I was wondering whether there is a formal proof that exists showing that some arbitrary set $A$ is not a subset of $\{A\}$ (though it would be an element)? It's not clear to me whether this is just a universally accepted axiom of set theory, or if there is a way to prove the relation (or lack thereof).

If it helps, I'd be happy with an answer in terms of the simple sets $\{1\}$ and $\{\{1\}\}$!

Edit: I had forgotten about the empty set! For my purposes, let's specifically exclude the case where $A$ = $\emptyset$.

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    $\begingroup$ If $A$ is the empty set then it is a subset of $\{A\}.$ $\endgroup$
    – David K
    Jun 2, 2022 at 0:50
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    $\begingroup$ @DavidK Conversely, if we have any $x \in A$, then we know that $A \notin A$ and therefore $x \neq A$. So $x \notin \{A\}$, and thus $A \nsubseteq \{A\}$. $\endgroup$ Jun 2, 2022 at 0:56
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    $\begingroup$ For another start, notice $\{A\}$ has exactly two subsets: $\emptyset$ and $\{A\}$. $\endgroup$
    – aschepler
    Jun 2, 2022 at 1:03
  • $\begingroup$ @MarkSaving Thanks for the response! I think my follow-up is essentially the same as my original question: while I think $A \notin A$ if $x \in A$ makes sense intuitively, is this something that can be formalized, or just taken for granted? Maybe I'm being a bit too dense here! $\endgroup$
    – hillard28
    Jun 2, 2022 at 1:19
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    $\begingroup$ The fact that $A \notin A$ is a consequence of the Axiom of Regularity. $\endgroup$ Jun 2, 2022 at 1:37

1 Answer 1


Thanks to Mark and Robert for the clarification here. Hopefully this self-answer suffices:

By the Axiom of Regularity, every non-empty set $A$ contains an element that is disjoint from $A$.

Extending this axiom to the set containing non-empty $A$, denoted as $\{A\}$, there exists an element of $\{A\}$ that is disjoint from $\{A\}$. Since the only element of $\{A\}$ is $A$, it must follow that $A$ is disjoint from $\{A\}$ and $A \cap \{A\} = \emptyset$. Since $A \in \{A\}$, it must be that $A \notin A$.

Drawing from the above, assume $x \neq \emptyset \in A$. It follows that $x \neq A$ and therefore $x \notin \{A\}$. Thus, $A \nsubseteq \{A\}$.

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jun 2, 2022 at 17:49
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    $\begingroup$ Except that if $A = \emptyset$ then $A \subseteq A$ is true. $\endgroup$
    – aschepler
    Jun 3, 2022 at 1:40

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