# Cp(omega 1 + 1) is not normal

In this and subsequent posts, we consider $C_p(X)$ where $X$ is a compact space. Recall that $C_p(X)$ is the space of all continuous real-valued functions defined on $X$ and that it is endowed with the pointwise convergence topology. One of the compact spaces we consider is $\omega_1+1$, the first compact uncountable ordinal. There are many interesting results about the function space $C_p(\omega_1+1)$. In this post we show that $C_p(\omega_1+1)$ is not normal. An even more interesting fact about $C_p(\omega_1+1)$ is that $C_p(\omega_1+1)$ does not have any dense normal subspace [1].

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$. The set $\omega_1$ is the first uncountable ordinal. Furthermore consider these ordinals as topological spaces endowed with the order topology. As mentioned above, the space $\omega_1+1$ is the first compact uncountable ordinal. In proving that $C_p(\omega_1+1)$ is not normal, a theorem that is due to D. P. Baturov is utilized [2]. 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 $C_p(\omega_1+1)$ is not normal is established by the following two points.

• If $C_p(\omega_1+1)$ is normal, then $C_p(\omega_1+1)$ has countable extent, i.e. every closed and discrete subspace of $C_p(\omega_1+1)$ is countable.
• There exists an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$.

We discuss each of the bullet points separately.

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The function space $C_p(\omega_1+1)$ is a dense subspace of $\mathbb{R}^{\omega_1}$, the product of $\omega_1$ many copies of $\mathbb{R}$. According to a result of D. P. Baturov [2], any dense normal subspace of the product of $\omega_1$ many separable metric spaces has countable extent (also see Theorem 1a in this previous post). Thus $C_p(\omega_1+1)$ cannot be normal if the second bullet point above is established.

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Now we show that there exists an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$. For each $\alpha$ with $0<\alpha<\omega_1$, define $h_\alpha:\omega_1 + 1 \rightarrow \left\{0,1 \right\}$ by:

$h_\alpha(\gamma) = \begin{cases} 1 & \mbox{if } \gamma \le \alpha \\ 0 & \mbox{if } \alpha<\gamma \le \omega_1 \end{cases}$

Clearly, $h_\alpha \in C_p(\omega_1 +1)$ for each $\alpha$. Let $H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\}$. We show that $H$ is a closed and discrete subspace of $C_p(\omega_1 +1)$. The fact that $H$ is closed in $C_p(\omega_1 +1)$ is establish by the following claim.

Claim 1
Let $h \in C_p(\omega_1 +1) \backslash H$. There exists an open subset $U$ of $C_p(\omega_1 +1)$ such that $h \in U$ and $U \cap H=\varnothing$.

First we get some easy cases out of the way. Suppose that there exists some $\alpha<\omega_1$ such that $h(\alpha) \notin \left\{0,1 \right\}$. Then let $U=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in \mathbb{R} \backslash \left\{0,1 \right\} \right\}$. Clearly $h \in U$ and $U \cap H=\varnothing$.

Another easy case: If $h(\alpha)=0$ for all $\alpha \le \omega_1$, then consider the open set $U$ where $U=\left\{f \in C_p(\omega_1 +1): f(0) \in \mathbb{R} \backslash \left\{1 \right\} \right\}$. Clearly $h \in U$ and $U \cap H=\varnothing$.

From now on we can assume that $h(\omega_1+1) \subset \left\{0,1 \right\}$ and that $h$ is not identically the zero function. Suppose Claim 1 is not true. Then $h \in \overline{H}$. Next observe the following:

Observation.
If $h(\beta)=1$ for some $\beta \le \omega_1$, then $h(\alpha)=1$ for all $\alpha \le \beta$.

To see this, if $h(\alpha)=0$, $h(\beta)=1$ and $\alpha<\beta$, then define the open set $V$ by $V=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in (-0.1,0.1) \text{ and } f(\beta) \in (0.9,1.1) \right\}$. Note that $h \in V$ and $V \cap H=\varnothing$, contradicting that $h \in \overline{H}$. So the above observation is valid.

Now either $h(\omega_1)=1$ or $h(\omega_1)=0$. We claim that $h(\omega_1)=1$ is not possible. Suppose that $h(\omega_1)=1$. Let $V=\left\{f \in C_p(\omega_1 +1): f(\omega_1) \in (0.9,1.1) \right\}$. Then $h \in V$ and $V \cap H=\varnothing$, contradicting that $h \in \overline{H}$. It must be the case that $h(\omega_1)=0$.

Because of the continuity of $h$ at the point $x=\omega_1$, of all the $\gamma<\omega_1$ for which $h(\gamma)=1$, there is the largest one, say $\beta$. Now $h(\beta)=1$. According to the observation made above, $h(\alpha)=1$ for all $\alpha \le \beta$. This means that $h=h_\beta$. This is a contradiction since $h \notin H$. Thus Claim 1 must be true and the fact that $H$ is closed is established.

Next we show that $H$ is discrete in $C_p(\omega_1 +1)$. Fix $h_\alpha$ where $0<\alpha<\omega_1$. Let $W=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in (0.9,1.1) \text{ and } f(\alpha+1) \in (-0.1,0.1) \right\}$. It is clear that $h_\alpha \in W$. Furthermore, $h_\gamma \notin W$ for all $\alpha < \gamma$ and $h_\gamma \notin W$ for all $\gamma <\alpha$. Thus $W$ is open such that $\left\{h_\alpha \right\}=W \cap H$. This completes the proof that $H$ is discrete.

We have established that $H$ is an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$. This implies that $C_p(\omega_1 +1)$ is not normal.

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Remarks

The set $H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\}$ as defined above is closed and discrete in $C_p(\omega_1 +1)$. However, the set $H$ is not discrete in a larger subspace of the product space. The set $H$ is also a subset of the following $\Sigma$-product:

$\Sigma(\omega_1)=\left\{x \in \mathbb{R}^{\omega_1}: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \omega_1 \right\}$

Because $\Sigma(\omega_1)$ is the $\Sigma$-product of separable metric spaces, it is normal (see here). By Theorem 1a in this previous post, $\Sigma(\omega_1)$ would have countable extent. Thus the set $H$ cannot be closed and discrete in $\Sigma(\omega_1)$. We can actually see this directly. Let $\alpha<\omega_1$ be a limit ordinal. Define $t:\omega_1 + 1 \rightarrow \left\{0,1 \right\}$ by $t(\beta)=1$ for all $\beta<\alpha$ and $t(\beta)=0$ for all $\beta \ge \alpha$. Clearly $t \notin C_p(\omega_1 +1)$ and $t \in \Sigma(\omega_1)$. Furthermore, $t \in \overline{H}$ (the closure is taken in $\Sigma(\omega_1)$).

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Reference

1. Arhangel’skii, A. V., Normality and Dense Subspaces, Proc. Amer. Math. Soc., 48, no. 2, 283-291, 2001.
2. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.

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$\copyright \ 2014 \text{ by Dan Ma}$