PlanetMath (more info)
 Math for the people, by the people.
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 (232)
Orphanage
Unclass'd
Unproven (519)
Corrections (23)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: Very high Entry average rating: No information on entry rating
completely normal (Definition)

Let $ X$ be a topological space. $ X$ is said to be completely normal if whenever $ A,B\subseteq X$ with $ A\cap\overline{B}=\overline{A}\cap B=\emptyset$, then there are disjoint open sets $ U$ and $ V$ such that $ A\subseteq U$ and $ B\subseteq V$.

Equivalently, a topological space $ X$ is completely normal if and only if every subspace is normal.



"completely normal" is owned by Mathprof. [ full author list (2) | owner history (1) ]
(view preamble | get metadata)

View style:

See Also: normal

Other names:  complete normality
Keywords:  topology
Log in to rate this entry.
(view current ratings)

Cross-references: normal, subspace, open sets, disjoint, topological space
There is 1 reference to this entry.

This is version 3 of completely normal, born on 2002-01-24, modified 2006-10-30.
Object id is 1606, canonical name is CompletelyNormal.
Accessed 2986 times total.

Classification:
AMS MSC: 54-00 (General topology :: General reference works )
Pending Errata and Addenda
None.
[ View all 1 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add derivation | add example | add (any)