K-topology
In mathematics, particularly in the field of topology, the K-topology,[1] also called Smirnov's deleted sequence topology,[2] is a topology on the set R of real numbers which has some interesting properties. Relative to the standard topology on R, the set is not closed since it doesn't contain its limit point 0. Relative to the K-topology however, the set K is declared to be closed by adding more open sets to the standard topology on R. Thus the K-topology on R is strictly finer than the standard topology on R. It is mostly useful for counterexamples in basic topology. In particular, it provides an example of a Hausdorff space that is not regular.
Formal definition
[edit]Let R be the set of real numbers and let The K-topology on R is the topology obtained by taking as a base the collection of all open intervals together with all sets of the form [1] The neighborhoods of a point are the same as in the usual Euclidean topology. The neighborhoods of are of the form , where is a neighborhood of in the usual topology.[3] The open sets in the K-topology are precisely the sets of the form with open in the usual Euclidean topology and [2]
Properties
[edit]Throughout this section, T will denote the K-topology and (R, T) will denote the set of all real numbers with the K-topology as a topological space.
1. The K-topology is strictly finer than the standard topology on R. Hence it is Hausdorff, but not compact.
2. The K-topology is not regular, because K is a closed set not containing , but the set and the point have no disjoint neighborhoods. And as a further consequence, the quotient space of the K-topology obtained by collapsing K to a point is not Hausdorff. This illustrates that a quotient of a Hausdorff space need not be Hausdorff.
3. The K-topology is connected. However, it is not path connected; it has precisely two path components: and
4. The K-topology is not locally path connected at and not locally connected at . But it is locally path connected and locally connected everywhere else.
5. The closed interval [0,1] is not compact as a subspace of (R, T) since it is not even limit point compact (K is an infinite closed discrete subspace of (R, T), hence has no limit point in [0,1]). More generally, no subspace A of (R, T) containing K is compact.
See also
[edit]Notes
[edit]- ^ a b Munkres 2000, p. 82.
- ^ a b Steen & Seebach 1995, Counterexample 64.
- ^ Willard 2004, Example 14.2.
References
[edit]- Munkres, James R. (2000). Topology (2nd ed.). Upper Saddle River, NJ: Prentice Hall, Inc. ISBN 978-0-13-181629-9. OCLC 42683260.
- Steen, Lynn Arthur; Seebach, J. Arthur Jr. (1995) [1978]. Counterexamples in Topology (Dover reprint of 1978 ed.). Berlin, New York: Springer-Verlag. ISBN 978-0-486-68735-3. MR 0507446.
- Willard, Stephen (2004) [1970]. General Topology. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-43479-7. OCLC 115240.