom system. For example ZF. However the formalism won't help someone who is lost and confused by the terminology. I was aiming to help intuition. $\endgroup$ – btilly Mar 22 '17 at 3:32
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    $\begingroup$ @btilly I agree that your accepted answer is most likely what the OP was looking for, however, that does not change that fact that it is not entirely correct. Set theory can be surprisingly subtle and often non-intuitive, and mistakes get made when we oversimplify. $\endgroup$ – ChocolateAndCheese Mar 22 '17 at 16:53
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Supppose $\;\;\mathcal{P}(A)\;\;\;$is the power set of $\;\;A\;.\;\;$ Generally, the elements of $\;\;A\;\;$ and that of $\;\;\mathcal{P}(A)\;\;\;$ are not comparable and they are of different types. For example, if $\;\;A=\{\;1\;,\;2\;\}\;\;\;$ then $\;\;\mathcal{P}(A)\;=\;\{\;\Phi\;,\;\{1\}\;,\;\{2\}\;,\;\{1,2\}\;\}\;\;\;$and in this case $\;A\;\not \subset\;\mathcal{P}(A)\;.\;\;$

But, we can easily construct an example of class in which an element can also be considered as a subset of the power set. Example, $\;A\;=\{\Phi\;,\;\{\Phi\}\;,\;\{\Phi\;,\;\{\Phi\}\}\;,...\}\;$

The indistinguishability of elements and subsets of classes leads to logical inconsistency, in general discussions, such classes are avoided by (restricting the class ) certain axioms like no set is an element of itself, no element of a set is subset of that set etc. The use of null set is also confusing for the beginners,as it is of different nature in different contexts, regarding the nature of its elements (though it has not a single element!)

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    $\begingroup$ How does this lead to logical inconsistencies? Also, such sets are not avoided by the usual axioms of ZFC, see Rob's answer. $\endgroup$ – Christian Matt Mar 22 '17 at 0:49
  • $\begingroup$ Maybe you want to use \empyset instead of \Phi to denote the empty set. $\endgroup$ – martin.koeberl Mar 22 '17 at 1:21

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