Michael Line Basics

Like the Sorgenfrey line, the Michael line is a classic counterexample that is covered in standard topology textbooks and in first year topology courses. This easily accessible example helps transition students from the familiar setting of the Euclidean topology on the real line to more abstract topological spaces. One of the most famous results regarding the Michael line is that the product of the Michael line with the space of the irrational numbers is not normal. Thus it is an important example in demonstrating the pathology in products of paracompact spaces. The product of two paracompact spaces does not even have be to be normal, even when one of the factors is a complete metric space. In this post, we discuss this classical result and various other basic results of the Michael line.

Let \mathbb{R} be the real number line. Let \mathbb{P} be the set of all irrational numbers. Let \mathbb{Q}=\mathbb{R}-\mathbb{P}, the set of all rational numbers. Let \tau be the usual topology of the real line \mathbb{R}. The following is a base that defines a topology on \mathbb{R}.

    \mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}

The real line with the topology generated by \mathcal{B} is called the Michael line and is denoted by \mathbb{M}. In essense, in \mathbb{M}, points in \mathbb{P} are made isolated and points in \mathbb{Q} retain the usual Euclidean open sets.

The Euclidean topology \tau is coarser (weaker) than the Michael line topology (i.e. \tau being a subset of the Michael line topology). Thus the Michael line is Hausdorff. Since the Michael line topology contains a metrizable topology, \mathbb{M} is submetrizable (submetrized by the Euclidean topology). It is clear that \mathbb{M} is first countable. Having uncountably many isolated points, the Michael line does not have the countable chain condition (thus is not separable). The following points are discussed in more details.

  1. The space \mathbb{M} is paracompact.
  2. The space \mathbb{M} is not Lindelof.
  3. The extent of the space \mathbb{M} is c where c is the cardinality of the real line.
  4. The space \mathbb{M} is not locally compact.
  5. The space \mathbb{M} is not perfectly normal, thus not metrizable.
  6. The space \mathbb{M} is not a Moore space, but has a G_\delta-diagonal.
  7. The product \mathbb{M} \times \mathbb{P} is not normal where \mathbb{P} has the usual topology.
  8. The product \mathbb{M} \times \mathbb{P} is metacompact.
  9. The space \mathbb{M} has a point-countable base.
  10. For each n=1,2,3,\cdots, the product \mathbb{M}^n is paracompact.
  11. The product \mathbb{M}^\omega is not normal.
  12. There exist a Lindelof space L and a separable metric space W such that L \times W is not normal.

Results 10, 11 and 12 are shown in some subsequent posts.

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Baire Category Theorem

Before discussing the Michael line in greater details, we point out one connection between the Michael line topology and the Euclidean topology on the real line. The Michael line topology on \mathbb{Q} coincides with the Euclidean topology on \mathbb{Q}. A set is said to be a G_\delta-set if it is the intersection of countably many open sets. By the Baire category theorem, the set \mathbb{Q} is not a G_\delta-set in the Euclidean real line (see the section called “Discussion of the Above Question” in the post A Question About The Rational Numbers). Thus the set \mathbb{Q} is not a G_\delta-set in the Michael line. This fact is used in Result 5.

The fact that \mathbb{Q} is not a G_\delta-set in the Euclidean real line implies that \mathbb{P} is not an F_\sigma-set in the Euclidean real line. This fact is used in Result 7.

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Result 1

Let \mathcal{U} be an open cover of \mathbb{M}. We proceed to derive a locally finite open refinement \mathcal{V} of \mathcal{U}. Recall that \tau is the usual topology on \mathbb{R}. Assume that \mathcal{U} consists of open sets in the base \mathcal{B}. Let \mathcal{U}_\tau=\mathcal{U} \cap \tau. Let Y=\cup \mathcal{U}_\tau. Note that Y is a Euclidean open subspace of the real line (hence it is paracompact). Then there is \mathcal{V}_\tau \subset \tau such that \mathcal{V}_\tau is a locally finite open refinement \mathcal{V}_\tau of \mathcal{U}_\tau and such that \mathcal{V}_\tau covers Y (locally finite in the Euclidean sense). Then add to \mathcal{V}_\tau all singleton sets \left\{ x \right\} where x \in \mathbb{M}-Y and let \mathcal{V} denote the resulting open collection.

The resulting \mathcal{V} is a locally finite open collection in the Michael line \mathbb{M}. Furthermore, \mathcal{V} is also a refinement of the original open cover \mathcal{U}. \blacksquare

A similar argument shows that \mathbb{M} is hereditarily paracompact.

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Result 2

To see that \mathbb{M} is not Lindelof, observe that there exist Euclidean uncountable closed sets consisting entirely of irrational numbers (i.e. points in \mathbb{P}). For example, it is possible to construct a Cantor set entirely within \mathbb{P}.

Let C be an uncountable Euclidean closed set consisting entirely of irrational numbers. Then this set C is an uncountable closed and discrete set in \mathbb{M}. In any Lindelof space, there exists no uncountable closed and discrete subset. Thus the Michael line \mathbb{M} cannot be Lindelof. \blacksquare

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Result 3

The argument in Result 2 indicates a more general result. First, a brief discussion of the cardinal function extent. The extent of a space X is the smallest infinite cardinal number \mathcal{K} such that every closed and discrete set in X has cardinality \le \mathcal{K}. The extent of the space X is denoted by e(X). When the cardinal number e(X) is e(X)=\aleph_0 (the first infinite cardinal number), the space X is said to have countable extent, meaning that in this space any closed and discrete set must be countably infinite or finite. When e(X)>\aleph_0, there are uncountable closed and discrete subsets in the space.

It is straightforward to see that if a space X is Lindelof, the extent is e(X)=\aleph_0. However, the converse is not true.

The argument in Result 2 exhibits a closed and discrete subset of \mathbb{M} of cardinality c. Thus we have e(\mathbb{M})=c. \blacksquare

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Result 4

The Michael line \mathbb{M} is not locally compact at all rational numbers. Observe that the Michael line closure of any Euclidean open interval is not compact in \mathbb{M}. \blacksquare

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Result 5

A set is said to be a G_\delta-set if it is the intersection of countably many open sets. A space is perfectly normal if it is a normal space with the additional property that every closed set is a G_\delta-set. In the Michael line \mathbb{M}, the set \mathbb{Q} of rational numbers is a closed set. Yet, \mathbb{Q} is not a G_\delta-set in the Michael line (see the discussion above on the Baire category theorem). Thus \mathbb{M} is not perfectly normal and hence not a metrizable space. \blacksquare

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Result 6

The diagonal of a space X is the subset of its square X \times X that is defined by \Delta=\left\{(x,x): x \in X \right\}. If the space is Hausdorff, the diagonal is always a closed set in the square. If \Delta is a G_\delta-set in X \times X, the space X is said to have a G_\delta-diagonal. It is well known that any metric space has G_\delta-diagonal. Since \mathbb{M} is submetrizable (submetrized by the usual topology of the real line), it has a G_\delta-diagonal too.

Any Moore space has a G_\delta-diagonal. However, the Michael line is an example of a space with G_\delta-diagonal but is not a Moore space. Paracompact Moore spaces are metrizable. Thus \mathbb{M} is not a Moore space. For a more detailed discussion about Moore spaces, see Sorgenfrey Line is not a Moore Space. \blacksquare

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Result 7

We now show that \mathbb{M} \times \mathbb{P} is not normal where \mathbb{P} has the usual topology. In this proof, the following two facts are crucial:

  • The set \mathbb{P} is not an F_\sigma-set in the real line.
  • The set \mathbb{P} is dense in the real line.

Let H and K be defined by the following:

    H=\left\{(x,x): x \in \mathbb{P} \right\}
    K=\mathbb{Q} \times \mathbb{P}.

The sets H and K are disjoint closed sets in \mathbb{M} \times \mathbb{P}. We show that they cannot be separated by disjoint open sets. To this end, let H \subset U and K \subset V where U and V are open sets in \mathbb{M} \times \mathbb{P}.

To make the notation easier, for the remainder of the proof of Result 7, by an open interval (a,b), we mean the set of all real numbers t with a<t<b. By (a,b)^*, we mean (a,b) \cap \mathbb{P}. For each x \in \mathbb{P}, choose an open interval U_x=(a,b)^* such that \left\{x \right\} \times U_x \subset U. We also assume that x is the midpoint of the open interval U_x. For each positive integer k, let P_k be defined by:

    P_k=\left\{x \in \mathbb{P}: \text{ length of } U_x > \frac{1}{k} \right\}

Note that \mathbb{P}=\bigcup \limits_{k=1}^\infty P_k. For each k, let T_k=\overline{P_k} (Euclidean closure in the real line). It is clear that \bigcup \limits_{k=1}^\infty P_k \subset \bigcup \limits_{k=1}^\infty T_k. On the other hand, \bigcup \limits_{k=1}^\infty T_k \not\subset \bigcup \limits_{k=1}^\infty P_k=\mathbb{P} (otherwise \mathbb{P} would be an F_\sigma-set in the real line). So there exists T_n=\overline{P_n} such that \overline{P_n} \not\subset \mathbb{P}. So choose a rational number r such that r \in \overline{P_n}.

Choose a positive integer j such that \frac{2}{j}<\frac{1}{n}. Since \mathbb{P} is dense in the real line, choose y \in \mathbb{P} such that r-\frac{1}{j}<y<r+\frac{1}{j}. Now we have (r,y) \in K \subset V. Choose another integer m such that \frac{1}{m}<\frac{1}{j} and (r-\frac{1}{m},r+\frac{1}{m}) \times (y-\frac{1}{m},y+\frac{1}{m})^* \subset V.

Since r \in \overline{P_n}, choose x \in \mathbb{P} such that r-\frac{1}{m}<x<r+\frac{1}{m}. Now it is clear that (x,y) \in V. The following inequalities show that (x,y) \in U.

    \lvert x-y \lvert \le \lvert x-r \lvert + \lvert r-y \lvert < \frac{1}{m}+\frac{1}{j} \le \frac{2}{j} < \frac{1}{n}

The open interval U_x is chosen to have length > \frac{1}{n}. Since \lvert x-y \lvert < \frac{1}{n}, y \in U_x. Thus (x,y) \in \left\{ x \right\} \times U_x \subset U. We have shown that U \cap V \ne \varnothing. Thus \mathbb{M} \times \mathbb{P} is not normal. \blacksquare

Remark
As indicated above, the proof of Result 7 hinges on two facts about \mathbb{P}, namely that it is not an F_\sigma-set in the real line and it is dense in the real line. We can modify the construction of the Michael line by using other partition of the real line (where one set is isolated and its complement retains the usual topology). As long as the set D that is isolated is not an F_\sigma-set in the real line and is dense in the real line, the same proof will show that the product of the modified Michael line and the space D (with the usual topology) is not normal. This will be how Result 12 is derived.

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Result 8

The product \mathbb{M} \times \mathbb{P} is not paracompact since it is not normal. However, \mathbb{M} \times \mathbb{P} is metacompact.

A collection of subsets of a space X is said to be point-finite if every point of X belongs to only finitely many sets in the collection. A space X is said to be metacompact if each open cover of X has an open refinement that is a point-finite collection.

Note that \mathbb{M} \times \mathbb{P}=(\mathbb{P} \times \mathbb{P}) \cup (\mathbb{Q} \times \mathbb{P}). The first \mathbb{P} in \mathbb{P} \times \mathbb{P} is discrete (a subspace of the Michael line) and the second \mathbb{P} has the Euclidean topology.

Let \mathcal{U} be an open cover of \mathbb{M} \times \mathbb{P}. For each a=(x,y) \in \mathbb{Q} \times \mathbb{P}, choose U_a \in \mathcal{U} such that a \in U_a. We can assume that U_a=A \times B where A is a usual open interval in \mathbb{R} and B is a usual open interval in \mathbb{P}. Let \mathcal{G}=\lbrace{U_a:a \in \mathbb{Q} \times \mathbb{P}}\rbrace.

Fix x \in \mathbb{P}. For each b=(x,y) \in \lbrace{x}\rbrace \times \mathbb{P}, choose some U_b \in \mathcal{U} such that b \in U_b. We can assume that U_b=\lbrace{x}\rbrace \times B where B is a usual open interval in \mathbb{P}. Let \mathcal{H}_x=\lbrace{U_b:b \in \lbrace{x}\rbrace \times \mathbb{P}}\rbrace.

As a subspace of the Euclidean plane, \bigcup \mathcal{G} is metacompact. So there is a point-finite open refinement \mathcal{W} of \mathcal{G}. For each x \in \mathbb{P}, \mathcal{H}_x has a point-finite open refinement \mathcal{I}_x. Let \mathcal{V} be the union of \mathcal{W} and all the \mathcal{I}_x where x \in \mathbb{P}. Then \mathcal{V} is a point-finite open refinement of \mathcal{U}.

Note that the point-finite open refinement \mathcal{V} may not be locally finite. The vertical open intervals in \lbrace{x}\rbrace \times \mathbb{P}, x \in \mathbb{P} can “converge” to a point in \mathbb{Q} \times \mathbb{P}. Thus, metacompactness is the best we can hope for. \blacksquare

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Result 9

A collection of sets is said to be point-countable if every point in the space belongs to at most countably many sets in the collection. A base \mathcal{G} for a space X is said to be a point-countable base if \mathcal{G}, in addition to being a base for the space X, is also a point-countable collection of sets. The Michael line is an example of a space that has a point-countable base and that is not metrizable. The following is a point-countable base for \mathbb{M}:

    \mathcal{G}=\mathcal{H} \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}

where \mathcal{H} is the set of all Euclidean open intervals with rational endpoints. One reason for the interest in point-countable base is that any countable compact space (hence any compact space) with a point-countable base is metrizable (see Metrization Theorems for Compact Spaces).

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

Alexandroff Double Circle

We discuss the Alexandroff double circle, which is a compact and non-metrizable space. A theorem about the hereditarily normality of a product space Y_1 \times Y_2 is also discussed.

Let C_1 and C_2 be the two concentric circles centered at the origin with radii 1 and 2, respectively. Specifically C_i=\left\{(x,y) \in \mathbb{R}^2: x^2 + y^2 =i \right\} where i=1,2. Let X=C_1 \cup C_2. Furthermore let f:C_1 \rightarrow C_2 be the natural homeomorphism. Figure 1 below shows the underlying set.

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Figure 1 – Underlying Set

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We define a topology on X as follows:

  • Points in C_2 are isolated.
  • For each x \in C_1 and for each positive integer j, let O(x,j) be the open arc in C_1 whose center contains x and has length \frac{1}{j} (in the Euclidean topology on C_1). For each x \in C_1, an open neighborhood is of the form B(x,j) where
      \text{ }

      B(x,j)=O(x,j) \cup (f(O(x,j))-\left\{f(x) \right\}).

    The following figure shows an open neighborhood at point in C_1.

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Figure 2 – Open Neighborhood

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A List of Results

It can be verified that the open neighborhoods defined above form a base for a topology on X. We discuss the following points about the Alexandroff double circle.

  1. X is a Hausdorff space.
  2. X is not separable.
  3. X is not hereditarily Lindelof.
  4. X is compact.
  5. X is sequentially compact.
  6. X is not metrizable.
  7. X is not perfectly normal.
  8. X is completely normal (and thus hereditarily normal).
  9. X \times X is not hereditarily normal.

The proof that X \times X is not hereditarily normal can be generalized. We discuss this theorem after presenting the proof of Result 9.
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Results 1, 2, 3

It is clear that the Alexandroff double circle is a Hausdorff space. It is not separable since the outer circle C_2 consists of uncountably many singleton open subsets. For the same reason, C_2 is a non-Lindelof subspace, making the Alexandroff double circle not hereditarily Lindelof. \blacksquare

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Result 4

The property that X is compact is closely tied to the compactness of the inner circle C_1 in the Euclidean topology. Note that the subspace topology of the Alexandroff double circle on C_1 is simply the Euclidean topology. Let \mathcal{U} be an open cover of X consisting of open sets as defined above. Then there are finitely many basic open sets B(x_1,j_1), B(x_2,j_2), \cdots, B(x_n,j_n) from \mathcal{U} covering C_1. These open sets cover the entire space except for the points f(x_1), f(x_2), \cdots,f(x_n), which can be covered by finitely many open sets in \mathcal{U}. \blacksquare

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Result 5

A space W is sequentially compact if every sequence of points of W has a subsequence that converges to a point in W. The notion of sequentially compactness and compactness coincide for the class of metric spaces. However, in general these two notions are distinct.

The sequentially compactness of the Alexandroff double circle X hinges on the sequentially compactness of C_1 and C_2 in the Euclidean topology. Let \left\{x_n \right\} be a sequence of points in X. If the set \left\{x_n: n=1,2,3,\cdots \right\} is a finite set, then \left\{x_n: n>m \right\} is a constant sequence for some large enough integer m. So assume that A=\left\{x_n: n=1,2,3,\cdots \right\} is an infinite set. Either A \cap C_1 is infinite or A \cap C_2 is infinite. If A \cap C_1 is infinite, then some subsequence of \left\{x_n \right\} converges in C_1 in the Euclidean topology (hence in the Alexandroff double circle topology). If A \cap C_2 is infinite, then some subsequence of \left\{x_n \right\} converges to x \in C_2 in the Euclidean topology. Then this same subsequence converges to f^{-1}(x) in the Alexandroff double circle topology. \blacksquare

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Result 6

Note that any compact metrizable space satisfies a long list of properties, which include separable, Lindelof, hereditarily Lindelof. \blacksquare

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Result 7

A space is perfectly normal if it is normal with the additional property that every closed set is a G_\delta-set. For the Alexandroff double circle, the inner circle C_1 is not a G_\delta-set, or equivalently the outer circle C_2 is not an F_\sigma-set. To see this, suppose that C_2 is the union of countably many sets, we show that the closure of at least one of the sets goes across to the inner circle C_1. Let C_2=\bigcup \limits_{i=1}^\infty T_n. At least one of the sets is uncountable. Let T_j be one such. Consider f^{-1}(T_j), which is also uncountable and has a limit point in C_1 (in the Euclidean topology). Let t be one such point (i.e. every Euclidean open set containing t contains points of f^{-1}(T_j)). Then the point t is a member of the closure of T_j (Alexandroff double circle topology). \blacksquare

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Result 8

We first discuss the notion of separated sets. Let T be a Hausdorff space. Let E \subset T and F \subset T. The sets E and F are said to be separated (are separated sets) if E \cap \overline{F}=\varnothing and F \cap \overline{E}=\varnothing. In other words, two sets are separated if each one does not meet the closure of the other set. In particular, any two disjoint closed sets are separated. The space T is said to be completely normal if T satisfies the property that for any two sets E and F that are separated, there are disjoint open sets U and V with E \subset U and F \subset V. Thus completely normality implies normality.

It is a well know fact that if a space is completely normal, it is hereditarily normal (actually the two notions are equivalent). Note that any metric space is completely normal. In particular, any Euclidean space is completely normal.

To show that the Alexandroff double circle X is completely normal, let E \subset X and F \subset X be separated sets. Thus we have E \cap \overline{F}=\varnothing and F \cap \overline{E}=\varnothing. Note that E \cap C_1 and F \cap C_1 are separated sets in the Euclidean space C_1. Let G_1 and G_2 be disjoint Euclidean open subsets of C_1 with E \cap C_1 \subset G_1 and F \cap C_1 \subset G_2.

For each x \in E \cap C_1, choose open U_x (Alexandroff double circle open) with x \in U_x, U_x \cap C_1 \subset G_1 and U_x \cap \overline{F}=\varnothing. Likewise, for each y \in F \cap C_1, choose open V_y (Alexandroff double circle open) with y \in V_y, V_y \cap C_1 \subset G_2 and V_y \cap \overline{E}=\varnothing. Then let U and V be defined by the following:

    U=\biggl(\bigcup \limits_{x \in E \cap C_1} U_x \biggr) \cup \biggl(E \cap C_2 \biggr)

    \text{ }

    V= \biggl(\bigcup \limits_{y \in F \cap C_1} V_y \biggr) \cup \biggl(F \cap C_2 \biggr)

Because G_1 \cap G_2 =\varnothing, the open sets U_x and V_y are disjoint. As a result, U and V are disjoint open sets in the Alexandroff double circle with E \subset U and F \subset V.

For the sake of completeness, we show that any completely normal space is hereditarily normal. Let T be completely normal. Let Y \subset T. Let H \subset Y and K \subset Y be disjoint closed subsets of Y. Then in the space T, H and K are separated. Note that H \cap cl_T(K)=\varnothing and K \cap cl_T(H)=\varnothing (where cl_T gives the closure in T). Then there are disjoint open subsets O_1 and O_2 of T such that H \subset O_1 and K \subset O_2. Now, O_1 \cap Y and O_2 \cap Y are disjoint open sets in Y such that H \subset O_1 \cap Y and K \subset O_2 \cap Y.

Thus we have established that the Alexandroff double circle is hereditarily normal. \blacksquare

For the proof that a space is completely normal if and only if it is hereditarily normal, see Theorem 2.1.7 in page 69 of [1],
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Result 9

We produce a subspace Y \subset X \times X that is not normal. To this end, let D=\left\{d_n:n=1,2,3,\cdots \right\} be a countable subset of X such that \overline{D}-D\ne \varnothing. Let y \in \overline{D}-D. Let Y=X \times X-C_1 \times \left\{y \right\}. We show that Y is not normal.

Let H=C_1 \times (X-\left\{y \right\}) and K=C_2 \times \left\{y \right\}. These are two disjoint closed sets in Y. Let U and V be open in Y such that H \subset U and K \subset V. We show that U \cap V \ne \varnothing.

For each integer j, let U_j=\left\{x \in X: (x,d_j) \in U \right\}. We claim that each U_j is open in X. To see this, pick x \in U_j. We know (x,d_j) \in U. There exist open A and B (open in X) such that (x,d_j) \in A \times B \subset U. It is clear that x \in A \subset U_j. Thus each U_j is open.

Furthermore, we have C_1 \subset U_j for each j. Based in Result 7, C_1 is not a G_\delta-set. So we have C_1 \subset \bigcap \limits_{j=1}^\infty U_j but C_1 \ne \bigcap \limits_{j=1}^\infty U_j. There exists t \in \bigcap \limits_{j=1}^\infty U_j but t \notin C_1. Thus t \in C_2 and \left\{t \right\} is open.

Since (t,y) \in K, we have (t,y) \in V. Choose an open neighborhood B(y,k) of y such that \left\{t \right\} \times B(y,k) \subset V. since y \in \overline{D}, there exists some d_j such that (t,d_j) \in \left\{t \right\} \times B(y,k). Hence (t,d_j) \in V. Since t \in U_j, (t,d_j) \in U. Thus U \cap V \ne \varnothing. \blacksquare

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Generalizing the Proof of Result 9

The proof of Result 9 requires that one of the factors has a countable set that is not discrete and the other factor has a closed set that is not a G_\delta-set. Once these two requirements are in place, we can walk through the same proof and show that the cross product is not hereditarily normal. Thus, the statement that is proved in Result 9 is the following.

Theorem
If Y_1 has a countable subset that is not closed and discrete and if Y_2 has a closed set that is not a G_\delta-set then Y_1 \times Y_2 has a subspace that is not normal.

The theorem can be restated as:

Theorem
If Y_1 \times Y_2 is hereditarily normal, then either every countable subset of Y_1 is closed and discrete or Y_2 is perfectly normal.

The above theorem is due to Katetov and can be found in [2]. It shows that the hereditarily normality of a cross product imposes quite strong restrictions on the factors. As a quick example, if both Y_1 and Y_2 are compact, for Y_1 \times Y_2 to be hereditarily normal, both Y_1 and Y_2 must be perfectly normal.

Another example. Let W=\omega_1+1, the succesor of the first uncountable ordinal with the order topology. Note that W is not perfectly normal since the point \omega_1 is not a G_\delta point. Then for any compact space Y, W \times Y is not hereditarily normal. Let C=\omega+1, the successor of the first infinite ordinal with the order topology (essentially a convergent sequence with the limit point). The product W \times C is the Tychonoff plank and based on the discussion here is not hereditarily normal. Usually the Tychonoff plank is shown to be not hereditarily normal by removing the cornor point (\omega_1,\omega). The resulting space is the deleted Tychonoff plank and is not normal (see The Tychonoff Plank).

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Przymusinski, T. C., Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
  3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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\copyright \ \ 2012

A Space with G-delta Diagonal that is not Submetrizable

The property of being submetrizable implies having a G_\delta-diagonal. There are several other properties lying between these two properties (see [1]). Before diving into these other properties, it may be helpful to investigate a classic example of a space with a G_\delta-diagonal that is not submetrizable.

The diagonal of a space X is the set \Delta=\left\{(x,x): x \in X \right\}, a subset of the square X \times X. An interesting property is when the diagonal of a space is a G_\delta-set in X \times X (the space is said to have a G_\delta-diagonal). Any compact space or a countably compact space with this property must be metrizable (see compact and countably compact space). A space (X,\tau) is said to be submetrizable if there is a topology \tau^* that can be defined on X such that (X,\tau^*) is a metrizable space and \tau^* \subset \tau. In other words, a submetrizable space is a space that has a coarser (weaker) metrizable topology. Every submetrizable space has a G_\delta-diagonal. Note that when X has a weaker metric topology, the diagonal \Delta is always a G_\delta-set in the metric square X \times X (hence in the square in the original topology). The property of having a G_\delta-diagonal is strictly weaker than the property of having a weaker metric topology. In this post, we discuss the Mrowka space, which is a classic example of a space with a G_\delta-diagonal that is not submetrizable.

The Mrowka space (also called Psi space) was discussed previously in this blog (see this post). For the sake of completeness, the example is defined here.

First, we define some basic notions. Let \omega be the first infinite ordinal (or more conveniently the set of all nonnegative integers). Let \mathcal{A} be a family of infinite subsets of \omega. The family \mathcal{A} is said to be an almost disjoint family if for each two distinct A,B \in \mathcal{A}, A \cap B is finite. An almost disjoint family \mathcal{A} is said to be a maximal almost disjoint family if B is an infinite subset of \omega such that B \notin \mathcal{A}, then B \cap A is infinite for some A \in \mathcal{A}. In other words, if you put one more set into a maximal almost disjoint family, it ceases to be almost disjoint.

A natural question is whether there is an uncountable almost disjoint family of subsets of \omega. In fact, there is one whose cardinality is continuum (the cardinality of the real line). To see this, identify \omega with \mathbb{Q}=\lbrace{r_0,r_1,r_2,...}\rbrace (the set of all rational numbers). Let \mathbb{P}=\mathbb{R}-\mathbb{Q} be the set of all irrational numbers. For each x \in \mathbb{P}, choose a subsequence of \mathbb{Q} consisting of distinct elements that converges to x (in the Euclidean topology). Then the family of all such sequences of rational numbers would be an almost disjoint family. By a Zorn’s Lemma argument, this almost disjoint family is contained within a maximal almost disjoint family. Thus we also have a maximal almost disjoint family of cardinality continuum. On the other hand, there is no countably infinite maximal almost disjoint family of subsets of \omega (see this post).

Let \mathcal{A} be an infinite almost disjoint family of subsets of \omega. We now define a Mrowka space (or \Psi-space), denoted by \Psi(\mathcal{A}). The underlying set is \Psi(\mathcal{A})=\mathcal{A} \cup \omega. Points in \omega are isolated. For A \in \mathcal{A}, a basic open set is of the form \lbrace{A}\rbrace \cup (A-F) where F \subset \omega is finite. It is straightforward to verify that \Psi(\mathcal{A}) is Hausdorff, first countable and locally compact. It has a countable dense set of isolated points. Note that \mathcal{A} is an infinite discrete and closed set in the space \Psi(\mathcal{A}). Thus \Psi(\mathcal{A}) is not countably compact.

We would like to point out that the definition of a Mrowka space \Psi(\mathcal{A}) only requires that the family \mathcal{A} is an almost disjoint family and does not necessarily have to be maximal. For the example discribed in the title, \mathcal{A} needs to be a maximal almost disjoint family of subsets of \omega.

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Example
Let \mathcal{A} be a maximal almost disjoint family of subsets of \omega. Then \Psi(\mathcal{A}) as defined above is a space in which there is a G_\delta-diagonal that is not submetrizable.

Note that \Psi(\mathcal{A}) is pseudocompact (proved in this post). Because there is no countable maximal almost disjoint family of subsets of \omega, \mathcal{A} must be an uncountable in addition to being a closed and discrete subspace of \Psi(\mathcal{A}) (thus the space is not Lindelof). Since \Psi(\mathcal{A}) is separable and is not Lindelof, \Psi(\mathcal{A}) is not metrizable. Any psuedocompact submetrizable space is metrizable (see Theorem 4 in this post). Thus \Psi(\mathcal{A}) must not be submetrizable.

On the other hand, any \Psi-space \Psi(\mathcal{A}) (even if \mathcal{A} is not maximal) is a Moore space. It is well known that any Moore space has a G_\delta-diagonal. The remainder of this post has a brief discussion of Moore space.

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Moore Space

A sequence \lbrace{\mathcal{D}_n}\rbrace_{n<\omega} of open covers of a space X is a development for X if for each x \in X and each open set U \subset X with x \in U, there is some n such that any open set in \mathcal{D}_n containing the point x is contained in U. A developable space is one that has a development. A Moore space is a regular developable space.

Suppose that X is a Moore space. We show that X has a G_\delta-diagonal. That is, we wish to show that \Delta=\left\{(x,x): x \in X \right\} is a G_\delta-set in X \times X.

Let \lbrace{\mathcal{D}_n}\rbrace_{n<\omega} be a development. For each n, let U_n=\bigcup \lbrace{V \times V:V \in \mathcal{D}_n}\rbrace. Clearly \Delta \subset \bigcap_{n<\omega} U_n. Let (x,y) \in \bigcap_{n<\omega} U_n. For each n, (x,y) \in V_n \times V_n for some V_n \in \mathcal{D}_n. We claim that x=y. Suppose that x \ne y. By the definition of development, there exists some m such that every open set in \mathcal{D}_m containing the point x has to be a subset of X-\left\{y \right\}. Then V_m \subset X-\left\{y \right\}, which contradicts y \in V_m. Thus we have \Delta = \bigcap_{n<\omega} U_n.

The remaining thing to show is that \Psi(\mathcal{A}) is a Moore space. For each positive integer n, let F_n=\left\{0,1,\cdots,n-1 \right\} and let F_0=\varnothing. The development is defined by \lbrace{\mathcal{E}_n}\rbrace_{n<\omega}, where for each n, \mathcal{E}_n consists of open sets of the form \lbrace{A}\rbrace \cup (A-F_n) where A \in \mathcal{A} plus any singleton \left\{j \right\} (j \in \omega) that has not been covered by the sets \lbrace{A}\rbrace \cup (A-F_n).

Reference

  1. Arhangel’skii, A. V., Buzyakova, R. Z., The rank of the diagonal and submetrizability, Commentationes Mathematicae Universitatis Carolinae, Vol. 47 (2006), No. 4, 585-597.
  2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

When is a Pseudocompact Space Metrizable?

Compactness, countably compactness and pseudocompactness are three successively weaker properties. It follows easily from definitions that

(A) \ \ \ \ \ \text{compact} \Rightarrow \text{countably compact} \Rightarrow \text{pseudocompact}

None of these arrows can be reversed. It is well known that either compactness or countably complactness plus having a G_\delta-diagonal implies metrizability. We have:

(B) \ \ \ \ \ \text{compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable}

(C) \ \ \ \ \ \text{countably compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable}

A question can be asked whether these results can be extended to pseudocompactness.

Question (D) \ \ \ \ \ \text{pseudocompact compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable?}

The answer to this question is no. The space defined using a maximal almost disjoint family of subsets of \omega is an example of a non-metrizable pseudocompact space with a G_\delta-diagonal (discussed in this post). In this post we show that if we strengthen “having a G_\delta-diagonal” to being submetrizable, we have a theorem. Specifically, we show:

(E) \ \ \ \ \ \text{pseudocompact} + \text{submetrizable} \Rightarrow \text{metrizable}

For the result of (B), see this post. For the result of (C), see this post. In this post, we discuss the basic properties of pseudocompactness that build up to the result of (E). All spaces considered here are at least Tychonoff (i.e. completely regular). For any basic notions not defined here, see [1] or [2].

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Pseudocompact Spaces

A space X is said to be pseudocompact if every real-valued continuous function defined on X is a bounded function. Any real-valued continuous function defined on a compact space must be bounded (and is thus pseudocomppact). If there were an unbounded real-valued continuous function defined on a space X, then X would have a countably infinite discrete set (thus not countably compact). Thus countably compact implies pseudocompact, as indicated by (A).

A space X is submetrizable if there is a coarser (i.e. weaker) topology that is a metrizable topology. Specifically the topological space (X,\tau) is submetrizable if there is another topology \tau^* that can be defined on X such that \tau^* \subset \tau and (X,\tau^*) is metrizable. The Sorgenfrey line is non-metrizable and yet the Sorgenfrey topology has a weaker topology that is metrizable, namely the Euclidean topology of the real line.

The following two theorems characterizes pseudocompact spaces in terms of locally finite open family of open sets (Theorem 1) and the finite intersection property (Theorem 2). Both theorems are found in Engelking (Theorem 3.10.22 and Theorem 3.10.23 in page 207 of [1]). Theorem 3 states that in a pseudocompact space, closed domains are pseudocompact (the definition of closed domain is stated before the theorem). Theorem 4 is the main theorem (result E stated above).

Theorem 1
Let X be a space. The following conditions are equivalent:

  1. The space X is pseudocompact.
  2. If \mathcal{V} is a locally finite family of non-empty open subsets of X, then \mathcal{V} is finite.
  3. If \mathcal{V} is a locally finite open cover of X, then \mathcal{V} is finite.
  4. If \mathcal{V} is a locally finite open cover of X, then \mathcal{V} has a finite subcover.

Proof
1 \Rightarrow 2
Suppose that condition 2 does not hold. Then there is an infinite locally finite family of non-empty open sets \mathcal{V} such that \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}. We wish to define an unbounded continuous function using \mathcal{V}.

This is where we need to invoke the assumption of complete regularity. For each n choose a point x_n \in V_n. Then for each n, there is a continuous function f_n:X \rightarrow [0,n] such that f_n(x_n)=n and f_n(X-V_n) \subset \left\{ 0 \right\}. Define f:X \rightarrow [0,\infty) by f(x)=f_1(x)+f_2(x)+f_3(x)+\cdots.

Because \mathcal{V} is locally finite, the function f is essentially pointwise the sum of finitely many f_n. In other words, for each x \in X, for some positive integer N, f_j(x)=0 for all j \ge N. Thus the function f is well defined and is continuous at each x \in X. Note that for each x_n, f(x_n) \ge n, showing that it is unbounded.

The directions 2 \Rightarrow 3 and 3 \Rightarrow 4 are clear.

4 \Rightarrow 1
Let g:X \rightarrow \mathbb{R} be a continuous function. We want to show that g is a bounded function. Consider the open family \mathcal{O}=\left\{\cdots,O_{-3},O_{-2},O_{-1},O_0,O_1,O_2,O_3,\cdots \right\} where each O_n=g^{-1}((n,n+2)). Note that \mathcal{O} is a locally finite family in X since its members O_n=g^{-1}((n,n+2)) are inverse images of members of a locally finite family in the range space \mathbb{R}. By condition 4, \mathcal{O} has a finite subcover, leading to the conclusion that g is a bounded function. \blacksquare

Theorem 2
Let X be a space. The following conditions are equivalent:

  1. The space X is pseudocompact.
  2. If \mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\} is a family of non-empty open subsets of X such that O_n \supset O_{n+1} for each n, then \bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing.
  3. If \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\} is a family of non-empty open subsets of X such that \mathcal{V} has the finite intersection property, then \bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing.

Proof
1 \Rightarrow 2
Suppose that X is pseudocompact. Suppose \mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\} satisfies the hypothesis of condition 2. If there is some positive integer m such that O_n=O_m for all n \ge m, then we are done. So assume that O_n are distinct for infinitely many n. According to condition 2 in Theorem 1, \mathcal{O} must not be a locally finite family. Then there exists a point x \in X such that every open set containing x must meet infinitely many O_n. This implies that x \in \overline{O_n} for infinitely many n. Thus x \in \bigcap \limits_{n=1}^\infty \overline{O_n}.

2 \Rightarrow 3
Suppose \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\} is a family of non-empty open sets with the finite intersection property as in the hypothesis of 3. Then let O_1=V_1, O_2=V_1 \cap V_2, O_3=V_1 \cap V_2 \cap V_3, and so on. By condition 2, we have \bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing, which implies \bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing.

3 \Rightarrow 1
Let g:X \rightarrow \mathbb{R} be a continuous function such that g is unbounded. For each positive integer n, let V_n=\left\{x \in X: \lvert g(x) \lvert > n \right\}. Clearly the open sets V_n have the finite intersection property. Because g is unbounded, it follows that \bigcap \limits_{n=1}^\infty \overline{V_n} = \varnothing. \blacksquare

Let X be a space. Let A \subset X. The interior of A, denoted by \text{int}(A), is the set of all points x \in X such that there exists an open set O with x \in O \subset A. Points of \text{int}(A) are called the interior points of A. A subset C \subset X is said to be a closed domain if C=\overline{\text{int}(C)}. It is clear that C is a closed domain if and only if C is the closure of an open set.

Theorem 3
The property of being a pseudocompact space is hereditary with respect to subsets that are closed domains.

Proof
Let X be a pseudocompact space. We show that \overline{U} is pseudocompact for any nonempty open set U \subset X. Let Y=\overline{U} where U is a non-empty open subset of X. Let S_1 \supset S_2 \supset S_3 \supset \cdots be a decreasing sequence of open subsets of Y. Note that each S_i contains points of the open set U. Let O_i=S_i \cap U for each i. Note that the open sets O_i form a decreasing sequence of open sets in the pseudocompact space X. By Theorem 2, we have \bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing (closure here is with respect to X). Note that points in \bigcap \limits_{n=1}^\infty \overline{O_n} are also points in \bigcap \limits_{n=1}^\infty \overline{S_n} (closure with respect to Y). By Theorem 2, Y=\overline{U} is pseudocompact. \blacksquare

Theorem 4 (Statement E above)
Let X be a pseudocompact submetrizable space. Then X is metrizable.

Proof
Let (X,\tau) be a pseudocompact submetrizable space. Then there exists topology \tau^* on X such (X,\tau^*) is metrizable and \tau^* \subset \tau. We show that \tau \subset \tau^*, leading to the conclusion that (X,\tau) is also metrizable. If A \subset X, we denote the closure of A in (X,\tau) by cl_{\tau}(A) and the closure of A in (X,\tau^*) by cl_{\tau^*}(A).

To show that \tau \subset \tau^*, we show any closed set with respect to the topology \tau is also a closed set with respect to the topology \tau^*. Let C be a closed set in (X,\tau). Consider the family \mathcal{W}=\left\{cl_{\tau}(U): U \in \tau \text{ and } C \subset U \right\}. We make the following claims.

Claim 1. C=\bigcap \left\{W: W \in \mathcal{W} \right\}.

Claim 2. Each W \in \mathcal{W} is pseudocompact in (X,\tau).

Claim 3. Each W \in \mathcal{W} is pseudocompact in (X,\tau^*).

Claim 4. Each W \in \mathcal{W} is compact in (X,\tau^*).

We now discuss each of these four claims. For Claim 1, it is clear that C \subset \bigcap \left\{W: W \in \mathcal{W} \right\}. The reverse set inclusion follows from the fact that X is a regular space. Claim 2 follows from Theorem 3. Note that sets in \mathcal{W} are closed domains in the pseudocompact space (X,\tau).

If sets in \mathcal{W} are pseudocompact in the larger topology \tau, they would be pseudocompact in the weaker topology \tau^* too. Thus Claim 3 is established. In a metrizable space, compactness and weaker notions such as countably compactness and pseudocompactness coincide. Because they are pseudocompact subsets, sets in \mathcal{W} are compact in the metrizable space (X,\tau^*). Thus Claim 4 is established.

It follows that C is closed in (X,\tau^*) since it is the intersection of compact sets in (X,\tau^*). Thus (X,\tau) is identical to (X,\tau^*), implying that (X,\tau) is metrizable. \blacksquare

Reference

  1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
  2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

A Note About Countably Compact Spaces

This is a discussion on several additional conditions that would turn a countably compact space into a compact space. For example, a countably compact space having a G_\delta- diagonal is compact (proved in this post). Each of the following properties, if possessed by a countably compact space, would lead to compactness: (1) having a G_\delta- diagonal, (2) being metrizable, (3) being a Moore space, (4) being paracompact, and (5) being metacompact. All spaces are at least Hausdorff. We have the following theorem. Some relevant definitions and links to posts in this blog are given below. For any terms that are not defined here, see Engelking ([1]).

Theorem. Let X be a countably compact space. If X possesses any one of the following conditions, then X is compact.

  1. Having a G_\delta- diagonal.
  2. Being a metrizable space.
  3. Being a Moore space.
  4. Being a paracompact space.
  5. Being a metacompact space.

The proof of 1 has already been presented in another post in this blog. Since metrizable spaces are Moore spaces, between 2 and 3 we only need to prove 3. Between 4 and 5, we only need to prove 5 (since paracompact compact spaces are metacompact).

Proof of 3. A Moore space is a regular space that has a development (see this post for the definition). In this post, I showed that a space X has a G_\delta-diagonal if and only it has a G_\delta-diagonal sequence. It is easy to verify that the development for a Moore space is a G_\delta-diagonal sequence. Thus any Moore space has a G_\delta-diagonal and any countably compact Moore space is compact (and metrizable). Saying in another way, in the class of Moore spaces, countably compactness is equivalent to compactness.

Proof of 5. A space X is metacompact if every open cover of X has a point-finite open refinement. Let X be metacompact. Let \mathcal{U} be an open cover of X. By the metacompactness, \mathcal{U} has a point-finite open refinement \mathcal{O}. We are done if we can show \mathcal{O} has a finite subcover. This finite subcover is obtained through the following claims.

Claim 1. There is a set M \subset X such that \lvert M \cap O \lvert \thinspace \leq 1 for each O \in \mathcal{O} and such that M is maximal. That is, by adding an additional point x \notin M, \lvert (M \cup \lbrace{x}\rbrace) \cap O \lvert \thinspace \ge 2 for some O \in \mathcal{O}.

Such a set can be obtained by using the Zorn’s Lemma.

Claim 2. Let \mathcal{W}=\lbrace{O \in \mathcal{O}:O \cap M \neq \phi}\rbrace. We claim that \mathcal{W} is an open cover of X.

To see this, let x \in X. If x \in M, then x \in O for some O \in \mathcal{W}. If x \notin M, then by the maximality of M, M \cup \lbrace{x}\rbrace intersects with some O \in \mathcal{O} with at least 2 points. This means that x and at least one point of M are in O. Then O \in \mathcal{W}.

Since each open set in \mathcal{W} contains at most one point of M, M is a closed and discrete set in X. By the countably compactness of X, M must be finite. Since each point of M is in at most finitely many open sets in \mathcal{O}, \mathcal{W} is finite. Thus \mathcal{W} is a finite subcover of \mathcal{O}.

Reference

  1. Engelking, R., General Topology, Revised and Completed Edition, 1989, Heldermann Verlag, Berlin.

Countably Compact Spaces with G-delta Diagonals

It is a classic result in general topology that any compact space with a G_\delta-diagonal is metrizable ([3]). This theorem also holds for countably compact spaces (due to Chaber in [2]). The goal of this post is to present a proof of this theorem. We prove that if X is countably compact and has a G_\delta-diagonal, then X is compact and thus metrizable. All spaces are at least Hausdorff. This post has a discussion on the theorem on compact spaces with G_\delta-diagonal. This post has a discussion on some metrizaton theorems for compact spaces.

If \mathcal{T} is a collection of subsets of a space X, then for each x \in X, define st(x,\mathcal{T})=\bigcup\lbrace{T \in \mathcal{T}:x \in T}\rbrace. A sequence of open covers \lbrace{\mathcal{T}_n:n \in \omega}\rbrace of the space X is a G_\delta-diagonal sequence for X if for each x \in X, we have \lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{T}_n). We use the following lemma (due to Ceder, [1]). This lemma was proved in this previous post.

Lemma. The space X has a G_\delta-diagonal if and only if it has a G_\delta-diagonal sequence.

Theorem. Let X be a countably compact space that has a G_\delta-diagonal. Then X is compact.

Proof. Let X be a countably compact space. Let \lbrace{\mathcal{T}_n:n \in \omega}\rbrace be a G_\delta-diagonal sequence for X. If X is Lindelof, then we are done. Suppose we have an open cover \mathcal{V} of X that has no countable subcover. From this open cover \mathcal{V}, we derive a contradiction.

We inductively, for each \alpha < \omega_1, choose a point x_\alpha \in X and an integer m(\alpha) \in \omega with the following properties:

For each \alpha < \omega_1,

  1. x_\alpha \in X-\bigcup\lbrace{st(x_\beta,\mathcal{T}_{m(\beta)}): \beta < \alpha}\rbrace, and
  2. the open cover \mathcal{V} does not have a countable subcollection that covers X-\bigcup_{\beta \leq \alpha} st(x_\beta,\mathcal{T}_{m(\beta)}).

To start off, choose x_0 \in X. There is an integer m(0) \in \omega such that no countable subcollection of \mathcal{V} covers X-st(x_0,\mathcal{T}_{m(0)}). Suppose this integer m(0) does not exist. Then for each n \in \omega, we have a countable \mathcal{V}_n \subset \mathcal{V} such that \mathcal{V}_n covers X-st(x_0,\mathcal{T}_n). Then \bigcup_{n<\omega} \mathcal{V}_n would be a countable subcollection of \mathcal{V} that covers X-\lbrace{x_0}\rbrace. This would mean that \mathcal{V} has a countable subcover of X.

Suppose that \lbrace{x_\beta:\beta<\alpha}\rbrace and \lbrace{m(\beta):\beta<\alpha}\rbrace have been chosen such that conditions (1) and (2) are satisfied for each \beta<\alpha. We have the following claim. Proving this claim allows us to choose x_\alpha and m(\alpha).

Claim. No countable subcollection of \mathcal{V} covers X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)}).

Suppose we do have a countable \mathcal{W} \subset \mathcal{V} such that \mathcal{W} covers X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)}). Then \mathcal{S}=\lbrace{st(x_\beta,\mathcal{T}_m(\beta)):\beta < \alpha}\rbrace \cup \mathcal{W} is a countable open cover of X and thus has a finite subcover \mathcal{F}. Let \delta be the largest ordinal <\alpha such that st(x_\delta,\mathcal{T}_m(\delta)) is in this finite subcover \mathcal{F}. Then \mathcal{W} is a counntable subcollection of \mathcal{V} that covers X-\bigcup_{\beta \leq \delta} st(x_\beta,\mathcal{T}_{m(\beta)}). This violates condition (2) above for the ordinal \delta. This proves the claim.

Now, pick x_\alpha \in X-\bigcup\lbrace{st(x_\beta,\mathcal{T}_{m(\beta)}): \beta < \alpha}\rbrace. There must be some integer m(\alpha) \in \omega such that conditon (2) above is satisfied for \alpha. If not, for each n \in \omega, there is some countable \mathcal{V}_n \subset \mathcal{V} such that \mathcal{V}_n covers X-\bigcup_{\beta \leq \alpha} st(x_\beta,\mathcal{T}_n). Then \bigcup_{n<\omega} \mathcal{V}_n would be a countable subcollection of \mathcal{V} that covers X-\biggl(\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)}) \biggr) \bigcup \lbrace{x_\alpha}\rbrace. This would mean that \mathcal{V} has a countable subcover of X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)}). This violates the above claim. Now the induction process is completed.

To conclude the proof of the theorem, note that there is some n \in \omega and there is some uncountable D \subset \omega_1 such that for each \alpha \in D, n=m(\alpha). Then Y=\lbrace{x_\alpha:\alpha \in D}\rbrace is an uncountable closed and discrete set in X. Note that each open set in \mathcal{T}_n contains at most one point of Y. Thus X must be Lindelof. With X being countably compact, X is compact.

Reference

  1. Ceder, J. G. Some generalizations of metric spaces, Pacific J. Math., 11 (1961), 105-125.
  2. Chaber, Conditions which imply compactness in countably compact spaces, Bull. Acad. Pol. Sci. Ser. Math., 24 (1976), 993-998.
  3. Sneider, V., Continuous images of Souslin and Borel sets: metrization theorems, Dokl. Acad. Nauk USSR, 50 (1945), 77-79.

Metrization Theorems for Compact Spaces

In this blog I have already presented two metrization theorems for compact spaces: (1) any compact space with a countable network is metrizable (see the post), (2) any compact space with a G_\delta-diagonal is metrizable (see the post). I now present another classic theorem: any countably compact space with a point-countable base is metrizable. This theorem is a classic result of Miscenko ([1]). All spaces are at least Hausdorff and regular. We have the following three metrization theorems for compact spaces. In subsequent posts, I will discuss generalizations of these theorems and discuss related concepts.

Thoerem 1. Any compact space with a countable network is metrizable.
The proof is in this post.

Thoerem 2. Any compact space with a G_\delta-\text{diagonal} is metrizable.
The proof is in this post.

Thoerem 3. Any countably compact space with a point-countable base is metrizable.

A base \mathcal{B} for a space X is a point-countabe base if every point in X belongs to at most countably elements of \mathcal{B}.

Proof of Theorem 3. Let \mathcal{B} be a point-countable base for the countably compact space X. We show that X is separable. Once we have a countable dense subset, the base \mathcal{B} has to be a countable base. So we inductively define a sequence of countable sets \lbrace{D_0,D_1,...}\rbrace such that D=\bigcup_{n<\omega}D_n is dense in X.

Let D_0=\lbrace{x_0}\rbrace be a one-point set to start with. For n>0, let E_n=\bigcup_{i<n}D_n. Let \mathcal{B}_n=\lbrace{B \in \mathcal{B}:B \cap E_n \neq \phi}\rbrace. For each finite T \subset \mathcal{B}_n such that X - \bigcap T \neq \phi, choose a point x(T) \in X - \bigcup T. Let D_n be the union of E_n and the set of all points x(T). Let D=\bigcup_{n<\omega}D_n.

We claim that \overline{D}=X. Suppose we have x \in X-\overline{D}. Let \mathcal{A}=\lbrace{B \in \mathcal{B}:B \cap D \neq \phi \phantom{X} \text{and} \thinspace x \notin B}\rbrace. We know that \mathcal{A} is countable since every element of \mathcal{A} contains points of the countable set D. We also know that \mathcal{A} is an open cover of \overline{D}. By the countably compactness of \overline{D}, we can find a finite T \subset \mathcal{A} such that \overline{D} \subset \bigcup T. The finite set T must have appeared during the induction process of selecting points for D_n for some n (i.e. T \subset \mathcal{B}_n). So a point x(T) has been chosen such that x(T) \notin \bigcup T (thus we have x(T) \in D_n \subset \overline{D}). On the other hand, since \overline{D} \subset \bigcup T, we observe that x(T) \notin \overline{D}, producing a contradiction. Thus the countable set D is dense in X, making the point-countable base \mathcal{B} a countable base.

Reference

  1. Miscenko, A., Spaces with a point-countable base, Dokl. Acad. Nauk SSSR, 144 (1962), 985-988. (English translation: Soviet Math. Dokl. 3 (1962), 1199-1202).