PlanetMath (more info)
 Math for the people, by the people. Sponsor PlanetMath
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (237)
Orphanage
Unclass'd
Unproven (550)
Corrections (54)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Copyright
About
[parent] Viewing Correction to 'axiom of foundation'
no set is a member of itself by CWoo

Correction id: 14548
Filed on: 2009-02-10 14:07:17
Status: Accepted on 2009-02-16 17:54:22
Type: Addendum

Correction text:
Hi, maybe this famous statement can be added as a remark somewhere in the entry.... perhaps a short proof of it would be very nice too!

Thanks!

No comment from object owner Henry.
Discussion
Style: Expand: Order:
forum policy
no set is element of itself by csabay on 2009-02-16 13:03:29
Suppose, there be a H which is element of itself. Because of the Axiom of pairing there must exist the {H}. So according to the Axiom of regularity {H} must have an element which is disjunct from it. Namely {H} has got merely one element, it must be the case that H and {H} are disjunct sets. The fact, however, is that H is an element of H and {H} as well. This contradiction proofs that H mustn't contain itself.
[ reply | up ]
Interact
new correction | post message