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 ) that “for every infinite cardinal , the product includes as a closed subspace” where is the discrete space of cardinality of . When (the first infinite cardinal), there is a closed and discrete subset of cardinality in the product space (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 . 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 . 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  and . I could not find these papers online. It turns out that Engelking, the author of , included this fact as an exercise in his general topology textbook (see Exercise 3.1.H (a) in ). This post presents a proof of this fact based on the hints that are given in . To make the argument easier to follow, the proof uses some of the hints in a slightly different form.
Let be the closed unit interval. Let be the unit interval with the discrete topology. Let be the set of all nonnegative integers with the discrete topology. Let where each . We can also denote by . The problem is to show that the discrete space can be embedded as a closed and discrete subspace of .
For each , choose a sequence of open intervals (in the usual topology of ) such that
- for each ,
- for each (the closure is in the usual topology of ),
For any , we can make open intervals of the form . For , have the form . For , have the form .
The above sequences of open intervals help define a homeomorphic embedding of the discrete space into the product . For each , define the function by letting:
for each . We now define the embedding by letting:
for each . For each point , is the point in the product space such that the coordinate of is the value of the function evaluated at . Let .
The mapping is an evaluation map (called diagonal map in ). It is a homeomorphism if the following three conditions are met:
- If each is continuous, then is continuous.
- If the family of functions separates points in , then is injective (i.e. a one-to-one function).
- If separates points from closed sets in , then the inverse is also continuous.
The first point is easily seen. Note that both the domain and the range of have the discrete topology. Thus these functions are continuous. To see the second point, let with . Note that while .
Since is discrete, any subset of is closed. To see the third point, let and . Once again, while for all . Clearly (closure in the discrete space ). Thus is also continuous. For more details about why is an embedding, see the previous post called The Evaluation Map.
Let . Since is a homeomorphism, is a discrete subspace of the product space . We only need to show is closed in . Let such that . We show that there is an open neighborhood of that misses the set . There are two cases to consider. One is that for all . The other is that for some . The first case is more involved.
Suppose that for all . First define a local base of the point . Let be finite and let be defined by:
Let be the set of all possible . For an arbitrary , consider . By definition, . To make the argument below easier to see, let’s further describe .
The last description above indicates that if and if , then . To wrap up Case 1, we would like to produce one particular . Consider the open cover of . Since is compact in the usual topology, there is a finite such that . This means that for this particular finite set , . Putting it in another way, the open neighborhood of contains no point of . The proof for Case 1 is completed.
Suppose that for some . Since , in particular . So for some , . Now define the following open set containing .
Note that since . Furthermore, for each , since . Thus is an open neighborhood of containing no point of . The proof for Case 2 is completed.
We have shown that the image of the discrete space under the homeomorphism is closed in the product space .
The preceding proof shows that the product space of continuum many copies of 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 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 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.
- Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
- Engelking, R., On the double circumference of Alexandroff, Bull. Acad. Polon. Sci., 16, 629-634, 1968.
- Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
- Juhasz, I., On closed discrete subspace of product spaces, Bull. Acad. Polon. Sci., 17, 219-223, 1969.