A pair of identical elements is a singleton

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Let [ilmath]t[/ilmath] be a set. By the axiom of pairing we may construct a unique (unordered) pair, which up until now we have denoted by [ilmath]\{t,t\} [/ilmath]. We now show that [ilmath]\{t,t\} [/ilmath] is a singleton, thus justifying the notation:

  • [ilmath]\{t\} [/ilmath] for a pair consisting of the same thing for both parts.

Formally we must show:

  • [ilmath]\exists x[x\in\{t,t\}\wedge\forall y(y\in\{t,t\}\rightarrow y\eq x)][/ilmath] (as per definition of singleton

Proof of claim

Recall the definition: for singleton

Let [ilmath]X[/ilmath] be a set. We call [ilmath]X[/ilmath] a singleton if[1]:

  • [ilmath]\exists t[t\in X\wedge\forall s(s\in X\rightarrow s\eq t)][/ilmath]Caveat:See:[Note 1]
    • In words: [ilmath]X[/ilmath] is a singleton if: there exists a thing such that ( the thing is in [ilmath]X[/ilmath] and for any stuff ( if that stuff is in [ilmath]X[/ilmath] then the stuff is the thing ) )

More concisely this may be written:

  • [ilmath]\exists t\in X\forall s\in X[t\eq s][/ilmath][Note 2]
TODO: When the paring axiom has a page, do the same thing
  • [ilmath]\forall A\forall B\exists C\forall x(x\in C\leftrightarrow x\eq A\vee x\eq B)[/ilmath] this is the pairing axiom, in this case [ilmath]A[/ilmath] and [ilmath]B[/ilmath] are [ilmath]t[/ilmath] and [ilmath]C[/ilmath] is the (it turns out unique) set [ilmath]\{t,t\} [/ilmath]
  • To show they are equivalent we must use the axiom of extensionality
    • TODO: until it has a page, use:
      [ilmath]\forall X\forall Y(\forall u(u\in X\leftrightarrow u\in Y)\rightarrow X\eq Y)[/ilmath] (to compare sets [ilmath]X[/ilmath] and [ilmath]Y[/ilmath]

Proof body

Grade: A*
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Would be good to have.

I did it on paper with paring given slightly differently:

  • [ilmath]\forall A\forall B\exists C\big[A\in C\wedge B\in C\wedge\forall x[x\in C\implies (c\eq A\vee c\eq B)]\big][/ilmath]
and that worked at least, I suspect this is equivalent to paring but I really want to move forward so haven't shown this

This proof has been marked as an page requiring an easy proof


  1. Note that:
    • [ilmath]\exists t[t\in X\rightarrow\forall s(s\in X\rightarrow s\eq t)][/ilmath]
    Does not work! As if [ilmath]t\notin X[/ilmath] by the nature of logical implication we do not care about the truth or falsity of the right hand side of the first [ilmath]\rightarrow[/ilmath]! Spotted when starting proof of "A pair of identical elements is a singleton"
  2. see rewriting for-all and exists within set theory


  1. Warwick lecture notes - Set Theory - 2011 - Adam Epstein - page 2.75.