Recently, I read a paper in which the author construct a space $X$ which is dense in a $\sigma$-product $S$ of closed unit intervals. The space $X$ is CCC (denotes countable chain condition); it is also, clearly Tychonoff (see its Theorem). But i want to verify it is not normal. How could I see quickly that this space is not normal?
Thanks very much.
Theorem. A dense subspace $X$ of the Tychonoff cube $[0,1]^\kappa$ is not normal if $X$ contains an uncountable closed discrete subspace $D$ such that each $x\in D$ has finite support $\mathrm{supp}(x)=\{i\in\kappa:x(i)\ne0\}$.
Proof. By the $\Delta$-lemma, the uncountable set $D$ contains an uncountable subset $D'\subset D$ such that $\{\mathrm{supp}(x):x\in D'\}$ is a $\Delta$-system, which means that for some finite set $F$ we get $F=\mathrm{supp}(x)\cap\mathrm{supp}(y)$ for any distinct points $x,y\in D'$.
Let $\mathrm{pr}_F:[0,1]^\kappa\to[0,1]^F$ be the coordinate projection. Let $\mathcal U$ be the family of all open subsets of $[0,1]^F$ such that the preimage $D'\cap \mathrm{pr}_F^{-1}(U)$ is countable. Since the subspace $\bigcup\mathcal U$ is Lindel\"of, the union $\bigcup\mathcal U$ belongs to the family $\mathcal U$. It follows that $K=[0,1]^F\setminus \bigcup\mathcal U$ is a non-empty closed subset of $[0,1]^F$ such that for any non-empty open subspace $W\subset K$ the set $D'\cap \mathrm{pr}_F^{-1}(W)$ is uncountable.
Replacing the set $D$ by $D'\cap \mathrm{pr}_F^{-1}(K)$, we can assume that $\mathrm{pr}_F(D)\subset K$ and for every non-empty open set $W\subset K$ the set $D\cap \mathrm{pr}_F^{-1}(W)$ is uncountable.
Since the space $K$ has countable base, we can choose a countable set $A\subset D$ such that for any non-empty open set $W\subset K$ the set $A\cap \mathrm{pr}_F^{-1}(W)$ is infinite. Next, take any uncountable subset $B\subset D\setminus A$.
We claim that the closed subsets $A,B$ of the space $X$ cannot be separated by disjoint open sets (so $X$ is not normal). To derive a contradiction, assume that $A\subset U_A$ and $B\subset U_B$ for some disjoint open sets $U_A,U_B\subset X$.
For a function $x\in\mathbb R^\kappa$ and a finite subset $E\subset \kappa$ put $$\|x\|_E=\max_{i\in E}|x(i)|.$$
For every $a\in A$ find a finite set $F_a\subset\kappa$ and a number $\varepsilon_a>0$ such that $X\cap O_a\subset U_A$ where $$O_a:=\{x\in [0,1]^\kappa:\|x-a\|_{F_a}<\varepsilon_a\}.$$ Replacing $F_a$ by a larger finite set, we can assume that $F\subset\mathrm{supp}(a)\subset F_a$ for all $a\in A$.
By analogy, for every $b\in B$ find a finite set $F_b\subset\kappa$ and $\varepsilon_b>0$ such that $F\subset\mathrm{supp}(b)\subset F_b$ and $X\cap O_b\subset U_B$ where $$O_b:=\{x\in [0,1]^\kappa:\|x-b\|_{F_b}<\varepsilon_b\}.$$
Fix a countable base $\mathcal B$ of the topology of the space $[0,1]^F$. By the Pigeonhole principle, there exists $W\in\mathcal B$ such that the set $$B'=\big\{b\in B:b|F\in W\subset \{x\in[0,1]^F:\|x-b\|_F<\varepsilon_b\}\big\}$$ is uncountable. By the $\Delta$-lemma, there exists an uncountable subset $B''\subset B'$ and a finite set $F'\subset\kappa$ such that $F_b\cap F_{b'}=F'$ for any distinct elements $b,b'\in B''$.
By the choice of $A$, the set $A'=\{a\in A:\mathrm{pr}_F(a)\in W\}$ is infinite. Since $(\mathrm{supp}(a))_{a\in A'}$ is a $\Delta$-system with kernel $F$, the set $\{a\in A':\mathrm{supp}(a)\cap (F'\setminus F)\ne\emptyset\}$ is finite. Consequently, there exists $a\in A'$ such that $\mathrm{supp}(a)\cap F'\subset F$. Since $(F_b)_{b\in B''}$ is a $\Delta$-system with kernel $F'$ and $(\mathrm{supp}(b))_{b\in B''}$ is a $\Delta$-system with kernel $F$, there exists $b\in B''$ such that $F_b\cap F_a\subset F'$ and $\mathrm{supp}(b)\cap F_a\subset F$. Observe that $$\mathrm{supp}(a)\cap F_b\subset \mathrm{supp}(a)\cap F_a\cap F_b\subset \mathrm{supp}(a)\cap F'\subset F.$$ Take any function $x\in[0,1]^\kappa$ such that $x|F_a=a|F_a$ and $x|F_b\setminus F_a=b|F_b\setminus F_a$. We claim that $x\in O_a\cap O_b$. The inclusion $x\in O_a$ follows from $x|F_a=a|F_a$.
To see that $x\in O_b$, we need to check that $|x(i)-b(i)|<\varepsilon_b$ for every $i\in F_b$. For $i\in F$ this inequality follows from $x|F=a|F\in W$ and the choice of $b\in B''$. For $i\in F_b\setminus F_a$ we get $x(i)=b(i)$ and hence $|x(i)-b(i)|=0<\varepsilon_b$. Finally, for $i\in F_b\cap F_a\setminus F$, we get $x(i)=a(i)=0=b(i)$, which follows from $\mathrm{supp}(a)\cap F_b\subset F$ and $\mathrm{supp}(b)\cap F_a\subset F$.
So, $x\in O_a\cap O_b$. Since the space $X$ is dense in $[0,1]^\kappa$, the intersection $X\cap O_a\cap O_b\subset U_A\cap U_B$ is not empty, which is a desired contradiction.
