In this and subsequent posts, we consider where is a compact space. Recall that is the space of all continuous real-valued functions defined on and that it is endowed with the pointwise convergence topology. One of the compact spaces we consider is , the first compact uncountable ordinal. There are many interesting results about the function space . In this post we show that is not normal. An even more interesting fact about is that does not have any dense normal subspace .
Let be the first uncountable ordinal, and let be the successor ordinal to . The set is the first uncountable ordinal. Furthermore consider these ordinals as topological spaces endowed with the order topology. As mentioned above, the space is the first compact uncountable ordinal. In proving that is not normal, a theorem that is due to D. P. Baturov is utilized . This theorem is also proved in this previous post.
For the basic working of function spaces with the pointwise convergence topology, see the post called Working with the function space Cp(X).
The fact that is not normal is established by the following two points.
- If is normal, then has countable extent, i.e. every closed and discrete subspace of is countable.
- There exists an uncountable closed and discrete subspace of .
We discuss each of the bullet points separately.
The function space is a dense subspace of , the product of many copies of . According to a result of D. P. Baturov , any dense normal subspace of the product of many separable metric spaces has countable extent (also see Theorem 1a in this previous post). Thus cannot be normal if the second bullet point above is established.
Now we show that there exists an uncountable closed and discrete subspace of . For each with , define by:
Clearly, for each . Let . We show that is a closed and discrete subspace of . The fact that is closed in is establish by the following claim.
Let . There exists an open subset of such that and .
First we get some easy cases out of the way. Suppose that there exists some such that . Then let . Clearly and .
Another easy case: If for all , then consider the open set where . Clearly and .
From now on we can assume that and that is not identically the zero function. Suppose Claim 1 is not true. Then . Next observe the following:
If for some , then for all .
To see this, if , and , then define the open set by . Note that and , contradicting that . So the above observation is valid.
Now either or . We claim that is not possible. Suppose that . Let . Then and , contradicting that . It must be the case that .
Because of the continuity of at the point , of all the for which , there is the largest one, say . Now . According to the observation made above, for all . This means that . This is a contradiction since . Thus Claim 1 must be true and the fact that is closed is established.
Next we show that is discrete in . Fix where . Let . It is clear that . Furthermore, for all and for all . Thus is open such that . This completes the proof that is discrete.
We have established that is an uncountable closed and discrete subspace of . This implies that is not normal.
The set as defined above is closed and discrete in . However, the set is not discrete in a larger subspace of the product space. The set is also a subset of the following -product:
Because is the -product of separable metric spaces, it is normal (see here). By Theorem 1a in this previous post, would have countable extent. Thus the set cannot be closed and discrete in . We can actually see this directly. Let be a limit ordinal. Define by for all and for all . Clearly and . Furthermore, (the closure is taken in ).
- Arhangel’skii, A. V., Normality and Dense Subspaces, Proc. Amer. Math. Soc., 48, no. 2, 283-291, 2001.
- Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.