Regular implies semiregular

From Topospaces
Revision as of 17:17, 28 January 2012 by Vipul (talk | contribs) (→‎Definitions used)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., regular space) must also satisfy the second topological space property (i.e., semiregular space)
View all topological space property implications | View all topological space property non-implications
Get more facts about regular space|Get more facts about semiregular space

Statement

Any regular space is a semiregular space.

Definitions used

Term Definition used
regular space A topological space X is termed regular if the following holds: given any point x∈X and open subset V⊆X such that x∈V, there exists an open subset U⊆X such that x∈U and the closure U¯ is contained in V.
semiregular space A topological space X is termed regular if the following holds: given any x∈X and any open subset V⊆X containing x, there exists a regular open subset U of X containing x and contained in V. Note that regular open does not mean an open subset that is regular in the subspace topology. Rather, it means a subset that is the interior of its closure.

Proof

Given: A regular space X. A point x∈X and any open subset V⊆X containing x,

To prove: There exists a regular open subset U of X containing x and contained in V.

Proof:

Step no. Assertion/construction Given data used Previous steps used Explanation
1 There exists an open subset U0 containing x such that U0¯ is contained in V. X is regular, x∈V, V open in X directly from regularity.
2 Let U be the interior of the closure U0¯. Step (1)
3 U contains U0 and hence contains x. Steps (1), (2) U is the largest open subset of U0¯, hence contains the open subset U0.
4 U is a regular open subset. Step (2) step-direct
5 U is contained in V. Steps (1), (2) By Step (2), U⊆U0¯. By Step (1), U0¯⊆V. Combining, U⊆V.
6 U is the desired open subset. Steps (3), (4), (5)