A closed subset $B$ of the Hilbert cube such that $\operatorname{Int}(B) = \emptyset$ and $B$ is not a z-set

Question

$\def\cube{\mathbf Q}$Let $\cube$ be the Hilbert cube. Give an example of a closed subset $B$ of $\cube$ such that $\operatorname{Int}(B) = \emptyset$ and $B$ is not a z-set.

If anyone has any idea to solve this problem, it would help me a lot. I write down some information:

Definition. Let $\cube= \prod_{n=1}^{\infty} [0,1]$ with product topology), with metric $d$ such that if $(x_n)_n, (y_n)_n \in \cube$, $d((x_n)_n,(y_n)_n) = \sum_{n=1}^{\infty} \frac{\lvert x_n - y_n\rvert}{2^n}$. Then $B \subseteq \cube$ is a z-set if, for each $\epsilon > 0$, there is a continuous function $f_\epsilon$ from $\cube$ to $\cube \setminus B$ such that $d(f_\epsilon(x),x) < \epsilon$ for all $x \in X$.

I have tried the following, where a $\sigma$z-set is a countable union of z-sets. (In $\cube$, I think that all $\sigma$z-sets are z-sets.)

Lemma. If $B \subseteq \cube$ is a $\sigma$z-set, then $\cube \setminus B$ is a topologically complete, separable metric AR (Absolute Retract).

Note. In $\cube$, $B'$ being a retract is equivalent to $B'$ being AR.

Then, I have tried to find an example of a $B \subseteq \cube$ such that $\cube \setminus B$ is not a retract. By the lemma, this $B$ would not be a $\sigma$z-set. Then check if this set $B$ has empty interior in $\cube$ and is closed. I could not find such a $B$.

References.

[1] Alejandro Wanes, Sam Nadler Jr. HYPERSPACES: Fundamental and Recent Advances.

[2] Macías - Una introducción a los retractos absolutos y a los retractos de vecindad absolutos.

Answer 1

$A=\{0\} \times [-1,1]^{\Bbb N}$ works as an example, e.g.

For more info on $Z$-sets in $Q$, see Infinite-dimensional Topology by Jan van Mill.