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1) Is there any set A such that A={A}?

2) Is there any set A such that {A} is a subset of A?

3) Is there any set A such that each of its elements is a subset of A?

4) Is there any set A such that each of its elements is a bijection from A to A?

Bach
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1 Answers1

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In the modern context of set theory, namely $\mathsf{ZFC}$, the answer to the first two is no. The reason is that we require $\in$ to be well-founded which forbids $A\in A$, and therefore forbids $\{A\}\subseteq A$ (which is really the same thing).

To the third question consider $\varnothing$, or $\{\varnothing\}$ or $\{\varnothing,\{\varnothing\}\}$. A set is called transitive if every element is also a subset. Transitive sets are used often in set theory, for example the von Neumann ordinals are transitive sets which are well-ordered by $\in$.

To the final question, the answer is again no if we require the set to be non-empty, and again this is a consequence of the axiom of regularity (one of the axioms of $\mathsf{ZFC}$). If $f\in A$ is a bijection from $A$ to $A$ then $f\in\operatorname{dom}(f)$, which means (in the standardn Kuratowski definition of an ordered pair) that $\{\{f\},\{f,x\}\}\in f$, which means that $$\dots\in f\in\{f\}\in\{\{f\},\{f,x\}\}\in f\in\dots$$ is a counterexample to the well-foundedness of $\in$.

In contexts outside of $\mathsf{ZFC}$ it is possible to have sets of the form $A=\{A\}$ or $A\in A$. For example in the set theory $\mathsf{NF}$ there is a universal set, i.e. the set of all sets. Call this universal set $\cal U$ then $\cal U\in U$, because $\cal U$ is a set, and every set is an element of $\cal U$.

Furthermore if we take a theory in which there is some $x=\{x\}$ then we have that $\langle x,x\rangle=\{\{x\},\{x,x\}\}=\{x,x\}=\{x\}$ and therefore $x=\{\langle x,x\rangle\}$. It follows that $x$ is a bijection of itself with itself, and it is also an element of itself.

Also related:

  1. When is $x=\{ x\}$?
  2. How to construct $\{\{\{...\}\}\}$ in ZF without axiom of foundation
Community
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Asaf Karagila
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  • Techincally, the trival case of $A = \emptyset$ is an example satisfying both 3) and 4). – ferson2020 Jan 30 '13 at 20:06
  • @ferson2020: I pointed that out for the third question, but I have added this to the fourth question as well. – Asaf Karagila Jan 30 '13 at 20:09
  • let A be a set. is there any transitive set B with same cardinality? –  Jan 30 '13 at 20:16
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    @CutieKrait: If we assume the axiom of choice then yes. Every set is equipotent with an ordinal, and ordinals are transitive sets. For finite sets, of course, we don't need the axiom of choice. – Asaf Karagila Jan 30 '13 at 20:18
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    @CutieKrait Yes. Let $B$ be the cardinality of $A$; then $B$ is an ordinal, since it is a cardinal, hence it is transitive. – ferson2020 Jan 30 '13 at 20:19
  • Is ZFC complete? so that any proposition about existence of a set is either true or false. –  Jan 30 '13 at 20:27
  • @CutieKrait: No, ZFC is far far from complete. But we can prove some of the properties are always true or always false. – Asaf Karagila Jan 30 '13 at 20:29
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    @CutieKrait Probably this wasn't what you were asking, but ZFC does prove that every proposition is either true or false, because the law of excluded middle (LEM) from the underlying classical logic proves this (and LEM would follow from the Axiom of Choice even if it weren't included in the underlying logic of the theory.) However, ZFC may prove a disjunction (e.g. $P \vee \neg P$) without proving either disjunct, so this is not very useful. – Trevor Wilson Jan 30 '13 at 21:59
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    @CutieKrait One thing you have, and does not require choice, is that any set $x$ is contained in a transitive set: Simply take $x\cup\bigcup x\cup\bigcup\bigcup x\cup\dots$ This is called the transitive closure of $x$, and any transitive set that contains $x$ must contain the transitive closure of $x$. – Andrés E. Caicedo Jan 31 '13 at 01:20
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    @CutieKrait: Is there a particular reason you unaccepted? – Asaf Karagila May 23 '13 at 01:28
  • @AsafKaragila: You answered completely this question. And some of my other questions. thanks. btw. I'll reaccept it tomorrow, if I can. –  May 23 '13 at 01:33
  • @CutieKrait: If you can? You're deleting your account for one reason or another... – Asaf Karagila May 23 '13 at 01:35
  • If I can, I'll reaccept it tomorrow. –  May 23 '13 at 01:37
  • Yeah, right... :| – Asaf Karagila May 24 '13 at 00:25