# Miscenko example of linearly Lindelof non Lindelof is not normal

In the paper of Norman Howes "A note on transfinite sequences" is mentioned that Miscenko space

$M = \{f \in \prod_{n \in \omega \smallsetminus \{0\}} \aleph_n+1 | \space \exists k \space \space \forall m \geq k \space f(m) < \aleph_m\}$

is not normal. The closed sets that support this are:

$H = \{f \in M | \space \forall k \space f(k) \neq 0 \}$.

$K = \{f \in M | \space \exists k \space f(k+1)=0 \space \space and \space \forall m \leq k \space f(m) = \aleph_m\}$

How to prove that for any open $U$, $V$ such that $H \subset U$ and $K \subset V$, $U \cap V \neq \emptyset$ ?

Assume $U$ and $V$ are open in $M$ and $H \subseteq V$ and $K \subseteq U$, such that $U \cap V = \emptyset$.
Lemma: There are $\{D_n : n >0\}$ such that $D_n \subseteq U$ and $D_n = \prod_{k\leq n}(\alpha_{k}^{n}, \aleph_k] \times \prod_{k>n}\{0\}$ and $\prod_{k \leq m} (\alpha_k^{n}, \aleph_k] \subseteq \prod_{k \leq m} (\alpha_k^m, \aleph_k]$ for $m < n$.
Proof: By induction on n>0. If n=1 then define $g^1$ as following: $g^1(1)=\aleph_1$ and $g^1(k)=0$ for $k>1$. Then $g^1 \in K \subseteq U$. Hence there is $\alpha_1^1<\aleph_1$ such that $(\alpha_1^1, \aleph_1] \times \prod_{k>1} \{0\} \subseteq U$. Assume $\{D_k : k \leq n\}$ are defined. We should define $D_{n+1}$. Define $g^{n+1}$ as following: $g^{n+1}(k) = \aleph_k$ for $k \leq n+1$ and $g^{n+1}(k)=0$ for $k>n+1$. Then $g^{n+1} \in K \subseteq U$. Then for $k \leq n$ there are $\alpha^{n+1}_k$ such that $\alpha^{n+1}_k > \alpha^n_k$ and there is $\alpha_{n+1}^{n+1} < \aleph_{n+1}$ such that $\prod_{k \leq n+1} (\alpha_k^{n+1}, \aleph_k] \times \prod_{k>n+1} \{0\} \subseteq U$. $\square$
Select $y \in H$ as following: $\aleph_k > y(k) > \sup \{\alpha_k^n : n \geq k\}$. It is possible since $cf(\aleph_k) > \aleph_0$ for $k>0$. $y \in H \subseteq V$ therefore there is $W \subseteq V$ such that $y \in W$ and $W = (\prod_{k \leq n} (\beta_k, y(k)] \times \prod_{k>n} \aleph_k +1) \cap M$. Then $W\cap D_n \neq \emptyset$. So $U \cap V \neq \emptyset$.