Products of compact spaces with countable tightness

In the previous two posts, we discuss the definitions of the notion of tightness and its relation with free sequences. This post and the next post are to discuss the behavior of countable tightness under the product operation. In this post, we show that countable tightness behaves well in products of compact space. In particular we show that countable tightness is preserved in finite products and countable products of compact spaces. In the next post we show that countable tightness is easily destroyed in products of sequential fans and that the tightness of such a product can be dependent on extra set theory assumptions. All spaces are Hausdorff and regular.

The following theorems are the main results in this post.

Theorem 1
Let X and Y be countably tight spaces. If one of X and Y is compact, then X \times Y is countably tight.

Theorem 2
The product of finitely many compact countably tight spaces is countably tight.

Theorem 3
Suppose that X_1, X_2, X_3, \cdots are countably many compact spaces such that each X_i has at least two points. If each X_i is a countably tight space, then the product space \prod_{i=1}^\infty X_i is countably tight.

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Finite products

Before proving Theorem 1 and Theorem 2, we prove the following results.

Theorem 4
Let f:Y_1 \rightarrow Y_2 be a continuous and closed map from the space Y_1 onto the space Y_2. Suppose that the space Y_2 is countably tight and that each fiber of the map f is countably tight. Then the space Y_1 is countably tight.

Proof of Theorem 4
Let x \in Y_1 and x \in \overline{A} where A \subset Y_1. We proceed to find a countable W \subset Y_1 such that x \in \overline{W}. Choose y \in Y_2 such that y=f(x).

Let M be the fiber of the map f at the point y, i.e. M=f^{-1}(y). By assumption, M is countably tight. Call a point w \in M countably reached by A if there is some countable C \subset A such that w \in \overline{C}. Let G be the set of all points in M that are countably reached by A. We claim that x \in \overline{G}.

Let U \subset Y_1 be open such that x \in U. Because the space Y_1 is regular, choose open V \subset U such that x \in V and \overline{V} \subset U. Then V \cap A \ne \varnothing. Furthermore, x \in \overline{V \cap A}. Let C=f(V \cap A). By the continuity of f, we have y \in \overline{C}. Since Y_2 is countably tight, there exists some countable D \subset C such that y \in \overline{D}. Choose a countable E \subset V \cap A such that f(E)=D. It follows that y \in \overline{f(E)}.

We show that that \overline{E} \cap M \ne \varnothing. Since E \subset \overline{E}, we have f(E) \subset f(\overline{E}). Note that f(\overline{E}) is a closed set since f is a closed map. Thus \overline{f(E)} \subset f(\overline{E}). As a result, y \in f(\overline{E}). Then y=f(t) for some t \in \overline{E}. We have t \in \overline{E} \cap M.

By the definition of the set G, we have \overline{E} \cap M \subset G. Furthermore, \overline{E} \cap M \subset \overline{V} \subset U. Note that the arbitrary open neighborhood U of x contains points of G. This establishes the claim that x \in \overline{G}.

Since M is a fiber of f, M is countably tight by assumption. Choose some countable T \subset G such that x \in \overline{T}. For each t \in T, choose a countable W_t \subset A with t \in \overline{W_t}. Let W=\bigcup_{t \in T} W_t. Note that W \subset A and W is countable with x \in \overline{W}. This establishes the space Y_1 is countably tight at x \in Y_1. \blacksquare

Lemma 5
Let f:X \times Y \rightarrow Y be the projection map. If X is a compact space, then f is a closed map.

Proof of Lemma 5
Let A be a closed subset of X \times Y. Suppose that f(A) is not closed. Let y \in \overline{f(A)}-f(A). It follows that no point of X \times \left\{y \right\} belongs to A. For each x \in X, choose open subset O_x of X \times Y such that (x,y) \in O_x and O_x \cap A=\varnothing. The set of all O_x is an open cover of the compact space X \times \left\{y \right\}. Then there exist finitely many O_x that cover X \times \left\{y \right\}, say O_{x_i} for i=1,2,\cdots,n.

Let W=\bigcup_{i=1}^n O_{x_i}. We have X \times \left\{y \right\} \subset W. Since X is compact, we can then use the Tube Lemma which implies that there exists open G \subset Y such that X \times \left\{y \right\} \subset X \times G \subset W. It follows that G \cap f(A) \ne \varnothing. Choose t \in G \cap f(A). Then for some x \in X, (x,t) \in A. Since t \in G, (x,t) \in W, implying that W \cap A \ne \varnothing, a contradiction. Thus f(A) must be a closed set in Y. This completes the proof of the lemma. \blacksquare

Proof of Theorem 1
Let X be the factor that is compact. let f: X \times Y \rightarrow Y be the projection map. The projection map is always continuous. Furthermore it is a closed map by Lemma 5. The range space Y is countably tight by assumption. Each fiber of the projection map f is of the form X \times \left\{y \right\} where y \in Y, which is countably tight. Then use Theorem 4 to establish that X \times Y is countably tight. \blacksquare

Proof of Theorem 2
This is a corollary of Theorem 1. According to Theorem 1, the product of two compact countably tight spaces is countably tight. By induction, the product of any finite number of compact countably tight spaces is countably tight. \blacksquare

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Countable products

Our proof to establish that the product space \prod_{i=1}^\infty X_i is countably tight is an indirect one and makes use of two non-trivial results. We first show that \omega_1 \times \prod_{i=1}^\infty X_i is a closed subspace of a \Sigma-product that is normal. It follows from another result that the second factor \prod_{i=1}^\infty X_i is countably tight. We now present all the necessary definitions and theorems.

Consider a product space Y=\prod_{\alpha<\kappa} Y_\alpha where \kappa is an infinite cardinal number. Fix a point p \in Y. The \Sigma-product of the spaces Y_\alpha with p as the base point is the following subspace of the product space Y=\prod_{\alpha<\kappa} Y_\alpha:

    \displaystyle \Sigma_{\alpha<\kappa} Y_\alpha=\left\{y \in \prod_{\alpha<\kappa} Y_\alpha: y_\alpha \ne p_\alpha \text{ for at most countably many } \alpha < \kappa \right\}

The definition of the space \Sigma_{\alpha<\kappa} Y_\alpha depends on the base point p. The discussion here is on properties of \Sigma_{\alpha<\kappa} Y_\alpha that hold regardless of the choice of base point. If the factor spaces are indexed by a set A, the notation is \Sigma_{\alpha \in A} Y_\alpha.

If all factors Y_\alpha are identical, say Y_\alpha=Z for all \alpha, then we use the notation \Sigma_{\alpha<\kappa} Z to denote the \Sigma-product. Once useful fact is that if there are \omega_1 many factors and each factor has at least 2 points, then the space \omega_1 can be embedded as a closed subspace of the \Sigma-product.

Theorem 6
For each \alpha<\omega_1, let Y_\alpha be a space with at least two points. Then \Sigma_{\alpha<\omega_1} Y_\alpha contains \omega_1 as a closed subspace. See Exercise 3 in this previous post.

Now we discuss normality of \Sigma-products. This previous post shows that if each factor is a separable metric space, then the \Sigma-product is normal. It is also well known that if each factor is a metric space, the \Sigma-product is normal. The following theorem handles the case where each factor is a compact space.

Theorem 7
For each \alpha<\kappa, let Y_\alpha be a compact space. Then the \Sigma-product \Sigma_{\alpha<\kappa} Y_\alpha is normal if and only if each factor Y_\alpha is countably tight.

Theorem 7 is Theorem 7.5 in page 821 of [1]. Theorem 7.5 in [1] is stated in a more general setting where each factor of the \Sigma-product is a paracompact p-space. We will not go into a discussion of p-space. It suffices to know that any compact Hausdorff space is a paracompact p-space. We also need the following theorem, which is proved in this previous post.

Theorem 8
Let Y be a compact space. Then the product space \omega_1 \times Y is normal if and only if Y is countably tight.

We now prove Theorem 3.

Proof of Theorem 3
Let \omega_1=\cup \left\{A_n: n \in \omega \right\}, where for each n, \lvert A_n \lvert=\omega_1 and that A_n \cap A_m=\varnothing if n \ne m. For each n=1,2,3,\cdots, let S_n=\Sigma_{\alpha \in A_n} X_n. By Theorem 7, each S_n is normal. Let S_0=\Sigma_{\alpha \in A_0} X_1, which is also normal. By Theorem 6, the space \omega_1 of countable ordinals is a closed subspace of S_0. Let T=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots. We have the following derivation.

    \displaystyle \begin{aligned} T&=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots \\&\subset S_0 \times S_1 \times S_2 \times S_3 \times \cdots \\&\cong W=\Sigma_{\alpha<\omega_1} W_\alpha \end{aligned}

Recall that \omega_1=\cup \left\{A_n: n \in \omega \right\}. The space W=\Sigma_{\alpha<\omega_1} W_\alpha is defined such that for each n \ge 1 and for each \alpha \in A_n, W_\alpha=X_n. Furthermore, for n=0, for each \alpha \in A_0, let W_\alpha=X_1. Thus W is a \Sigma-product of compact countably tight spaces and is thus normal by Theorem 7. The space T=\omega_1 \times \prod_{n=1}^\infty X_n is a closed subspace of the normal space W. By Theorem 8, the product space \prod_{n=1}^\infty X_n must be countably tight. \blacksquare

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Remarks

Theorem 2, as indicated above, is a corollary of Theorem 1. We also note that Theorem 2 is also a corollary of Theorem 3 since any finite product is a subspace of a countable product. To see this, let X=X_1 \times X_2 \times \cdots \times X_n.

    \displaystyle \begin{aligned} X&=X_1 \times X_2 \times \cdots \times X_n \\&\cong X_1 \times X_2 \times \cdots \times X_n \times \left\{t_{n+1} \right\} \times \left\{t_{n+2} \right\} \times \cdots \\&\subset  X_1 \times X_2 \times \cdots \times X_n \times X_{n+1}  \times X_{n+2} \times \cdots  \end{aligned}

In the above derivation, t_m is a point of X_m for all m >n. When the countable product space is countably tight, the finite product, being a subspace of a countably tight space, is also countably tight.

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Exercise

Exercise 1
Let f:X \times Y \rightarrow Y be the projection map. If X is a countably compact space and Y is a Frechet space, then f is a closed map.

Exercise 2
Let X and Y be countably tight spaces. If one of X and Y is a countably compact space and the other space is a Frechet space, then X \times Y is countably tight.

Exercise 2 is a variation of Theorem 1. One factor is weakened to “countably compact”. However, the other factor is strengthened to “Frechet”.

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Reference

  1. Przymusinski, T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.

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\copyright \ 2015 \text{ by Dan Ma}

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