Regularity is not refining-preserved - Revision history

Vipul: /* Examples */

2008-07-20T20:59:20Z

Examples

← Older revision		Revision as of 20:59, 20 July 2008
Line 19:		Line 19:
	One example is the space <math>\mathbb{R}_K</math> constructed by Munkres. The underlying set is the reals, and the basis is chosen as the usual open intervals, along with all sets of the form <math>(a,b) \setminus K</math> where <math>K = \{ 1/n \mid n \in \mathbb{N} \}</math>. This topology is clearly finer than the usual topology on the reals, and the reals form a regular space under the usual topology. However, <math>\mathbb{R}_K</math> is not a regular space.		One example is the space <math>\mathbb{R}_K</math> constructed by Munkres. The underlying set is the reals, and the basis is chosen as the usual open intervals, along with all sets of the form <math>(a,b) \setminus K</math> where <math>K = \{ 1/n \mid n \in \mathbb{N} \}</math>. This topology is clearly finer than the usual topology on the reals, and the reals form a regular space under the usual topology. However, <math>\mathbb{R}_K</math> is not a regular space.

	~~However~~, the closed subset <math>K</math> in this space (closed because its complement is open by construction in this topology) and the point 0 cannot be separated by disjoint open subsets, so this space is not a regular space.		For instance, the closed subset <math>K</math> in this space (closed because its complement is open by construction in this topology) and the point 0 cannot be separated by disjoint open subsets, so this space is not a regular space.

	==References==		==References==

Vipul at 20:55, 20 July 2008

2008-07-20T20:55:35Z

← Older revision		Revision as of 20:55, 20 July 2008
Line 20:		Line 20:

	However, the closed subset <math>K</math> in this space (closed because its complement is open by construction in this topology) and the point 0 cannot be separated by disjoint open subsets, so this space is not a regular space.		However, the closed subset <math>K</math> in this space (closed because its complement is open by construction in this topology) and the point 0 cannot be separated by disjoint open subsets, so this space is not a regular space.

			==References==

			===Textbook references===

			* {{booklink\|Munkres}}, Page 197, Example 1, Chapter 4, Section 31 (Munkres is using this example to construct a Hausdorff non-regular space, and does not mention here that this also shows that passing to a finer topology fails to preserve regularity)

Vipul: New page: {{topospace metaproperty dissatisfaction}} ==Statement== ===Property-theoretic statement=== The property of topological spaces of being a regular space does ''not'' satisfy the ...

2008-07-20T20:54:11Z

New page: {{topospace metaproperty dissatisfaction}} ==Statement== ===Property-theoretic statement=== The property of topological spaces of being a regular space does ''not'' satisfy the ...

New page

{{topospace metaproperty dissatisfaction}}

==Statement==

===Property-theoretic statement===

The [[property of topological spaces]] of being a [[regular space]] does ''not'' satisfy the [[metaproperty of topological spaces]] of being [[refining-preserved property of topological spaces|refining-preserved]].

===Verbal statement===

It is possible to have a [[regular space]] such that choosing a [[finer topology]] yields a space that is not regular.

==Related facts==

* [[Hausdorff not implies regular]]: The fact that regularity is not preserved on passing to finer topologies clearly shows that regularity cannot be the same as Hausdorffness, because [[Hausdorffness is refining-preserved]]. More explicitly, the examples we construct to show that regularity is not refining-preserved, also serve as examples of Hausdorff non-regular spaces.

==Examples==

One example is the space <math>\mathbb{R}_K</math> constructed by Munkres. The underlying set is the reals, and the basis is chosen as the usual open intervals, along with all sets of the form <math>(a,b) \setminus K</math> where <math>K = \{ 1/n \mid n \in \mathbb{N} \}</math>. This topology is clearly finer than the usual topology on the reals, and the reals form a regular space under the usual topology. However, <math>\mathbb{R}_K</math> is not a regular space.

However, the closed subset <math>K</math> in this space (closed because its complement is open by construction in this topology) and the point 0 cannot be separated by disjoint open subsets, so this space is not a regular space.