# A Note on Topological Properties of  Non-Hausdorff Manifolds

Authors: Steven L. Kent; Roy A. Mimna; Jamal K. Tartir
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## 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 dimensionn is a Hausdorff topological space which has a countable base of open sets and is locally Euclidean of dimension n. We begin with the definition of a non-Hausdorff topological manifold.


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## Body

## 1. Topological Properties of Non-Hausdorff Manifolds

Definition 1.1.
A non-Hausdorff manifold of dimensionn is a topological space which has a countable base of open sets and is locally Euclidean of dimension n.Since every point of a non-Hausdorff manifold has a Euclidean neighborhood, it is easy to show that every non-Hausdorff manifold isT1.We now briefly review some of the well-known properties of non-Hausdorff manifolds. Sinceℝn is locally compact, a non-Hausdorff manifold of dimension n 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 T1-spaces, however, need not to be closed. This remains true for non-Hausdorff manifolds (Example 1.2). A non-Hausdorff manifold of dimension n must be locally connected. Since a non-Hausdorff manifold M has a countable base of open sets, M is Lindelöf; that is, every open cover of M has a countable subcover. Further, since locally compact Lindelöf spaces are sigma-compact, it follows that a non-Hausdorff manifold M of dimension n is sigma-compact. Finally, we note that when M is not Hausdorff, it is not regular.We now consider the property of paracompactness. A Hausdorff spaceX is paracompact if every open covering 𝒰 of X has a locally finite refinement 𝒱. That is, each V∈𝒱 is contained in some U∈𝒰 and each x∈X has a neighborhood N which meets only finitely many sets in 𝒱. Paracompactness can be defined for T1-spaces as follows. A T1-space X is paracompact if and only if each open covering of X has an open barycentric refinement, where 𝒱 is a barycentric refinement of 𝒰 if the collection {St(x,𝒱):x∈X} refines 𝒰, where St(x,𝒱)=∪{V∈𝒱:x∈V}. 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 manifoldM need not to be metacompact.LetM=ℝ∪(ℚ×{1}) and define a topology on M as follows.(i)
For eachx∈ℝ, a basic open neighborhood of x is open in ℝ with the usual topology.(ii)
For each(q,1)∈ℚ×{1}, a basic open neighborhood of (q,1) is of the form [{(q,1)}∪U]∖{q} where U is an open neighborhood of q in ℝ with the usual topology.Claim 1.
The non-Hausdorff manifoldM is not metacompact.Proof.
Let𝒰={{(q,1)}∪ℝ:q∈ℚ}. To see that 𝒰 has no point finite refinement, let 𝒱 be a refinement of 𝒰. Let q0∈ℚ and ε0>0 such that (q0-ε0,q0+ε0) is a subset of some element of 𝒱. For each n∈ℕ, choose qn∈ℚ,εn>0, and Vn∈𝒱 such that [qn-εn,qn+εn]⊆(qn-1-εn-1,qn-1+εn-1)∖{qn-1},εn<1/n, and ({(qn,1)}∪[qn-εn,qn+εn])∖{qn}⊆Vn. By the way 𝒰 is defined, no element of 𝒰 contains more than one element of ℚ×{1}. Since 𝒱 is a refinement of 𝒰, no element of 𝒱 contains more than one element of ℚ×{1}. Hence, Vj≠Vk whenever j≠k. By Cantor's Intersection theorem, there exists x∈ℝ such that {x}=⋂n=1∞[qn-εn,qn+εn]⊆⋂n=1∞Vn. Therefore, 𝒱 is not point finite.Remark 1.3.
In the above example,[0,1] is compact and Hausdorff but not closed.Remark 1.4.
For eachn∈ℕ,ℝn is a complete metric space and ℚn is a countable dense subset of ℝn. Therefore, a construction similar to the one above can be used to create a non-Hausdorff manifold of dimension n that is not metacompact.


## 2. Compatible Apparition Points

If a manifoldM of dimension n is non-Hausdorff, there exist at least two points x and y which cannot be separated by disjoint open sets. Also, the points x and y cannot be contained in the same Euclidean neighborhood since Euclidean neighborhoods are Hausdorff.Definition 2.1.
LetM be a non-Hausdorff manifold and let x and y be distinct points of M. Then x and y are compatible apparition points if there do not exist disjoint open sets U and V with x∈U and y∈V. By a “set of compatible apparition points,’’ 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.
LetC denote the Cantor ternary set and define X=ℝ∪(C×{0}). Define a topology on X as follows.(i)
For eachx∈ℝ, a basic open neighborhood of x is open in ℝ with the usual topology.(ii)
For eachx∈C, a basic open neighborhood of (x,0) is of the form [(x-ε,x+ε)∩C]×{0}∪(x-ε,x+ε)∖C.Note that for eachx∈C,{x,(x,0)} is a set of compatible apparition points. Also, note that since each ε can be chosen to be rational, X 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.
LetS be a set of compatible apparition points in a non-Hausdorff manifold M. Then S is nowhere dense in M.Proof.
SinceS is closed and discrete and every element of M has a Euclidean neighborhood, S is the frontier of M∖S which is open. Hence, S is nowhere dense by [2, 4G part 2 on page 37].Proposition 2.6.
LetM be an n-dimensional non-Hausdorff manifold. Suppose that M contains a nonempty set S of compatible apparition points. Then every continuous function from M to a Hausdorff space X is constant on S.Proof.
Suppose thatf:M→X is continuous. Attempting a contradiction, suppose that x1,x2∈S such that f(x1)≠f(x2). Since X is Hausdorff, there are disjoint open sets U1,U2⊆X such that f(x1)∈U1 and f(x2)∈U2. Then f-1(U1) and f-1(U2) are disjoint open subsets of M with x1∈f-1(U1) and x2∈f-1(U2), a contradiction.Theorem 2.7.
In a non-Hausdorff manifold, the set of points which are not apparition points is  dense.Proof.
Suppose thatM is a non-Hausdorff manifold. Since M is locally Hausdorff, Lemma  4.2 of [3] implies that each x∈M has a dense open Hausdorff neighborhood Ux. Since M is Lindelöf, the cover {Ux}x∈M has a countable subcover 𝒞. Since M is Baire, ∩𝒞 is dense in M. Since the elements of 𝒞 are Hausdorff, any point in ∩𝒞 can be separated from any other point in M. Therefore, ∩𝒞 is a dense set of nonapparition points.



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*Source: 891785-2009-07-27.xml*