A pair of identical elements is a singleton

Statement

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\}\rightarrow\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:

• [ilmath]\exists t[t\in X\wedge\forall s(s\in X\rightarrow s\eq t)][/ilmath]^Caveat:See:
  • 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]

TODO: When the paring axiom has a page, do the same thing

• Template:M\forall A\forall B\exists C\forall x(x\in C\leftrightarrow x=A\vee x=B) 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]

Proof body

• Choose [ilmath]x:\eq t[/ilmath]

TODO: This is wrong, saving work and switching computer

• • [ilmath]x\in\{t,t\} [/ilmath] as a result.

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