Difference between revisions of "A pair of identical elements is a singleton"
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Revision as of 16:03, 8 March 2017
Contents
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[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
- 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.
- [ilmath]x\in\{t,t\} [/ilmath] as a result.
Cite error: <ref> tags exist for a group named "Note", but no corresponding <references group="Note"/> tag was found, or a closing </ref> is missing
