Lindelof Exercise 1

A space X is called a \sigma-compact space if it is the union of countably many compact subspaces. Clearly, any \sigma-compact space is Lindelof. It is well known that the product of Lindelof spaces does not need to be Lindelof. The most well known example is perhaps the square of the Sorgenfrey line. In certain cases, the Lindelof property can be productive. For example, the product of countably many \sigma-compact spaces is a Lindelof space. The discussion here centers on the following theorem.

Theorem 1
Let X_1,X_2,X_3,\cdots be \sigma-compact spaces. Then the product space \prod_{i=1}^\infty X_i is Lindelof.

Theorem 1 is Exercise 3.8G in page 195 of General Topology by Engelking [1]. The reference for Exercise 3.8G is [2]. But the theorem is not found in [2] (it is not stated directly and it does not seem to be an obvious corollary of a theorem discussed in that paper). However, a hint is provided in Engelking for Exercise 3.8G. In this post, we discuss Theorem 1 as an exercise by giving expanded hint. Solutions to some of the key steps in the expanded hint are given at the end of the post.

Expanded Hint

It is helpful to first prove the following theorem.

Theorem 2
For each integer i \ge 1, let C_{i,1},C_{i,2},\cdots be compact spaces and let C_i be the topological sum:

    C_i=C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots=\oplus_{j=1}^\infty C_{i,j}

Then the product \prod_{i=1}^\infty C_i is Lindelof.

Note that in the topological sum C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots, the spaces C_{i,1},C_{i,2},C_{i,3},\cdots are considered pairwise disjoint. The open sets in the sum are simply unions of the open sets in the individual spaces. Another way to view this topology: each of the C_{i,j} is both closed and open in the topological sum. Theorem 2 is essentially saying that the product of countably many \sigma-compact spaces is Lindelof if each \sigma-compact space is the union of countably many disjoint compact spaces. The hint for Exercise 3.8G can be applied much more naturally on Theorem 2 than on Theorem 1. The following is Exercise 3.8F (a), which is the hint for Exercise 3.8G.

Lemma 3
Let Z be a compact space. Let X be a subspace of Z. Suppose that there exist F_1,F_2,F_3,\cdots, closed subsets of Z, such that for all x and y where x \in X and y \in Z-X, there exists F_i such that x \in F_i and y \notin F_i. Then X is a Lindelof space.

The following theorem connects the hint (Lemma 3) with Theorem 2.

Theorem 4
For each integer i \ge 1, let Z_i be the one-point compactification of C_i in Theorem 2. Then the product Z=\prod_{i=1}^\infty Z_i is a compact space. Furthermore, X=\prod_{i=1}^\infty C_i is a subspace of Z. Prove that Z and X satisfy Lemma 3.

Each C_i in Theorem 2 is a locally compact space. To define the one-point compactifications, for each i, choose p_i \notin C_i. Make sure that p_i \ne p_j for i \ne j. Then Z_i is simply

    Z_i=C_i \cup \{ p_i \}=C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots \cup \{ p_i \}

with the topology defined as follows:

  • Open subsets of C_i continue to be open in Z_i.
  • An open set containing p_i is of the form \{ p_i \} \cup (C_i - \overline{D}) where D is open in C_i and D is contained in the union of finitely many C_{i,j}.

For convenience, each point p_i is called a point at infinity.

Note that Theorem 2 follows from Lemma 3 and Theorem 4. In order to establish Theorem 1 from Theorem 2, observe that the Lindelof property is preserved by any continuous mapping and that there is a natural continuous map from the product space in Theorem 2 to the product space in Theorem 1.

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Proofs of Key Steps

Proof of Lemma 3
Let Z, X and F_1,F_2,F_3,\cdots be as described in the statement for Lemma 3. Let \mathcal{U} be a collection of open subsets of Z such that \mathcal{U} covers X. We would like to show that a countable subcollection of \mathcal{U} is also a cover of X. Let O=\cup \mathcal{U}. If Z-O=\varnothing, then \mathcal{U} is an open cover of Z and there is a finite subset of \mathcal{U} that is a cover of Z and thus a cover of X. Thus we can assume that Z-O \ne \varnothing.

Let F=\{ F_1,F_2,F_3,\cdots \}. Let K=Z-O, which is compact. We make the following claim.

Claim. Let Y be the union of all possible \cap G where G \subset F is finite and \cap G \subset O. Then X \subset Y \subset O.

To establish the claim, let x \in X. For each y \in K=Z-O, there exists F_{n(y)} such that x \in F_{n(y)} and y \notin F_{n(y)}. This means that \{ Z-F_{n(y)}: y \in K \} is an open cover of K. By the compactness of K, there are finitely many F_{n(y_1)}, \cdots, F_{n(y_k)} such that F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} misses K, or equivalently F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} \subset O. Note that x \in F_{n(y_1)} \cap \cdots \cap F_{n(y_k)}. Further note that F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} \subset Y. This establishes the claim that X \subset Y. The claim that Y \subset O is clear from the definition of Y.

Each set F_i is compact since it is closed in Z. The intersection of finitely many F_i is also compact. Thus the \cap G in the definition of Y in the above claim is compact. There can be only countably many \cap G in the definition of Y. Thus Y is a \sigma-compact space that is covered by the open cover \mathcal{U}. Choose a countable \mathcal{V} \subset \mathcal{U} such that \mathcal{V} covers Y. Then \mathcal{V} is a cover of X too. This completes the proof that X is Lindelof.

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Proof of Theorem 4
Recall that Z=\prod_{i=1}^\infty Z_i and that X=\prod_{i=1}^\infty C_i. Each Z_i is the one-point compactification of C_i, which is the topological sum of the disjoint compact spaces C_{i,1},C_{i,2},\cdots.

For integers i,j \ge 1, define K_{i,j}=C_{i,1} \oplus C_{i,2} \oplus \cdots \oplus C_{i,j}. For integers n,j \ge 1, define the product F_{n,j} as follows:

    F_{n,j}=K_{1,j} \times \cdots \times K_{n,j} \times Z_{n+1} \times Z_{n+2} \times \cdots

Since F_{n,j} is a product of compact spaces, F_{n,j} is compact and thus closed in Z. There are only countably many F_{n,j}.

We claim that the countably many F_{n,j} have the property indicated in Lemma 3. To this end, let f \in X=\prod_{i=1}^\infty C_i and g \in Z-X. There exists an integer n \ge 1 such that g(n) \notin C_{n}. This means that g(n) \notin C_{n,j} for all j, i.e. g(n)=p_n (so g(n) must be the point at infinity). Choose j \ge 1 large enough such that

    f(i) \in K_{i,j}=C_{i,1} \oplus C_{i,2} \oplus \cdots \oplus C_{i,j}

for all i \le n. It follows that f \in F_{n,j} and g \notin F_{n,j}. Thus the sequence of closed sets F_{n,j} satisfies Lemma 3. By Lemma 3, X=\prod_{i=1}^\infty C_i is Lindelof.

Reference

  1. Engelking R., General Topology, Revised and Completed edition, Elsevier Science Publishers B. V., Heldermann Verlag, Berlin, 1989.
  2. Hager A. W., Approximation of real continuous functions on Lindelof spaces, Proc. Amer. Math. Soc., 22, 156-163, 1969.

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Helly Space

This is a discussion on a compact space called Helly space. The discussion here builds on the facts presented in Counterexample in Topology [2]. Helly space is Example 107 in [2]. The space is named after Eduard Helly.

Let I=[0,1] be the closed unit interval with the usual topology. Let C be the set of all functions f:I \rightarrow I. The set C is endowed with the product space topology. The usual product space notation is I^I or \prod_{t \in I} W_t where each W_t=I. As a product of compact spaces, C=I^I is compact.

Any function f:I \rightarrow I is said to be increasing if f(x) \le f(y) for all x<y (such a function is usually referred to as non-decreasing). Helly space is the subspace X consisting of all increasing functions. This space is Example 107 in Counterexample in Topology [2]. The following facts are discussed in [2].

  • The space X is compact.
  • The space X is first countable (having a countable base at each point).
  • The space X is separable.
  • The space X has an uncountable discrete subspace.

From the last two facts, Helly space is a compact non-metrizable space. Any separable metric space would have countable spread (all discrete subspaces must be countable).

The compactness of X stems from the fact that X is a closed subspace of the compact space C.

Further Discussion

Additional facts of concerning Helly space are discussed.

  1. The product space \omega_1 \times X is normal.
  2. Helly space X contains a copy of the Sorgenfrey line.
  3. Helly space X is not hereditarily normal.

The space \omega_1 is the space of all countable ordinals with the order topology. Recall C is the product space I^I. The product space \omega_1 \times C is Example 106 in [2]. This product is not normal. The non-normality of \omega_1 \times C is based on this theorem: for any compact space Y, the product \omega_1 \times Y is normal if and only if the compact space Y is countably tight. The compact product space C is not countably tight (discussed here). Thus \omega_1 \times C is not normal. However, the product \omega_1 \times X is normal since Helly space X is first countable.

To see that X contains a copy of the Sorgenfrey line, consider the functions h_t:I \rightarrow I defined as follows:

    \displaystyle  h_t(x) = \left\{ \begin{array}{ll}           \displaystyle  0 &\ \ \ \ \ \ 0 \le x \le t \\            \text{ } & \text{ } \\          \displaystyle  1 &\ \ \ \ \ \ t<x \le 1 \\                                 \end{array} \right.

for all 0<t<1. Let S=\{ h_t: 0<t<1 \}. Consider the mapping \gamma: (0,1) \rightarrow S defined by \gamma(t)=h_t. With the domain (0,1) having the Sorgenfrey topology and with the range S being a subspace of Helly space, it can be shown that \gamma is a homeomorphism.

With the Sorgenfrey line S embedded in X, the square X \times X contains a copy of the Sorgenfrey plane S \times S, which is non-normal (discussed here). Thus the square of Helly space is not hereditarily normal. A more interesting fact is that Helly space is not hereditarily normal. This is discussed in the next section.

Finding a Non-Normal Subspace of Helly Space

As before, C is the product space I^I where I=[0,1] and X is Helly space consisting of all increasing functions in C. Consider the following two subspaces of X.

    Y_{0,1}=\{ f \in X: f(I) \subset \{0, 1 \} \}

    Y=X - Y_{0,1}

The subspace Y_{0,1} is a closed subset of X, hence compact. We claim that subspace Y is separable and has a closed and discrete subset of cardinality continuum. This means that the subspace Y is not a normal space.

First, we define a discrete subspace. For each x with 0<x<1, define f_x: I \rightarrow I as follows:

    \displaystyle  f_x(y) = \left\{ \begin{array}{ll}           \displaystyle  0 &\ \ \ \ \ \ 0 \le y < x \\           \text{ } & \text{ } \\          \displaystyle  \frac{1}{2} &\ \ \ \ \ y=x \\            \text{ } & \text{ } \\          \displaystyle  1 &\ \ \ \ \ \ x<y \le 1 \\                                 \end{array} \right.

Let H=\{ f_x: 0<x<1 \}. The set H as a subspace of X is discrete. Of course it is not discrete in X since X is compact. In fact, for any f \in Y_{0,1}, f \in \overline{H} (closure taken in X). However, it can be shown that H is closed and discrete as a subset of Y.

We now construct a countable dense subset of Y. To this end, let \mathcal{B} be a countable base for the usual topology on the unit interval I=[0,1]. For example, we can let \mathcal{B} be the set of all open intervals with rational endpoints. Furthermore, let A be a countable dense subset of the open interval (0,1) (in the usual topology). For convenience, we enumerate the elements of A and \mathcal{B}.

    A=\{ a_1,a_2,a_3,\cdots \}

    \mathcal{B}=\{B_1,B_2,B_3,\cdots \}

We also need the following collections.

    \mathcal{G}=\{G \subset \mathcal{B}: G \text{ is finite and is pairwise disjoint} \}

    \mathcal{A}=\{F \subset A: F \text{ is finite} \}

For each G \in \mathcal{G} and for each F \in \mathcal{A} with \lvert G \lvert=\lvert F \lvert=n, we would like to arrange the elements in increasing order, notated as follow:

    F=\{t_1,t_2,\cdots,t_n \}

    G=\{E_1,E_2,\cdots,E_n \}

For the set F, we have 0<t_1<t_2< \cdots <t_n<1. For the set G, E_i is to the left of E_j for i<j. Note that elements of G are pairwise disjoint. Furthermore, write E_i=(p_i,q_i). If 0 \in E_1, then E_1=[p_1,q_1)=[0,q_1). If 1 \in E_n, then E_n=(p_n,q_n]=(p_n,1].

For each F and G as detailed above, we define a function L(F,G):I \rightarrow I as follows:

    \displaystyle  L(F,G)(x) = \left\{ \begin{array}{ll}                     \displaystyle  t_1 &\ \ \ \ \ 0 \le x < q_1 \\           \text{ } & \text{ } \\          \displaystyle  t_2 &\ \ \ \ \ q_1 \le x < q_2 \\           \text{ } & \text{ } \\          \displaystyle  \vdots &\ \ \ \ \ \vdots \\           \text{ } & \text{ } \\          \displaystyle  t_{n-1} &\ \ \ \ \ q_{n-2} \le x < q_{n-1} \\           \text{ } & \text{ } \\          \displaystyle  t_n &\ \ \ \ \ q_{n-1} \le x \le 1 \\                                             \end{array} \right.

The following diagram illustrates the definition of L(F,G) when both F and G have 4 elements.

Figure 1 – Member of a countable dense set

Let D be the set of L(F,G) over all F \in \mathcal{A} and G \in \mathcal{G}. The set D is a countable set. It can be shown that D is dense in the subspace Y. In fact D is dense in the entire Helly space X.

To summarize, the subspace Y is separable and has a closed and discrete subset of cardinality continuum. This means that Y is not normal. Hence Helly space X is not hereditarily normal. According to Jones’ lemma, in any normal separable space, the cardinality of any closed and discrete subspace must be less than continuum (discussed here).

Remarks

The preceding discussion shows that both Helly space and the square of Helly space are not hereditarily normal. This is actually not surprising. According to a theorem of Katetov, for any compact non-metrizable space V, the cube V^3 is not hereditarily normal (see Theorem 3 in this post). Thus a non-normal subspace is found in V, V \times V or V \times V \times V. In fact, for any compact non-metric space V, an excellent exercise is to find where a non-normal subspace can be found. Is it in V, the square of V or the cube of V? In the case of Helly space X, a non-normal subspace can be found in X.

A natural question is: is there a compact non-metric space V such that both V and V \times V are hereditarily normal and V \times V \times V is not hereditarily normal? In other words, is there an example where the hereditarily normality fails at dimension 3? If we do not assume extra set-theoretic axioms beyond ZFC, any compact non-metric space V is likely to fail hereditarily normality in either V or V \times V. See here for a discussion of this set-theoretic question.

Reference

  1. Kelly, J. L., General Topology, Springer-Verlag, New York, 1955.
  2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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Dan Ma topology

Daniel Ma topology

Dan Ma math

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\copyright 2019 – Dan Ma