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I'm reading the section on power sets in Book of Proof, and the chapter includes this statement (Example 1.4 #13) of what isn't included in the power set:

$$P(\{1,\{1,2\}\})=\{\emptyset,\{\{1\}\},\{\{1,2\}\},\{\emptyset,\{1,2\}\}\}...\text{wrong because }\{\{1\}\}\subsetneq\{1,\{1,2\}\}$$

I understand $\{\{1\}\}\subsetneq\{1,\{1,2\}\}$, but why is the last element, $\{\emptyset,\{1,2\}\}$, in the power set if the empty set is not an element of the original set?

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    $\begingroup$ Why do you think $\{\emptyset,\{1,2\}\} \in P(\{1,\{1,2\}\})$? The statement absolutely never makes this claim. It isn't true. So why do you think the statement is claiming such? $\endgroup$ – fleablood Sep 7 '16 at 20:29
  • $\begingroup$ The empty set is a subset of every set. The power set is the set of all subsets. So the empty set is a member of every power set. $\endgroup$ – Doug M Sep 7 '16 at 20:31
  • $\begingroup$ @DougM That's not what the OP asked. The OP asked about $\{\emptyset, \{1,2\}\}$. That is not an element of the power set because (as the OP correctly argued) the empty set is not a member of the original set (and hence can not be a member of a subset). In short, the OP is 100% correct. But the author of the book never claimed it was. $\endgroup$ – fleablood Sep 7 '16 at 20:38
  • $\begingroup$ To be thorough. $\emptyset \in P$. $\{\{1\}\} \not \in P$. $\{\{1,2\}\} \in P$. $\{\emptyset,\{1,2\}\} \not \in P$. And finally $\{1,\{1,2\}\} \in P$ but $\{1,\{1,2\}\}$ wasn't listed in the set claimed to be the power set. So that set is not the Power set for three reasons. The book gave one. The op gave another. I gave a third.... and maybe I missed a 4th.... who knows.... $\endgroup$ – fleablood Sep 7 '16 at 20:43
  • $\begingroup$ Yes, a fourth would be $\{1\} \in P$ which wasn't listed in the set. Basically the given set over bracketed almost consistently. $\endgroup$ – fleablood Sep 7 '16 at 20:46
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Suppose that $A$ is the set $\{0,1,2\}$, then $A=\{3,4,5\}$ is wrong because $3\notin A$.

It's true that neither $4$ nor $5$ are elements of $A$ as well, but one counterexample is enough to disprove a statement.

In a nutshell, you're right, but so is the book. Both statements are valid counterexamples, but one is enough.

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It isn't: $$ \mathscr P(\{1,\{1,2\}\})=\{\emptyset,\{1\},\{\{1,2\}\},\{1,\{1,2\}\}\}. $$ If the author didn't comment on why $\{\emptyset,\{1,2\}\}$ is not in the set, it is likely because the author intended to comment on a particular reason why the proposed power set wasn't the correct power set. That is, he decided to be explicit about why $\{\{1\}\}$ isn't in the power set. Perhaps to be more complete, the author could have commented on why $\{\emptyset,\{1,2\}\}$ is also not in the power set, but it only takes one counterexample to do the job.

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  • $\begingroup$ I disagree that it is necessarily a typo. $\endgroup$ – Asaf Karagila Sep 7 '16 at 20:30
  • $\begingroup$ @AsafKaragila good point. $\endgroup$ – Alex Ortiz Sep 7 '16 at 20:30
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    $\begingroup$ It's not a typo. In showing {{1}} was not a subset the author didn't have to say anything about {emptyset,{1,2}} and the author didn't say anything about {emptyset,{1,2}}. If they author had said anything it'd be that it is not in the power set either. $\endgroup$ – fleablood Sep 7 '16 at 20:31
  • $\begingroup$ @fleablood noted. Thanks. $\endgroup$ – Alex Ortiz Sep 7 '16 at 20:32
  • $\begingroup$ It's not a "typo" but it is an omission. However it is an omission that the author was allowed to omit. $\endgroup$ – fleablood Sep 7 '16 at 20:47

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