# 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 (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

for all $0. Let $S=\{ h_t: 0. 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, 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

Let $H=\{ f_x: 0. 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. For the set $G$, $E_i$ is to the left of $E_j$ for $i. 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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