Difference between revisions of "A pair of identical elements is a singleton"

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(Created page with "__TOC__ ==Statement== Let {{M|t}} be a set. By ''the axiom of pairing'' we may construct a unique (unordered) pair, which up until now we have denoted by {{M|\{t,t\} }...")
 
m (Fixing and leaving for later)
 
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* {{M|\{t\} }} for a pair consisting of the same thing for both parts.
 
* {{M|\{t\} }} for a pair consisting of the same thing for both parts.
 
Formally we must show:
 
Formally we must show:
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* {{M|\exists x[x\in\{t,t\}\rightarrow\forall y(y\in\{t,t\}\rightarrow y\eq x)]}} (as per definition of {{link|singleton|set theory}}
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* {{M|\exists x[x\in\{t,t\}\wedge\forall y(y\in\{t,t\}\rightarrow y\eq x)]}} (as per definition of {{link|singleton|set theory}}
 
==Proof of claim==
 
==Proof of claim==
 
{{Begin Notebox}}
 
{{Begin Notebox}}
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{{End Notebox Content}}{{End Notebox}}
 
{{End Notebox Content}}{{End Notebox}}
 
{{XXX|When the paring axiom has a page, do the same thing}}
 
{{XXX|When the paring axiom has a page, do the same thing}}
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* {{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 {{M|A}} and {{M|B}} are {{M|t}} and {{M|C}} is the (it turns out unique) set {{M|\{t,t\} }}
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* {{M|\forall A\forall B\exists C\forall x(x\in C\leftrightarrow x\eq A\vee x\eq B)}} this is the pairing axiom, in this case {{M|A}} and {{M|B}} are {{M|t}} and {{M|C}} is the (it turns out unique) set {{M|\{t,t\} }}
 +
* To show they are equivalent we must use [[the axiom of extensionality]]
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** {{XXX|until it has a page, use: }} {{M|\forall X\forall Y(\forall u(u\in X\leftrightarrow u\in Y)\rightarrow X\eq Y)}} (to compare sets {{M|X}} and {{M|Y}}
 
===Proof body===
 
===Proof body===
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* Choose {{M|x:\eq t}}
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{{Requires proof|grade=A*|msg=Would be good to have.
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{{XXX|This is wrong, saving work and switching computer}}
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** {{M|x\in\{t,t\} }} as a result.
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I did it on paper with paring given slightly differently:
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* {{M|\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]}}
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and that worked at least, I suspect this is equivalent to paring but I really want to move forward so haven't shown this|easy=true}}
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==Notes==
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<references group="Note"/>
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==References==
 +
<references/>
 +
{{Theorem Of|Set Theory|Elementary Set Theory}}

Latest revision as of 23:35, 8 March 2017

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[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*
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
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

Notes

  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

References

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


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