ON DENSE METRIZABLE SUBSPACES OF TOPOLOGICAL SPACES

Ollie Nanyes

Bradley University

\underline{Abstract}. In this note we investigate the question: when does the metrizability of a dense subspace of a topological space imply the metrizability of the whole space? We show that certain conditions always fail to be sufficient and then we examine some elementary examples. We conclude with a theorem which states that a first countable, regular, Hausdorff space Y which has an open metrizable (in the subspace topology) subspace X is metrizable provided Y − X is scattered in Y. Our investigation is conducted on an elementary level.

1. Introduction. A question once asked by a teacher of a topology class (who was a non–specialist) was, “If T is a topological space and X is a dense subset of Y such that X is metrizable in the subspace topology, when can we conclude that Y is metrizable?” I have found no elementary topology textbook that deals with this question. In this note we will investigate this question and then prove a theorem which gives sufficient conditions for Y and X − Y to imply that Y is metrizable.

2. Counterexamples to Some Natural Conjectures. In this section, we will show that certain kinds of metric spaces can always be embedded as a dense subset of a non–metric space Y. Hence, any conjectures about the metrizability must always include certain conditions on X and on Y. We then use certain well–known examples to provide some counterexamples to various “natural” conjectures. Throughout this section, Y is the topological space in question, X is the dense subset of Y which is metrizable in the subspace topology of Y.

\underline{Theorem 1}. If X is any metric space, X can be embedded as a dense subset of a compact, non–Hausdorff space.

\underline{Proof}. We use the “open extension” example [1]. Let (X, τ) be a metrizable topological space and p a point not in X. We define a topology τ′ for X ∪ {q} by declaring a set U ∈ τ′ to be open if U ∈ τ or if U = X ∪ {q}. (X ∪ {q}, τ′) is then a compact non–Hausdorff space with a dense metrizable subspace X.

\underline{Theorem 2.}. If X is a metric space which has a non–open sequence which has no limit point then X can be embedded as an open dense subspace of a Hausdorff non–regular space.

\underline{Proof}. Let \{x_{i}\} be the non–open sequence which has no limit point. Let p be a point not in X and consider Y = (X ∪ {p}). We define a topology for Y as follows: the basic open neighborhoods of all x ≠ {p} will be the same basic open neighborhoods for x in X and the basic open neighborhoods for p will consist of p ∪ (X − {x_{j}}) and all

W_{i}^{n} = p ∪ ( ⋃_{j=n}^{∞} B_{j}(\frac{1}{2^{i}} ))

where B_{j}(ϵ) is the ϵ–ball centered at x_{j}. Note that {x_{j}} remains a closed set in Y. It is easy to see that Y is Hausdorff, given x_{k} ∈ {x_{j}} and p note that there exists ϵ > 0 such that B_{k}(ϵ) ∩ ({x_{j}} − x_{k}) = ∅ (by the regularity of X). Choose n > k and i such that 1/2^{i} < ϵ/3. Then x ∈ B_{k}(ϵ/3), p ∈ W_{i}^{n}, and B_{k}(ϵ/3) ∩ W_{i}^{n} = ∅. If x ∉ {x_{j}}, there exists ϵ > 0 such that B_{x}(ϵ) ∩ {x_{j}} = ∅. Choosing i as before, we have W_{i}^{n} ∩ B_{x}(ϵ/3) = ∅. Hence, Y is Hausdorff. Note that the collection of W_{i}^{n} − {x_{i}} is a local basis for p. But any open set containing {x_{j}} must necessarily intersect each {W_{i}^{n} − {x_{j}}} since each B_{i} is a local basis for x_{i}. Hence, there are no open sets separating p from {x_{j}}. Hence, Y is not regular.

\underline{Theorem 3}. If X is a non–second countable metric space then X can be embedded as an open dense subspace of a Hausdorff space that is not first countable.

\underline{Proof}. Since X is metric and not second countable there exists an uncountable collection of disjoint open sets

⋃_{α∈I} U_{α}.

(We assume that we have some well ordering on the uncountable index set I.) Let {x_{α}} be a net where x_{α} ∈ U_{α}. For each α there exists a ϵ_{α} (we can assume that for all α that ϵ_{α} < ϵ for some fixed ϵ > 0) such that x_{α} ∈ B_{α}(ϵ_{α}) ⊂ U_{α}. Let Y = {X ∪ {p}} and define the open sets to be the open sets of X together with

W_{i}^{β} = ( ⋃_{α>β} B_{α}(\frac{ϵ_{α}}{i} )) ∪ \{p\}, where i ∈ ℕ.

It is clear that Y is not first countable at p. To check that Y is Hausdorff we need only consider p and x where x ∉ {x_{α}} and x ∈ {x_{α}}. If x ∈ {x_{α}}, then x = x_{η} and B_{η}(ϵ_{η}) ∩ W_{1}^{η} = ∅. If x ∉ {x_{α}}, then by the regularity of X, there is some δ < ϵ/3 such that B_{x}(δ) ∩ {x_{α}} = ∅. So, p ∈ W_{3}^{β} and W_{3}^{β} ∩ B_{x}(δ) = ∅ for some β ∈ I.

We will now show, by using some well–known examples, that certain conditions fail to be sufficient.

\underline{Example 4}. The following is an example of a Hausdorff, regular, first countable, non– metrizable space Y which has a countable dense subspace which is metrizable in the subspace topology.

Let ℝ_{l} denote the real line ℝ^{1} with the topology generated by basis elements [x, y). Let ℚ_{l} denote the rationals in the subspace topology of ℝ_{l}. Since ℝ_{l} is Hausdorff, regular and first countable, so is ℚ_{l}. Since [q, p) (q, p ∈ ℚ) forms a basis for the subspace topology of ℚ_{l}, ℚ_{l} is second countable and therefore metrizable by the Urysohn metrization theorem. Let q_{i} be an enumeration of ℚ. Here is a metric for ℚ_{l}:

d(x, y) = |x − y| + ∑_{i=1}^{∞} (\frac{1}{2^{i}} ) |f_{i}(x) − f_{i}(y)|,

where f_{i}(x) = 1 if x ∈ [q_{i}, ∞) and f_{i}(x) = 0, otherwise. Note that ℚ_{l} with the above metric embeds isometrically into ℝ_{q}, with the topology of the real line generated by elements of the form [p, q) where p and q are rational. It is a challenging exercise to show that ℝ_{q} is homeomorphic to the irrationals in the standard Euclidean topology.

\underline{Example 5}. An example of a Hausdorff, regular, first countable, separable, non– metrizable space Y which has an open, dense, connected and second countable subset X which is metrizable in the subspace topology and whose complement, Y − X, is also metrizable in the subspace topology.

Consider the closed upper half plane ℝ^{2+}{(x, y) | y ≥ 0} in Niemytzki’s tangent disk topology ([1], Example 82). The basic open neighborhoods for points x not on the x–axis are open disks whose boundaries miss the x–axis. The basic open neighborhoods for points on the x–axis together with the tangent point (i.e. {(a, 0)} ∪ {(x, y) | x^{2} + (y − a)^{2} < a^{2}}). Let X = {(x, y) | y > 0} (the upper open half plane). Note that X is homeomorphic to the open upper half plane in the standard topology via the identity map. Y − X is also metrizable as Y − X in the subspace topology is merely the real line with the discrete topology. But Y is not metrizable as Y is separable but not second countable.

\underline{Example 6}. An example of a Hausdorff, regular, first countable, locally 1–Euclidean non–metrizable space Y which has an open dense subset X which is metrizable in the subspace topology.

Consider the long line L which is constructed from the ordinal space [0, Ω) (the set of ordinals 0, 1, 2, …, ω_{0}, ω_{0} + 1, … Ω in the standard order topology where ω_{0} denotes the least countable ordinal and Ω denotes the least uncountable ordinal) by placing between each ordinal α and its successor α + 1 a copy of the standard unit interval (0, 1). See [1], Example 45 or [2], p. 3. The subspace L − [0, Ω) is an open dense subset of L which is homeomorphic to an uncountable disjoint union of open intervals

⋃_{α∈[0,Ω)}(0, 1)_{α}.

L−[0, Ω) is metrizable in the subspace topology via a standard bounded metric. L is locally 1–Euclidean because every point x of L is contained in an open neighborhood which is homeomorphic to ℝ^{1}. If x ∈ L − [0, Ω) or if x has an immediate predecessor this is clear. If x ∈ [0, Ω) and x has no immediate predecessor x is contained in some neighborhood of the form (ω, x + 1) where ω is some countably infinite ordinal. But (ω, x) is homeomorphic to ℝ^{1}.

The reader is invited to investigate these concepts and prove theorems of the following type. If X is a dense subset of Y which is metrizable in the subspace topology, then X having properties P and Y having properties Q (and possibly X − Y having properties S) imply that Y is metrizable (or that the metric on X extends to a metric on Y which gives the subspace topology). Here is such a theorem.

\underline{Theorem 7}. Let Y be a regular, first countable Hausdorff space. Suppose that Y has a dense open subset X which is metrizable in the subspace topology. Furthermore, assume that the set Y −X is a scattered subset of Y (that is, there exist a mutually disjoint collection of open subsets of Y, each of which contains exactly one element of Y − X). Then Y is metrizable.

\underline{Proof}. We assume some well ordering of the elements of Y −X with index set I. Because Y − X is scattered and Y is regular, for each y_{α} ∈ Y − X, α ∈ I, we get an open set G_{α} containing y_{α} where \overline{G_{α}} ∩ \overline{G_{β}} = ∅ for α ≠ β. By the regularity and first countability of Y, for each y_{α} we get a countable local basis U^{α} such that for each local basis element U_{i}^{α} ∈ U^{α} and i > j, \overline{U_{i}^{α}} ⊂ U_{j}^{α} and \overline{U_{1}^{α}} ⊂ G_{α}.

Because X is metrizable, by the Bing–Nagata–Smirnov Metrization Theorem, X has a basis that is countably locally discrete in X. (Recall that a collection of sets C is said to be locally discrete in Y if every point in Y has an open neighborhood which intersects at most one element of C. C is said to be countably locally discrete if

C = ⋃_{i=1}^{∞} C_{i}

where each C_{i} is a locally discrete collection, (see chapter 7 of [3]).

Let

B = ⋃_{i=1}^{∞} B_{i}

be a countably locally discrete basis for X. Note that each basis element of X is open in Y. Let

B_{n}^{j} = \{B − ⋃_{α∈I} \overline{U_{j}^{α}} | B ∈ B_{n}, U_{j}^{α} ∈ U^{α}\}.

\underline{Claim}.

T = ⋃_{n=1}^{∞} ( ⋃_{j=1}^{∞} B_{n}^{j} ∪ ( ⋃_{α∈I} U_{j}^{α}))

is a countably locally discrete basis for Y. Proof of the claim will prove the Theorem by the Bing–Nagata–Smirnov Metrization Theorem.

\underline{Proof of the Claim}. It is an easy exercise to see that T is a countably locally discrete collection of open sets in Y. We need only show that T is a basis for Y. Let V be an open subset of Y and let y ∈ V. If y ∈ Y − X, we choose the appropriate U_{j}^{α}. Suppose now that y ∈ X. If y ∉ G_{α} for any α then y ∉ ∪_{α∈I} \overline{U_{1}^{α}}. Because ∪_{α∈I} \overline{U_{1}^{α}} is closed in Y and Y is regular there is some open set W in Y such that y ∈ W ⊂ Y − ∪_{α∈I} \overline{U_{1}^{α}}. So y ∈ W ∩ V ⊂ Y − ∪_{α∈I} \overline{U_{1}^{α}} ⊂ X. So W ∩ V is open in X so there exists an n and a B ∈ B_{n} such that y ∈ B ⊂ W ∩ V. Necessarily B ∈ B_{n}^{1}. Now suppose y ∈ G_{α} for some α. Because Y is Hausdorff, we can get an open set W and a U_{i}^{α} such that y ∈ W and W ∩ U_{i}^{α} = ∅. Note that W ∩ G_{α} is an open set in X that contains y. We can then find some n and B ∈ B_{n} such that y ∈ B ⊂ W ∩ G_{α}. Necessarily B ∈ B_{n}^{i}. There T is a basis for Y.

Examples 5 and 6 show that if Y − X is uncountable, Y − X cannot be allowed to be non–scattered, even when X is assumed to be open in Y. The reader is invited to investigate the case in which Y − X is countable but non–scattered and X is open.

\underline{References}

1. L. Steen and J. Seebach, Counterexamples in Topology, Springer–Verlag, New York, 1978.

2. B. Rushing, Topological Embeddings, Academic Press, New York, 1973.

3. J. R. Munkres, Topology: A First Course, Prentice–Hall Inc., New Jersey, 1975. æ