Difference between revisions of "Singleton (set theory)"
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__TOC__ | __TOC__ | ||
=={{subpage|Definition}}== | =={{subpage|Definition}}== | ||
| − | {{/Definition}} | + | {{/Definition}}<!-- |
| − | + | -->** For proof see '''Claim 1'''. | |
| − | * | + | |
==Significance== | ==Significance== | ||
Notice that we have manage to define a set containing one thing without any notion of the number 1. | Notice that we have manage to define a set containing one thing without any notion of the number 1. | ||
==See next== | ==See next== | ||
* [[A pair of identical elements is a singleton]] | * [[A pair of identical elements is a singleton]] | ||
| + | ==Proof of claims== | ||
| + | # {{M|\big(\exists t[t\in X\wedge\forall s(s\in X\rightarrow s\eq t)]\big)\iff\big(\exists t\in X\forall s\in X[t\eq s]\big)}} | ||
| + | #* By "''[[rewriting for-all and exists within set theory]]''" we see: | ||
| + | #** {{M|\big(\exists t[t\in X\wedge\forall s(s\in X\rightarrow s\eq t)]\big)\iff\big(\exists t\in X[\forall s(s\in X\rightarrow s\eq t)]\big)}} | ||
| + | #**: {{M|\iff\big(\exists t\in X\forall s(s\in X\rightarrow s\eq t)\big)}} by simplification | ||
| + | #**: {{M|\iff\big(\exists t\in X\forall s[s\in X\rightarrow s\eq t]\big)}} by changing the bracket style | ||
| + | #**: {{M|\iff\big(\exists t\in X\forall s\in X[s\eq t]\big)}} by re-writing again. | ||
| + | ==Notes== | ||
| + | <references group="Note"/> | ||
==References== | ==References== | ||
<references/> | <references/> | ||
{{Definition|Elementary Set Theory|Set Theory}} | {{Definition|Elementary Set Theory|Set Theory}} | ||
Latest revision as of 17:38, 8 March 2017
Definition
Let X be a set. We call X a singleton if[1]:
- ∃t[t∈X∧∀s(s∈X→s=t)]Caveat:See:[Note 1]
- In words: X is a singleton if: there exists a thing such that ( the thing is in X and for any stuff ( if that stuff is in X then the stuff is the thing ) )
More concisely this may be written:
- ∃t∈X∀s∈X[t=s][Note 2]
- For proof see Claim 1.
Significance
Notice that we have manage to define a set containing one thing without any notion of the number 1.
See next
Proof of claims
- (∃t[t∈X∧∀s(s∈X→s=t)])⟺(∃t∈X∀s∈X[t=s])
- By "rewriting for-all and exists within set theory" we see:
- (∃t[t∈X∧∀s(s∈X→s=t)])⟺(∃t∈X[∀s(s∈X→s=t)])
- ⟺(∃t∈X∀s(s∈X→s=t)) by simplification
- ⟺(∃t∈X∀s[s∈X→s=t]) by changing the bracket style
- ⟺(∃t∈X∀s∈X[s=t]) by re-writing again.
- (∃t[t∈X∧∀s(s∈X→s=t)])⟺(∃t∈X[∀s(s∈X→s=t)])
- By "rewriting for-all and exists within set theory" we see:
Notes
- Jump up ↑ Note that:
- ∃t[t∈X→∀s(s∈X→s=t)]
- Jump up ↑ see rewriting for-all and exists within set theory
