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I know that the power set is the set of all subsets, including the empty set.

The set in question is $\{\emptyset, \{\emptyset\}\}$. I know that there are $2^n$ subsets of a finite set, so that means there are 4 subsets then since the set above only contains 2 elements. The question asks to write out the powerset.

But if I attempt to write the subsets out, I get: $\{\emptyset, \{\emptyset\}\}; \{\emptyset\}; \{\{\emptyset\}\}$

But I'm missing one. I know that the emptyset is a subset of all sets, but I already have the emptyset as a subset, which was an element in the original set anyway. I also know that a set can't contain duplicates.

What is the remaining subset I'm missing?

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migrated from mathematica.stackexchange.com Sep 6 '17 at 14:23

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    $\begingroup$ A reminder that the power set of $\{a,b\}$ is $\{\emptyset,\{a\},\{b\},\{a,b\}\}$. Now, replace $a$ with $\emptyset$ and $b$ with $\{\emptyset\}$ $\endgroup$ – JMoravitz Sep 6 '17 at 14:30
  • $\begingroup$ Ohh that puts it into better perspective. Thanks! $\endgroup$ – Jose Ramirez Sep 6 '17 at 15:37
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I already have the emptyset as a subset, which was an element in the original set anyway.

No -- you have written $\{\emptyset\}$ as a subset (which it is), but $\{\emptyset\}$ is not the same as $\emptyset$ -- so $\emptyset$ is a different subset from $\{\emptyset\}$.

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  • $\begingroup$ But the emptyset is already an element of the set, though? $\endgroup$ – Jose Ramirez Sep 6 '17 at 15:36
  • $\begingroup$ @JozemiteApps: Yes: $\varnothing$ is an element of $\{\varnothing,\{\varnothing\}\}$ and therefore $\{\varnothing\}$ is in its power set. Why would that mean that $\varnothing$ itself should not be in the power set? Do you disagree that $\varnothing\subseteq\{\varnothing,\{\varnothing\}\}$? Or do you disagree that $\varnothing\ne\{\varnothing\}$? Or do you disagree that these two facts imply that $\varnothing$ must be listed among the elements of the power set? $\endgroup$ – Henning Makholm Sep 6 '17 at 15:56
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You are mising $\emptyset$, which is a subset of every set.

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  • $\begingroup$ That's where I was confused because it's already an element of the original set. $\endgroup$ – Jose Ramirez Sep 6 '17 at 15:35

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