Looking for a closed and discrete subspace of a product space

I had long suspected that there probably is an uncountable closed and discrete subset of the product space of uncountably many copies of the real line. Then I found the statement in the Encyclopedia of General Topology (page 76 in [3]) that “for every infinite cardinal \mathcal{K}, the product D(\mathcal{K})^{2^{\mathcal{K}}} includes D(2^{\mathcal{K}}) as a closed subspace” where D(\tau) is the discrete space of cardinality of \tau. When \mathcal{K}=\aleph_0 (the first infinite cardinal), there is a closed and discrete subset of cardinality c=2^{\aleph_0} in the product space \mathbb{N}^{c} (the product space of continuum many copies of a countable discrete space). Despite the fact that this product space is a separable space, a closed and discrete set of cardinality continuum is hiding in the product space \mathbb{N}^{c}. What is more amazing is that this result gives us a glimpse into the working of the product topology with uncountably many factors. There are easily defined discrete subspaces of \mathbb{N}^{c}. But these discrete subspaces are not closed in the product space, making the result indicated here a remarkable one.

The Encyclopedia of General Topology points to two references [2] and [4]. I could not find these papers online. It turns out that Engelking, the author of [2], included this fact as an exercise in his general topology textbook (see Exercise 3.1.H (a) in [1]). This post presents a proof of this fact based on the hints that are given in [1]. To make the argument easier to follow, the proof uses some of the hints in a slightly different form.

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The Exercise

Let I=[0,1] be the closed unit interval. Let X=I be the unit interval with the discrete topology. Let \omega be the set of all nonnegative integers with the discrete topology. Let Y=\prod_{t \in I} Y_t where each Y_t=\omega. We can also denote Y by \omega^I. The problem is to show that the discrete space X can be embedded as a closed and discrete subspace of Y.

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A Solution

For each t \in I, choose a sequence O_{t,1},O_{t,2},O_{t,3},\cdots of open intervals (in the usual topology of I) such that

  • t \in O_{t,j} for each j,
  • \overline{O_{t,j+1}} \subset O_{t,j} for each j (the closure is in the usual topology of I),
  • \left\{t \right\}=\bigcap \limits_{j=1}^\infty O_{t,j}.

For any t \in I-\left\{0,1 \right\}, we can make O_{t,j} open intervals of the form (a,b). For t=0, O_{0,j} have the form [0,b). For t=1, O_{1,j} have the form (a,1].

The above sequences of open intervals help define a homeomorphic embedding of the discrete space X into the product Y. For each t \in I, define the function f_t:I \rightarrow \omega by letting:

    f_t(x) = \begin{cases} 0 & \mbox{if } x=t \\ 1 & \mbox{if } x \in I-O_{t,1} \\ 2 & \mbox{if } x \in I-O_{t,2} \text{ and } x \in O_{t,1} \\ 3 & \mbox{if } x \in I-O_{t,3} \text{ and } x \in O_{t,2} \\ \cdots \\ j & \mbox{if } x \in I-O_{t,j} \text{ and } x \in O_{t,j-1} \\ \text{etc} \end{cases}

for each x \in X. We now define the embedding E:X \rightarrow Y=\omega^I by letting:

    E(x)=< f_t(x) >_{t \in I}

for each x \in X. For each point x \in X, E(x) is the point in the product space such that the t coordinate of E(x) is the value of the function f_t evaluated at x. Let \mathcal{F}=\left\{f_t: t \in I \right\}.

The mapping E is an evaluation map (called diagonal map in [1]). It is a homeomorphism if the following three conditions are met:

  • If each f_t \in \mathcal{F} is continuous, then E is continuous.
  • If the family of functions \mathcal{F} separates points in X, then E is injective (i.e. a one-to-one function).
  • If \mathcal{F} separates points from closed sets in X, then the inverse E^{-1} is also continuous.

The first point is easily seen. Note that both the domain and the range of f_t:X \rightarrow \omega have the discrete topology. Thus these functions are continuous. To see the second point, let p,q \in X with p \ne q. Note that f_p(p)=0 while f_p(q) \ne 0.

Since X is discrete, any subset of X is closed. To see the third point, let C \subset X and p \notin C. Once again, f_p(p)=0 while f_p(x) \ne 0 for all x \in C. Clearly f_p(p) \notin \overline{f(C)} (closure in the discrete space \omega). Thus E^{-1} is also continuous. For more details about why E is an embedding, see the previous post called The Evaluation Map.

Let W=E(X). Since E is a homeomorphism, W is a discrete subspace of the product space Y. We only need to show W is closed in Y. Let k \in Y such that k=< k_t >_{t \in I} \ \notin W. We show that there is an open neighborhood of k that misses the set W. There are two cases to consider. One is that k_r \ne 0 for all r \in I. The other is that k_r=0 for some r \in I. The first case is more involved.

Case 1
Suppose that k_r \ne 0 for all r \in I. First define a local base \mathcal{B}_{k} of the point k. Let H \subset I be finite and let G_{H} be defined by:

    G_{H}=\left\{b \in Y: \forall \ h \in H, b_h=k_h \right\}

Let \mathcal{B}_{k} be the set of all possible G_{H}. For an arbitrary G_{H}, consider E^{-1}(G_H). By definition, E^{-1}(G_H)=\left\{c \in I: E(c) \in G_H \right\}. To make the argument below easier to see, let’s further describe E^{-1}(G_H).

    \displaystyle \begin{aligned} E^{-1}(G_H)&=\left\{c \in I: E(c) \in G_H \right\} \\&=\left\{c \in I: \forall \ h \in H, f_h(c)=k_h \ne 0 \right\} \\&=\left\{c \in I: \forall \ h \in H, c \in I-O_{h,k_h} \right\} \\&=\bigcap \limits_{h \in H} I-O_{h,k_h} \\&=I-\bigcup \limits_{h \in H} O_{h,k_h} \end{aligned}

The last description above indicates that if x \in X and if x \in E^{-1}(G_H), then x \notin \bigcup \limits_{h \in H} O_{h,k_h}. To wrap up Case 1, we would like to produce one particular H. Consider the open cover \left\{O_{t,k_t}: t \in I  \right\} of I. Since I is compact in the usual topology, there is a finite H \subset I such that I=\bigcup \limits_{h \in H} O_{h,k_h}. This means that for this particular finite set H, E^{-1}(G_H)=\varnothing. Putting it in another way, the open neighborhood B_H of k contains no point of E(x). The proof for Case 1 is completed.

Case 2
Suppose that k_r=0 for some r \in I. Since k \notin W=E(X), in particular k \ne E(r)=< f_t(r) >_{t \in I}. So for some q \in I, k_q \ne f_q(r). Now define the following open set containing k.

    G=\left\{b \in Y: b_r=0 \text{ and } b_q=k_q \right\}

Note that E(r) \notin G since k_q \ne f_q(r). Furthermore, for each p \in I-\left\{r \right\}, E(p) \notin G since f_r(p) \ne 0. Thus G is an open neighborhood of k containing no point of E(X). The proof for Case 2 is completed.

We have shown that the image of the discrete space X under the homeomorphism E is closed in the product space Y=\prod_{t \in I} Y_t=\omega^I.

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Comments

The preceding proof shows that the product space of continuum many copies of \omega contains a closed and discrete subspace of cardinality continuum. This is a remarkable result. At a glance, it is not entirely clear that a closed and discrete set of this large size can be found in the product space in question.

One immediate consequence is that the product space Y=\prod_{t \in I} Y_t=\omega^I is not normal since it is a separable space. By Jones’ lemma, any separable normal space cannot have a closed and discrete subset of cardinality continuum. However, if the goal is only to show non-normality, we only need to show that \omega^{\omega_1} is not normal (a proof is found in this post). Thus the value of the preceding proof is to demonstrate how to produce a closed and discrete subspace of cardinality continuum in the product space in question.

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Engelking, R., On the double circumference of Alexandroff, Bull. Acad. Polon. Sci., 16, 629-634, 1968.
  3. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
  4. Juhasz, I., On closed discrete subspace of product spaces, Bull. Acad. Polon. Sci., 17, 219-223, 1969.

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\copyright \ 2014 \text{ by Dan Ma}

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