Let be the first uncountable ordinal. In a previous post called Looking for a closed and discrete subspace of a product space, it was shown that the product space , the product of continuum many copies of the real line , contains a closed and discrete subset of cardinality continuum. This example shows that a product space of uncountably many copies of a “nice” space is “big and wide” enough to hide uncountable closed and discrete sets even when the product space is separable. This post reinforces this same fact by showing that contains a closed and discrete subset of cardinality . It follows that for any uncountable cardinal , the product space contains an uncountable closed and discrete subset, i.e., the product of uncountably many copies of the real line has uncountable extent.
Let be the first infinite ordinal, i.e., the set of all nonnegative integers. Consider , the product of many copies of with the discrete topology. Since is a closed subspace of , it suffices to show that has an uncountable closed and discrete subset.
We now construct an uncountable closed and discrete subset of . Let be an infinite ordinal such that . Let . For each , let . We can also use interval notations: and . Consider as a space with the discrete topology. Then it is clear that is homeomorphic to the product space . Thus the focus is now on finding an uncountable closed and discrete subset of .
One interesting fact about the space is that every function is a pressing down function. That is, for every , for all . Note that is defined on , a closed and unbounded subset of (hence a stationary set). It follows that for each , there is a stationary set and there exists such that for all . This fact is called the pressing down lemma and will be used below. See this post for more information about the pressing down lemma.
For each , let be a one-to-one function. For each , define as follows:
Note that each is a pressing down function. Thus each . Let . Clearly if . Thus has cardinality . We claim that is a closed and discrete subset of . It suffices to show that for each , there exists an open set with such that contains at most one .
Let . As discussed above, there is a stationary set and there exists such that for all . In particular, choose such that . Thus . Let be the open set defined by:
Clearly, . We show that if , then . Suppose . Then . Consider two cases: Case 1: ; Case 2: one of and . The definition of indicates that on the interval . Note that is a one-to-one function. Since , it cannot be that , i.e., Case 1 is not possible. Thus Case 2 holds, say . Then by definition, . Putting everything together, . Thus . This concludes the proof that the set is a closed and discrete subset of .