Is the product space $[0,1]^\kappa$ hereditarily Lindelöf? Topological space $X$ is called Lindelöf is every open cover of $X$ has a countable subcover, moreover it is hereditarily Lindelöf if it is Lindelöf and so is ever of its subspaces. 
I'm trying to see whether the product space $[0,1]^\kappa$ ($[0,1]$ real unit interval with the standard topology and $\kappa$ arbitrary cardinal) is hereditarily Lindelöf.
In Engelking's General topology book there is a characterization saying:
$X$ Lidelöf is hereditarily Lindelöf if and only if $X$ is perfectly normal, 
where $X$ is perfectly normal e.g. if every two disjoint closed subsets $E,F$ of $X$ are precisely separated by continuous map $f$ from $X$ to the real line $R$, i.e. $E,F$ are $f$-preimages of $\{1\}$, resp. $\{0\}$.
So there is an equivalent question whether the space  $[0,1]^\kappa$ is perfectly normal.
 A: Certainly not. By Tychonoff's theorem, every completely regular space of weight $\kappa$ embeds into $[0,1]^\kappa$. So the answer is strongly negative. 
Actually, every perfectly normal compact Hausdorff space has cardinality at most $\mathfrak{c}$.
A: If $\kappa\le\omega$, the space is metrizable and hence perfectly normal.
If $\kappa>\omega$, no singleton in $[0,1]^\kappa$ is a $G_\delta$ set, so $[0,1]^\kappa$ is not perfectly normal.
In particular, let $z\in[0,1]^\kappa$ be, and for $n\in\omega$ let $U_n$ be an open nbhd of $z$. For each $n\in\omega$ there is a finite $F_n\subseteq\kappa$ such that 
$$\{x\in[0,1]^\kappa:x_\xi=z_\xi\text{ for each }\xi\in F_n\}\subseteq U_n\;.$$
Let $F=\bigcup_{n\in\omega}F_n$; then
$$G=\{x\in[0,1]^\kappa:x_\xi=z_\xi\text{ for each }\xi\in F\}\subseteq\bigcap_{n\in\omega}U_n\;.$$
Clearly, however, $G\ne\{z\}$, since $x\in G$ places no restriction on $x_\xi$ for $\xi\in\kappa\setminus F$.
A: (1).For $i,j\in k$ let $f_I(j)=0 $ if $i\ne j,$ and $f_i(i)=1.$ Then $F=\{f_i:i\in k\}$ is a  subspace of $[0,1]^k$ with $|F|=k.$ 
For $i,j\in k$ let $S_{i,j}=[0,1]$ if $i\ne j$ and $S_{i,i}=(1/2,1].$  Then $\prod_{j\in k}S_{i,j}$ is a nbhd of $f_i$ which is disjoint from $F$ \ $\{f_i\}.$ So  $F$ is discrete. So the Lindelof number of $F$ is $|F|=k.$
(2). For $f\in [0,1]^k$ the set $\{f\}$ is closed. Let $G$ be an open family with $f\in g$ for all $g\in G.$ For each $g\in G$ take $g'=\prod_{i\in k}g'_i \subset g$  where (i) each $g'_i$ is open in $[0,1],$ and  (ii) $f\in g',$ and (iii) $g'_i=[0,1]$ for all but finitely many $i.$  
Then  $f\in\{f|_{k\backslash b}\} \times \prod_{i\in b}[0,1]\subset \cap G$ where $b=k$ \ $\cup_{g\in G}\{i:g'_i\ne [0,1]\}.$  If $|G|<k>\omega$ then $|b|=k$ and $\cap G\ne \{f\}.$
