# 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\}\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:

• [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]