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Currently, I am self studying Enderton's Elements of Set Theory and would like to clarify if my approach to The question in Chapter $2$ Question $8$ is valid, or could be improved.

  • Question

Show that there is no set to which every singleton (that is, every set of the form $\{x\}$) belongs. [Suggestion: Show that from such a set we could construct a set to which every set belonged.]

  • Answer

Let $A$ be the set that contains all singletons. Since there is no set to which every sets belong (Theorem $2A$), there should exist some set $x \notin \cup A$. However, this would imply that $\{x\} \notin A$, which is a contradiction. Hence, there is no such set to which every singleton belongs.

Thank you.

nmasanta
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Joshua
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2 Answers2

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As it stands right now, you need to improve your argument. The theorem you are using refers to the set of all sets, not just the set of all singletons.

However, take a look at Why does the set of all singleton sets not exist? and that might help improve your argument

Dashi
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  • Thank you for the link provided. I found that the second answer in that question used the same approach as mine, while the first answer was more tailored to the circumstances of the OP's question. – Joshua May 08 '19 at 03:03
  • You can write this as $\exists A, \forall x,({x}\in A)\implies$ $ \exists A,\exists B , (B=\cup A \land \forall x,({x}\in A)) \implies$ $\exists A,\exists B (B=\cup A \land \forall x,(x\in B))\implies$ $ \exists B,\forall x,(x\in B)$. And the last one is false so the 1st one is false. – DanielWainfleet May 08 '19 at 09:34
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As far as I can see, your answer is fine: Since $\bigcup A$ can't contain all sets, $A$ can't contain all singletions. You could make that more explicit by noting that, if $\{x\}\in A$ then $x\in\bigcup A$, but that's pretty obvious.

Andreas Blass
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