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The space is this:

Let $D=\{0,1\}$, and let $Y=D^\mathfrak{c}$. For $y\in Y\;$ let $\operatorname{supp}(y)=\{\xi<\mathfrak{c}:y(\xi)=1\}$, the support of $y$, and let $X=\{x\in Y:0<|\operatorname{supp}(x)|\le\omega_1\}$.

PS: We know $\Sigma$-product of metric spaces is normal, see here.

The question is: is this space normal?

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The link doesn't work for me. –  Cihan Jul 3 '12 at 11:24

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$X$ is not normal.

Basic open sets in $X$ are of the form

$$B(F_0,F_1)=\{x\in X:x(\xi)=0\text{ for }\xi\in F_0\text{ and }x(\xi)=1\text{ for }\xi\in F_1\}\;,$$

where $F_0$ and $F_1$ are disjoint finite sets of indices; let $\mathscr{B}$ be the set of such basic open sets.

Assume first that $2^\omega=\omega_1$, i.e., that CH holds. Let the index set for the product $D^\mathfrak{c}$ be $\omega_1\cdot2$. For $\alpha\le\omega_1$ and $\beta<\omega_1$ define the point $x_{\alpha,\beta}\in X$ as follows:

$$x_{\alpha,\beta}(\xi)=\begin{cases} 0,&\text{if }\xi<\alpha\text{ or }\omega_1\le\xi<\omega_1+\beta\\ 1,&\text{if }\alpha\le\xi<\omega_1\text{ or }\xi\ge\omega_1+\beta\;. \end{cases}$$

Let $Z$ be the set of such points $x_{\alpha,\beta}$. If $x\in X\setminus Z$, then either there are $\alpha<\beta<\omega_1$ such that $x(\alpha)=1$ and $x(\beta)=0$, or there are $\alpha<\beta$ such that $\omega_1\le\alpha$, $x(\alpha)=1$, and $x(\beta)=0$; in either case $B\big(\{\beta\},\{\alpha\}\big)$ is a nbhd of $x$ disjoint from $Z$, so $Z$ is closed in $X$.

Let $H=\{x_{\alpha,\alpha}:\alpha<\omega_1\}$ and $K=\{x_{\omega_1,\alpha}:\alpha<\omega_1\}$; clearly $H$ and $K$ are disjoint, and I claim that they are closed in $Z$ (and hence in $X$).

If $x\in X\setminus H$, at least one of the following must be the case:

  1. there is an $\alpha<\omega_1$ such that $x(\alpha)=0$ and $x(\omega_1+\alpha)=1$;
  2. there is an $\alpha<\omega_1$ such that $x(\alpha)=1$ and $x(\omega_1+\alpha)=0$;

In case (1) $B\big(\{\alpha\},\{\omega_1+\alpha\}\big)$ is a nbhd of $x$ disjoint from $H$; in case (2), $B\big(\{\omega_1+\alpha\},\{\alpha\}\big)$ is.

If $x\in Z\setminus K$, there is an $\alpha<\omega_1$ such that $x(\alpha)=1$, in which case $B\big(\varnothing,\{\alpha\}\big)$ is a nbhd of $x$ disjoint from $K$.

Now suppose that $H\subseteq U$ and $K\subseteq V$, where $U$ and $V$ are disjoint open sets in $X$. For each $x_{\alpha,\alpha}\in H$ there is a $B(F_\alpha,G_\alpha)\in\mathscr{B}$ such that $x_{\alpha,\alpha}\in B(F_\alpha,G_\alpha)\subseteq U$. Now

$$x_{\alpha,\alpha}(\xi)=\begin{cases} 0,&\text{if }\xi<\alpha\\ 1,&\text{if }\alpha\le\xi<\omega_1\\ 0,&\text{if }\omega_1\le\xi<\omega_1+\alpha\\ 1,&\text{if }\xi\ge\omega_1+\alpha\;, \end{cases}$$

so without loss of generality we may assume that $\alpha,\omega_1+\alpha\in G_\alpha$. For $\alpha\ne 0$ we may further assume that $F_\alpha\cap\alpha\ne\varnothing\ne F_\alpha\setminus\omega_1$, and if $\mu_\alpha=\max(F_\alpha\cap\alpha)$, then $\max(F_\alpha\setminus\omega_1)=\omega_1+\mu_\alpha$. It follows that

$$Z\cap B(F_\alpha,G_\alpha)=\{x_{\beta,\gamma}\in Z:\eta_\alpha<\beta,\gamma\le\alpha\}\;.$$

By the pressing-down lemma there are an $\eta<\omega_1$ and a cofinal subset $S$ of $\omega_1$ such that $\eta_\alpha=\eta$ for each $\alpha\in S$. But then

$$U\supseteq Z\cap\bigcup_{\alpha\in S}B(F_\alpha,G_\alpha)=\{x_{\beta,\gamma}\in Z:\eta<\beta,\gamma<\omega_1\}\;,$$

and it’s easy to see that if $\eta<\alpha<\omega_1$, then $x_{\omega_1,\alpha}\in K\cap\operatorname{cl}U=\varnothing$, an obvious contradiction. It follows that $H$ and $K$ cannot be separated by disjoint open sets and hence that $X$ is not normal.

This is essentially just the argument that $(\omega_1+1)\times\omega_1$ isn’t normal, because $\{\langle\alpha,\alpha\rangle:\alpha<\omega_1\}$ and $\{\omega_1\}\times\omega_1$ can’t be separated by disjoint open sets. It works because there is no point $x_{\omega_1,\omega_1}$: it would have empty support.

If $2^\omega>\omega_1$, only minor changes are necessary. First, we use $2^\omega$ as the index set for the product $D^\mathfrak{c}$. Then we define $x_{\alpha,\beta}$ by

$$x_{\alpha,\beta}(\xi)=\begin{cases} 0,&\text{if }\xi<\alpha\omega_1\le\xi<\omega_1+\beta\\ 1,&\text{if }\alpha\le\xi<\omega_1\\ 0,&\text{if }\omega_1\le\xi<\omega_1+\beta\\ 1,&\text{if }\omega_1+\beta\le\xi<\omega_1\cdot2\\ 0,&\text{if }\xi\ge\omega_1\cdot2\;. \end{cases}$$

$Z,H$, and $K$ are as before; the only change is that a point of $x\in X$ might fail to be in $Z$ because $x(\xi)=1$ for some $\xi\ge\omega_1\cdot2$, in which case $B\big(\varnothing,\{\xi\}\big)$ is a nbhd of $x$ disjoint from $Z$.

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@Brain M Scott can u prove that (w1+1)x w1 is not normal –  math Feb 21 '13 at 16:21
    
@math: Let $\Delta=\{\langle\xi,\xi\rangle:\xi<\omega_1\}$. If $U$ is an open nbhd of $\Delta$, then for each $\xi<\omega_1$ there is an $\eta_\xi<\xi$ such that $(\eta_\xi,\xi]\times(\eta_\xi,\xi]\subseteq U$. By the pressing-down lemma there are $A\subseteq\omega_1$ and $\eta<\omega_1$ such that $|A|=\omega_1$ and $\eta_xi=\eta$ for each $\xi\in A$. It follows that $U\supseteq(\eta,\omega_1)\times(\eta,\omega_1)$. Thus, for any $\alpha\in(\eta,\omega_1)$ every open nbhd of $\langle\omega_1,\alpha\rangle$ meets $U$. This shows that the closed sets ... –  Brian M. Scott Feb 21 '13 at 16:28
    
... $\Delta$ and $\{\omega_1\}\times\omega_1$ cannot be separated by disjoint open sets. –  Brian M. Scott Feb 21 '13 at 16:28
    
@Brain M Scott do u mean Δ={⟨ξ,ξ⟩:ξ<ω 1 } is closed set –  math Feb 21 '13 at 16:40
    
@Brain M Scott I mean is granted or we have to prove it –  math Feb 21 '13 at 16:47

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