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Steven L. Kent, Roy A. Mimna, Jamal K. Tartir, "A Note on Topological Properties of Non-Hausdorff Manifolds", *International Journal of Mathematics and Mathematical Sciences*, vol. 2009, Article ID 891785, 4 pages, 2009. https://doi.org/10.1155/2009/891785

# A Note on Topological Properties of Non-Hausdorff Manifolds

**Academic Editor:**Francois Goichot

#### Abstract

The notion of compatible apparition points is introduced for non-Hausdorff manifolds, and properties of these points are studied. It is well known that the Hausdorff property is independent of the other conditions given in the standard definition of a topological manifold. In much of literature, a topological manifold of dimension is a Hausdorff topological space which has a countable base of open sets and is locally Euclidean of dimension . We begin with the definition of a non-Hausdorff topological manifold.

#### 1. Topological Properties of Non-Hausdorff Manifolds

*Definition 1.1. *A non-Hausdorff manifold of dimension is a topological space which has a countable base of open sets and is locally Euclidean of dimension

Since every point of a non-Hausdorff manifold has a Euclidean neighborhood, it is easy to show that every non-Hausdorff manifold is

We now briefly review some of the well-known properties of non-Hausdorff manifolds. Since is locally compact, a non-Hausdorff manifold of dimension is locally compact. In some of literature, compactness is only defined in Hausdorff spaces. In such cases, compact subsets must be closed. Compact subsets of -spaces, however, need not to be closed. This remains true for non-Hausdorff manifolds (Example 1.2). A non-Hausdorff manifold of dimension must be locally connected. Since a non-Hausdorff manifold has a countable base of open sets, is Lindelรถf; that is, every open cover of has a countable subcover. Further, since locally compact Lindelรถf spaces are sigma-compact, it follows that a non-Hausdorff manifold of dimension n is sigma-compact. Finally, we note that when is not Hausdorff, it is not regular.

We now consider the property of paracompactness. A Hausdorff space is paracompact if every open covering of has a locally finite refinement . That is, each is contained in some and each has a neighborhood which meets only finitely many sets in . Paracompactness can be defined for -spaces as follows. A -space is paracompact if and only if each open covering of has an open barycentric refinement, where is a barycentric refinement of if the collection refines where . A space is metacompact if every open cover has a point finite refinement. Since Hausdorff second countable manifolds are metrizable, they are paracompact and hence metacompact. In [1], an example of a non-Hausdorff manifold which is not metacompact is given. We present another one.

*Example 1.2. *A non-Hausdorff manifold need not to be metacompact.

Let and define a topology on as follows.

(i)For each a basic open neighborhood of is open in with the usual topology.(ii)For each a basic open neighborhood of is of the form where is an open neighborhood of in with the usual topology.*Claim 1. *The non-Hausdorff manifold is not metacompact.

*Proof. *Let . To see that has no point finite refinement, let be a refinement of . Let and such that is a subset of some element of . For each choose and such that and . By the way is defined, no element of contains more than one element of . Since is a refinement of no element of contains more than one element of . Hence, whenever . By Cantor's Intersection theorem, there exists such that . Therefore, is not point finite.

*Remark 1.3. *In the above example, is compact and Hausdorff but not closed.

*Remark 1.4. *For each is a complete metric space and is a countable dense subset of . Therefore, a construction similar to the one above can be used to create a non-Hausdorff manifold of dimension that is not metacompact.

#### 2. Compatible Apparition Points

If a manifold of dimension is non-Hausdorff, there exist at least two points and which cannot be separated by disjoint open sets. Also, the points and cannot be contained in the same Euclidean neighborhood since Euclidean neighborhoods are Hausdorff.

*Definition 2.1. *Let be a non-Hausdorff manifold and let and be distinct points of . Then and are compatible apparition points if there do not exist disjoint open sets and with and . By a โset of compatible apparition points,โโ we will mean that any pair of distinct points in the set are compatible apparition points.

*Remark 2.2. *Since a non-Hausdorff manifold is locally Hausdorff, then no more than one element of a set of compatible apparition points can be contained in a single Euclidean neighborhood. Hence, a set of compatible apparition points is a closed discrete set.

*Remark 2.3. *Since a non-Hausdorff manifold has a countable base and each point is contained in its own Euclidean neighborhood, any set of compatible apparition points must be countable.

A non-Hausdorff manifold can have an uncountable collection of sets of compatible apparition points.

*Example 2.4. *Let denote the Cantor ternary set and define . Define a topology on as follows.(i)For each a basic open neighborhood of is open in with the usual topology.(ii)For each a basic open neighborhood of is of the form .

Note that for each is a set of compatible apparition points. Also, note that since each can be chosen to be rational, is second countable.

Recall that a subset of a topological space is nowhere dense if the interior of its closure is empty.

Proposition 2.5. *Let be a set of compatible apparition points in a non-Hausdorff manifold . Then is nowhere dense in *

*Proof. *Since is closed and discrete and every element of has a Euclidean neighborhood, is the frontier of which is open. Hence, is nowhere dense by [2, 4G part 2 on page 37].

Proposition 2.6. *Let be an -dimensional non-Hausdorff manifold. Suppose that contains a nonempty set of compatible apparition points. Then every continuous function from to a Hausdorff space is constant on .*

*Proof. *Suppose that is continuous. Attempting a contradiction, suppose that such that . Since is Hausdorff, there are disjoint open sets such that and . Then and are disjoint open subsets of with and a contradiction.

Theorem 2.7. *In a non-Hausdorff manifold, the set of points which are not apparition points is dense.*

*Proof. *Suppose that is a non-Hausdorff manifold. Since is locally Hausdorff, Lemmaโโ4.2 of [3] implies that each has a dense open Hausdorff neighborhood . Since is Lindelรถf, the cover has a countable subcover . Since is Baire, is dense in . Since the elements of are Hausdorff, any point in can be separated from any other point in . Therefore, is a dense set of nonapparition points.

#### Acknowledgments

The authors would like to thank the referee for numerous helpful suggestions. The referee was particularly helpful in improving Theorem 2.7.

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#### Copyright

Copyright © 2009 Steven L. Kent et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.