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regular space (Definition)
Definition 1   A topological space is a regular space if it is both a $T_0$ space and a $T_3$ space.

Example. Consider the set $ \mathbbmss{R}$ with the topology $\sigma$ generated by the basis $$\beta=\{ U=V-C: V \mbox{ is open with the standard topology and\ } C \mbox{ is (infinite) numerable}\}.$$

Since $ \mathbbmss{Q}$ is numerable and $ \mathbbmss{R}$ open, the set of irrational numbers $ \mathbbmss{R}-\mathbbmss{Q}$ is open and therefore $ \mathbbmss{Q}$ is closed. It can be shown that $ \mathbbmss{R}-\mathbbmss{Q}$ is an open set with this topology and $ \mathbbmss{Q}$ is closed.

Take any irrational number $x$ . Any open set $V$ containing all $ \mathbbmss{Q}$ must contain also $x$ , so the regular space property cannot be satisfied. Therefore, $ (\mathbbmss{R},\sigma)$ is not a regular space.

Note

In topology, the terminology for separation axioms is not standard. Therefore there are also other meanings of regular. In some references (e.g. [2]) the meanings of regular and $T_3$ is exchanged. That is, $T_3$ is a stronger property than regular.

Bibliography

1
L.A. Steen, J.A.Seebach, Jr., Counterexamples in topology, Holt, Rinehart and Winston, Inc., 1970.
2
J.L. Kelley, General Topology, D. van Nostrand Company, Inc., 1955.




"regular space" is owned by drini. [ full author list (2) | owner history (1) ]
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See Also: separation axioms, T0 space, Hausdorff space, T3 space, Hausdorff space not completely Hausdorff, T1 space

Other names:  regular
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Cross-references: stronger, references, separation axioms, property, contain, open set, closed, irrational numbers, open, basis, generated by, topological space
There are 7 references to this entry.

This is version 8 of regular space, born on 2002-02-08, modified 2005-03-26.
Object id is 1863, canonical name is RegularSpace.
Accessed 5900 times total.

Classification:
AMS MSC54D10 (General topology :: Fairly general properties :: Lower separation axioms )

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Example is confusing by Linas on 2006-11-24 14:34:41
The example is confusing. Surely, I can explicitly build an open set that does not contain a given irrational x, and so therefore, it is not true that an irrational must be an element of any open set containing the rationals.
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Duplicate entry by igor on 2002-05-18 21:08:56
There's a duplicate of this entry under
"regular" in the same subject category.
[ reply | up ]

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