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 $K \subset V$, $U \cap V \neq \emptyset$ ?

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 $K \subset V$, $U \cap V \neq \emptyset$ ?