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

Sorgenfrey Line is not a Moore Space

We found an incorrect statement about the Sorgenfrey line in an entry in Wikipedia about Moore space (link). This statement opens up a discussion on the question of whether the Sorgenfrey line is a Moore space as well as a discussion on Moore space. The following is the incorrect statement found in Wikipedia by the author.

The Sorgenfrey line is the space whose underlying set is the real line S=\mathbb{R} where the topology is generated by a base consisting the half open intervals of the form [a,b). The Sorgenfrey plane is the square S \times S.

Even though the Sorgenfrey line is normal, the Sorgenfrey plane is not normal. In fact, the Sorgenfrey line is the classic example of a normal space whose square is not normal. Both the Sorgenfrey line and the Sorgenfrey plane are not Moore space but not for the reason given. The statement seems to suggest that any normal Moore space is second countable. But this flies in the face of all the profound mathematics surrounding the normal Moore space conjecture, which is also discussed in the Wikipedia entry.

The statement indicated above is only a lead-in to a discussion of Moore space. We are certain that it will be corrected. We always appreciate readers who kindly alert us to errors found in this blog.

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Moore Spaces

Let X be a regular space. A development for X is a sequence \mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots of open covers of X such that for each x \in X, and for each open subset U of X with x \in U, there exists one cover \mathcal{G}_n satisfying the condition that for any open set V \in \mathcal{G}_n, x \in V \Rightarrow V \subset U. When X has a development, X is said to be a Moore space (also called developable space). A Note On The Sorgenfrey Line is an introductory note on the Sorgenfrey line.

Moore spaces can be viewed as a generalization of metrizable spaces. Moore spaces are first countable (having a countable base at each point). For a development \mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots, the open sets in \mathcal{G}_n are considered “smaller” as the index n increases. In fact, this is how a development is defined for a metric space, where \mathcal{G}_n consists of all open balls with diameters less than \frac{1}{n}. Thus metric spaces are developable. There are plenty of non-metrizable Moore space. One example is the Niemytzki’s Tangent Disc space.

In a Moore space, every closed set is a G_\delta-set. Thus if a Moore space is normal, it is perfectly normal. Any Moore space has a G_\delta-diagonal (the diagonal \Delta=\left\{(x,x): x \in X \right\} is a G_\delta-set in X \times X). It is a well known theorem that every compact space with a G_\delta-diagonal is metrizable. Thus any compact Moore space is metrizable.

The last statement can be shown more directly. Suppose that X is compact and has a development \mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots. Then each \mathcal{G}_n has a finite subcover \mathcal{H}_n. Then \bigcup_{n=1}^\infty \mathcal{H}_n is a countable base for X. Thus any compact Moore space is second countable and hence metrizable.

What about paracompact Moore space? Suppose that X is paracompact and has a development \mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots. Then each \mathcal{G}_n has a locally finite open refinement \mathcal{H}_n. Then \bigcup_{n=1}^\infty \mathcal{H}_n is a \sigma-locally finite base for X. The Smirnov-Nagata metrization theorem states that a space is metrizable if and only if it has a \sigma-locally finite base (see Theorem 23.9 on page 170 of [2]). Thus any paracompact Moore space has a \sigma-locally finite base and is thus metrizable (after using the big gun of the Smirnov-Nagata metrization theorem).

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Sorgenfrey Line

The Sorgenfrey line is regular and Lindelof. Hence it is paracompact. Since the Sorgenfrey line is not metrizable, by the above discussion it cannot be a Moore space. The Sorgenfrey plane is also not a Moore space. Note that being a Moore space is a hereditary property. So if the Sorgenfrey plane is a Moore space, then every subspace of the Sorgenfrey plane (including the Sorgenfrey line) is a Moore space.

The following theorem is another way to show that the Sorgenfrey line is not a Moore space.

    Bing’s Metrization Theorem
    A topological space is metrizable if and only if it is a collectionwise normal Moore space.

Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [1]). Thus the Sorgenfrey line is collectionwise normal and hence cannot be a Moore space. A space X is said to be collectionwise normal if X is a T_1-space and for every discrete collection \left\{W_\alpha: \alpha \in A \right\} of closed sets in X, there exists a discrete collection \left\{V_\alpha: \alpha \in A \right\} of open subsets of X such that W_\alpha \subset V_\alpha. For a proof of Bing’s metrization theorem, see page 329 of [1].

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Remark

The normal Moore space conjecture is the statement that every normal Moore space is metrizable. This conjecture had been one of the key motivating questions for many set theorists and topologists during a large part of the twentieth century. The bottom line is that this statement cannot not be decided just on the basis of the set of generally accepted axioms called Zermelo–Fraenkel set theory with the axiom of choice, commonly abbreviated ZFC. But Bing’s metrization theorem states that if we strengthen normality to collectionwise normality, we have a definite answer.

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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

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.

Thinking about the Space of Irrationals Topologically

Let \mathbb{R} denote the real number line and \mathbb{P} denote the set of all irrational numbers. The irrational numbers and the set \mathbb{P} occupy an important place in mathematics. The set \mathbb{P} with the Euclidean topology inherited from the full real line is a topological space in its own right. Thus the space \mathbb{P} has some of the same properties inherited from the Euclidean real line, e.g., just to name a few, it is hereditarily separable, hereditarily Lindelof, paracompact and metrizable. The space of the irrational numbers \mathbb{P} has many properties apart from the full real line (e.g. \mathbb{P} is totally disconnected). In this post, we look at a topological description of the space \mathbb{P}.

Let \omega be the set of all nonnegative integers. Then the space \mathbb{P} of irrational numbers is topologically equivalent (i.e. homeomorphic to) the product space \prod \limits_{i=0}^\infty X_i where each X_i=\omega has the discrete topology. We will also denote the product space \prod \limits_{i=0}^\infty X_i by \omega^\omega. We have the following theorem.

Theorem
The space \mathbb{P} of irrational numbers is homeomorphic to the product space \omega^\omega.

Because of this theorem, we can look at irrational numbers as sequences of nonnegative integers. Specifically each irrational number can be identified by a unique sequence of nonnegative integers. We can think of each such unique sequence as an address to locate an irrational number. In the remainder of the post, we describe at a high level how to define the unique addresses, which will also give us a homeomorphic map between the space \mathbb{P} and the product space \omega^\omega.

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Step 0
Put the rational numbers in a sequence r_0,r_1,r_2,r_3,\cdots such that r_0=0. We divide the real line into countably many non-overlapping intervals. Specifically, let A_0=[0,1], A_1=[-1,0], A_2=[1,2], etc (see the following figure).

To facilitate the remaining construction, we denote the left endpoint of the interval A_i by L_i and denote the right endpoint by R_i.

Step 1
In each of the interval A_i, we further divide it into non-overlapping intervals. The left endpoint and the right endpoint of the interval A_i are L_i and R_i, respectively. We choose a sequence x_{i,0}, x_{i,1}, x_{i,2},\cdots of rational numbers converging to the right endpoint R_i. Then let A_{i,0}=[L_i,x_{i,0}], A_{i,1}=[x_{i,0},x_{i,1}], A_{i,2}=[x_{i,1},x_{i,2}], etc (see the following figure).

Two important points to consider in Step 1. One is that we make sure the rational number r_1 is chosen as an endpoint of some interval in Step 1. The second is that the length of each A_{i,j} is less than \displaystyle \frac{1}{2^1}.

Step 2
In each of the interval A_{i,j}, we further divide it into non-overlapping intervals. The left endpoint and the right endpoint of the interval A_{i,j} are L_{i,j} and R_{i,j}, respectively. We choose a sequence x_{i,j,0}, x_{i,j,1}, x_{i,j,2},\cdots of rational numbers converging to the left endpoint L_{i,j}. Then let A_{i,j,0}=[x_{i,j,0},R_{i,j}], A_{i,j,1}=[x_{i,j,1},x_{i,j,0}], A_{i,j,2}=[x_{i,j,2},x_{i,j,1}], etc (see the following figure).

As in the previous step, two important points to consider in Step 2. One is that we make sure the rational number r_2 is chosen as an endpoint of some interval in Step 2. The second is that the length of each A_{i,j,k} is less than \displaystyle \frac{1}{2^2}.

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Remark

In the process described above, the endpoints of the intervals A_{f(0),\cdots,f(n)} are rational numbers and we make sure that all the rational numbers are used as endpoints. We also make sure that the intervals from the successive steps are nested closed intervals with lengths \rightarrow 0. The consequence of this point is that the nested decreasing closed intervals will collapse to one single point (since the real line is a complete metric space) and this single point must be an irrational number (since all the rational numbers are used up as endpoints of the nested closed intervals).

In Step i where i is an odd integer, we make the endpoints of the new intervals converge to the right. In Step i where i>1 is an even integer, we make the endpoints of the new intervals converge to the left. This manipulation is to ensure that the nested closed intervals will never share the same endpoint from one step all the way to the end of the process.

Thus if we have f \in \omega^\omega, then \bigcap \limits_{n=0}^\infty A_{f(0),\cdots,f(n)} \ne \varnothing and in fact has only one point that is an irrational number.

On the other hand, for each point x \in \mathbb{P}, we can locate inductively a sequence of intervals, A_{f(0)}, A_{f(0),f(1)}, A_{f(0),f(1),f(2)}, \cdots, containing the point x. This sequence of nested closed intervals must collapse to a single point and the single point must be the irrational number x.

The process described above gives us a one-to-one mapping from \mathbb{P} onto the product space \omega^\omega. This mapping is also continuous in both directions, making it a homeomorphism. the nested intervals defined above form a base for the Euclidean topology on \mathbb{P}. These basic open sets have a natural correspondance with basic open sets in the product space \omega^\omega.

For example, for f \in \omega^\omega, \left\{ A_{f(0),\cdots,f(n)} \cap \mathbb{P}: n \in \omega \right\} is a local base at the point x \in \bigcap \limits_{n=0}^\infty A_{f(0),\cdots,f(n)}. One the other hand, each A_{f(0),\cdots,f(n)} \cap \mathbb{P} has a natural counterpart in a basic open set in the product space, namely the following set:

\displaystyle . \ \ \ \ \left\{g \in \omega^\omega: g \restriction n = f \restriction n \right\}

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The above process establishes that the countable product of the integers, \omega^\omega, is equivalent topologically to the Euclidean space \mathbb{P}.

Elementary Examples of Lindelof Spaces and Separable Spaces

The Euclidean spaces \mathbb{R} and \mathbb{R}^n are both Lindelof and separable. In fact these two properties are equivalent in the class of metrizable spaces. A space is metrizable if its topology can be induced by a metric. In a metrizable space, having one of these properties implies the other one. Any students in beginning topology courses who study basic notions such as the Lindelof property and separability must venture outside the confine of Euclidean spaces or metric spaces. The goal of this post is to present some elementary examples showing that these two notions are not equivalent.

All topological spaces under consideration are Hausdorff. Let X be a space. Let D \subset X. The set D is said to be dense in X if every nonempty open subset of X contains some point of D. The space X is said to be separable if there is countable subset of X that is also dense in X. All Euclidean spaces are separable. For example, in the real line \mathbb{R}, every open interval contains a rational number. Thus the set of all rational numbers \mathbb{Q} is dense in \mathbb{R}.

Let \mathcal{U} be a collection of subsets of the space X. The collection \mathcal{U} is said to be a cover of X if every point of X is contained in some element of \mathcal{U}. The collection \mathcal{U} is said to be an open cover of X if, in addition it being a cover, \mathcal{U} consists of open sets in X.

Let \mathcal{U} be a cover of the space X. Let \mathcal{V} \subset \mathcal{U}. If the collection \mathcal{V} is also a cover of X, we say that \mathcal{V} is a subcover of \mathcal{U}. The space X is a Lindelof space (or has the Lindelof property) if every open cover of X has a countable subcover.

The real \mathbb{R} is Lindelof. Both the Lindelof property and the separability of \mathbb{R} follows from the fact that the Euclidean topology on \mathbb{R} can be generated by a countable base (e.g. one countable base consists of all open intervals with rational endpoints). Now some non-Euclidean (and non-metrizable) examples.

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Example 1 – A Lindelof space that is not separable
Let X be any uncountable set. Let p be a point that is not in X, e.g., let p=\left\{ X \right\}. Define the space Y = \left\{p\right\} \cup X as follows. Let every point in X be isolated, meaning any singleton set \left\{ x \right\} is declared open for any x \in X. An open neighborhood of the point p is of the form \left\{p\right\} \cup W where X-W is a countable subset of X.

It is clear that the resulting space Y is Lindelof since every open set containing p contains all but countably many points of X. It is also clear that no countable set can be dense in Y.

Even though this example Y is Lindelof, it is not hereditarily Lindelof since the subspace X is uncountable discrete space.

In a previous post, we showed that the space Y defined in this example is a productively Lindelof space (meaning that its product with every Lindelof space is Lindelof).

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Remark

A space is said to have the countable chain condition (CCC) if there are no uncountable family of pairwise disjoint open subsets. It is clear that any separable space has the CCC. It follows that the space Y in Example 1 does not have the CCC, since the singleton sets \left\{ x \right\} (with x \in X) forms a pairwise disjoint collection of open sets, showing that the Lindelof property does not even imply the weaker property of having the CCC.

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Example 2 – A separable space that is not Lindelof
The example here is the Tangent Disc Space (Niemytzki’s Tangent Disc Topology in [2]). The underlying set is the upper half plane (the x-axis and the plane above the x-axis). In other words, consider the following set:

\displaystyle . \ \ \ \ \ X=\left\{(x,y) \in \mathbb{R}^2: y \ge 0 \right\}

Let \displaystyle X_u=\left\{(x,y) \in \mathbb{R}^2: y>0 \right\} and T=\left\{(x,0): x \in \mathbb{R} \right\}. The line T is the x-axis and X_u is the upper plane without the x-axis. We define a topology on X such that X_u as a subspace in this topology is Euclidean. The open neighborhoods of a point p=(x,0) \in T are of the form \left\{p \right\} \cup D where D is an open disc tangent to the x-axis at the point p. The figure below illustrates how open neighborhoods at the x-axis are defined.

It is clear that the points with rational coordinates in the upper half plane X_u form a dense set in the tangent disc topology. Thus X is separable. In any Lindelof space, there are no uncountable closed and discrete subsets. Note that the x-axis T is a closed and discrete subspace in the tangent disc space. Thus X is not Lindelof.

Though separable, the Tangent Disc Space is not hereditarily separable since the x-axis T is uncountable and discrete.

The Tangent Disc Space is an interesting example. For example, it is a completely regular space that is an example of a Moore space that is not normal. For these and other interesting facts about the Tangent Disc Space, see [2].

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For the Lindelof property and the property of being separable, there are plenty of examples of spaces that possess only one of the properties. All three references indicated below are excellent places to look. The book by Steen and Seebach ([2]) is an excellent catalog of interesting spaces (many of them are elementary).
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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
  2. Steen, L. A., Seebach, J. A.,Counterexamples in Topology, 1995, Dover Edition, Dover Publications, New York.
  3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

The product of first countable spaces

All spaces under consideration are Hausdorff. First countable spaces are those spaces where there is a countable local base at every point in the space. This is quite a strong property. For example, every first countable space that is also compact has a cap on its cardinality and the cap is the cardinality of the real line (the continuum). See The cardinality of compact first countable spaces, I in this blog. In fact, if the compact and first countable space is uncountable, it has cardinality continuum (see The cardinality of compact first countable spaces, III). Any metric space (or metrizable space) is first countable. In this post, we discuss the product of first countable spaces. In this regard, first countable spaces and metrizable spaces behave similarly. We show that the product of countably many first countable spaces is first countable while the product of uncountably many first countable is not first countable. For more information on the product topology, see [2].

The Product Space
Consider a collection of sets A_\alpha where \alpha \in S. Let W=\bigcup \limits_{\alpha \in S} A_\alpha. The product \prod \limits_{\alpha \in S} A_\alpha is the set of all functions f:S \mapsto W such that for each \alpha \in S, f(\alpha) \in A_\alpha. If the index set S=\left\{1,2,\cdots,n\right\} is finite, the functions f can be regarded as sequences (f_1,f_2,\cdots,f_n) where each f_i \in A_i. If the index set S=\mathbb{N}, we can think of elements f of the product as the sequence (f_1,f_2,\cdots) where each f_i \in A_i. In general we can regard f \in \prod \limits_{\alpha \in S} A_\alpha as functions f:S \mapsto W or as sequences f=(f_\alpha)_{\alpha \in S}.

Consider the topological spaces X_\alpha where \alpha \in S. Let X=\prod \limits_{\alpha \in S} X_\alpha be the product as defined above. The product space of the spaces X_\alpha is X with the topology defined in the following paragraph.

Let \tau_\alpha be the topology of each space X_\alpha, \alpha \in S. Consider Y=\prod \limits_{\alpha \in S} O_\alpha where for each \alpha \in S, O_\alpha \in \tau_\alpha (i.e. O_\alpha is open in X_\alpha) and O_\alpha=X_\alpha for all but finitely many \alpha \in S. The set of all such sets Y is a base for a topology on the product X=\prod \limits_{\alpha \in S} X_\alpha. This topology is called the product topology of the spaces X_\alpha, \alpha \in S.

To more effectively work with product spaces, we consider a couple of equivalent bases that we can define for the product topology. Let \mathcal{B}_\alpha be a base for the space X_\alpha. Consider B=\prod \limits_{\alpha \in S} B_\alpha such that there is a finite set F \subset S where B_\alpha \in \mathcal{B}_\alpha for each \alpha \in F and B_\alpha=X_\alpha for all \alpha \in S-F. The set of all such sets B is an equivalent base for the product topology.

Another equivalent base is defined using the projection maps. For each \alpha \in S, consider the map \pi_\alpha:\prod \limits_{\beta \in S} X_\beta \mapsto X_\alpha such that \pi_\alpha(f)=f_\alpha for each f in the product. In words, the function \pi_\alpha maps each point in the product space to its \alpha^{th} coordinate. The mapping \pi_\alpha is called the \alpha^{th} projection map. For each set U \subset X_\alpha, \pi_\alpha^{-1}(U) is the following set:

\pi_\alpha^{-1}(U)=\left\{f \in \prod \limits_{\beta \in S} X_\beta: \pi_\alpha(f)=f_\alpha \in U\right\}

Consider sets of the form \bigcap \limits_{\alpha \in F} \pi_\alpha^{-1}(U_\alpha) where F \subset S is finite and U_\alpha is open in X_\alpha for each \alpha \in F. The set of all such sets is another equivalent base for the product topology. If we only require that each U_\alpha \in \mathcal{B}_\alpha, a predetermined base for the coordinate space X_\alpha, we also obtain an equivalent base for the product topology.

Countable Product of First Countable Spaces
For i=1,2,3,\cdots, let X_i be a first countable space. We show that X=\prod \limits_{i=1}^{\infty}X_i is a first countable space.

Let \mathbb{N} be the set of positive integers. For each n \in \mathbb{N}, let [n]=\left\{1,2,\cdots,n\right\} and let \mathbb{N}^{[n]} be the set of all functions t:[n] \mapsto \mathbb{N}. For each i and each x \in X_i, let \mathcal{B}_x(i)=\left\{B_x(i,j): j \in \mathbb{N}\right\} be a countable local base at x.

Let f \in X=\prod \limits_{i=1}^{\infty}X_i. We wish to define a countable local base at f. For each n \in \mathbb{N}, define W_n to be:

W_n=\left\{\prod \limits_{i=1}^n B_{f(i)}(i,t(i)):t \in \mathbb{N}^{[n]}\right\}

Let W be the set of all subsets of the product space X of the following form:

\prod \limits_{j=1}^{\infty}V_j where there is some n \in \mathbb{N} such that \prod \limits_{j=1}^{n}V_j \in W_n and for all j>n, V_j=X_j.

Each W_n is countable and W is essentially the union of all the W_n. Thus W is countable. We claim that W is a local base at f. Let O \subset X be an open set containing f. We can assume that O=\prod \limits_{i=1}^\infty O_i where there is some n \in \mathbb{N} such that for each i \le n, O_i is open in X_i and for i>n, O_i=X_i.

For each i \le n, f(i) \in O_i. Choose some B_{f(i)}(i,t(i)) such that f(i) \in B_{f(i)}(i,t(i)) \subset O_i. Let V=\prod \limits_{i=1}^{\infty} V_i such that \prod \limits_{i=1}^{n} V_i=\prod \limits_{i=1}^n B_{f(i)}(i,t(i)) and V_i=X_i for all i>n. Then V \in W and f \in V \subset O. This completes the proof that X=\prod \limits_{i=1}^{\infty}X_i is a first countable space.

Uncountable Product
Let S be an uncountable index set. For \alpha \in S, let X_\alpha. We want to avoid the situation that all but countably many X_\alpha are one-point space. So we assume each coordinate space X_\alpha has at least two points, say, p_\alpha and q_\alpha with p_\alpha \ne q_\alpha. We show that X=\prod \limits_{\alpha \in S}X_\alpha is not first countable.

Let f \in \prod \limits_{\alpha \in S}X_\alpha. Let U_1,U_2, \cdots be open subsets of the product space such that for each i, f \in U_i. We show that there is some open set O such that f \in O and each U_i \nsubseteq O. For each i, there is a basic open set B_i=\bigcap \limits_{\alpha \in F_i} \pi_\alpha^{-1}(U_{\alpha,i}) such that f \in B_i \subset U_i.

Let F=F_1 \cup F_2 \cup \cdots. Since S is uncountable and F is countable, choose \gamma \in S-F. Since X_\gamma has at least two points p_\gamma and q_\gamma, choose one of them that is different from f_\gamma, say, p_\gamma. Choose two disjoint open subsets M_1 and M_2 of X_\gamma such that f_\gamma \in M_1 and p_\gamma \in M_2 of X_\gamma. Let O=\prod \limits_{\alpha \in S}O_\alpha such that O_\gamma=M_1 and O_\alpha=X_\alpha for all \alpha \ne \gamma. We have f \in O. For each i, there is g_i \in B_i \subset U_i such that g_i(\gamma)=p_\gamma. Thus each g_i \notin O. Thus there is no countable local base at f. Thus any product space with uncountably many factors, each of which has at least two points, is never first countable.

Reference

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