Be $\mathbb{Q}$ Hilbert cube. Give an example of a closed subset $B$ of $\mathbb{Q}$ such that $Int(B) = \emptyset$ 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:

$\mathbf{Def.}$ Let $\mathbb{Q}$ Hilbert Cube ($\mathbb{Q}= \prod_{n=1}^{\infty} [0,1]$ with product topology) with metrid $d$ such that if $(x_n)_n, (y_n)_n \in \mathbb{Q}$, $d((x_n)_n,(y_n)_n) = \sum_{n=1}^{\infty} \frac{|x_n - y_n|}{2^n}$. $B \subseteq \mathbb{Q}$ is a z-set of $\mathbb{Q}$ if for each $\epsilon > 0$, there is a continuous function, $f_\epsilon$ from $\mathbb{Q}$ to $\mathbb{Q} \setminus B$ such that $d(f_\epsilon(x),x) \ < \epsilon$ for all $x \in X$.

I have tried the following:

$\mathbf{Lemma.}$ If $B \subseteq \mathbb{Q}$ is a $\sigma$z-set ($\sigma$z-set is a countable union of Z-sets. In $\mathbb{Q}$ I think that all $\sigma$z-set is a z-set) then $\mathbb{Q} \setminus B$ is a topologically complete, separable metric AR (Absolute Retract).

$\mathbf{Note.}$ In $\mathbb{Q}$, $B \subseteq \mathbb{Q}$ Retract is equivalent to $B$ is AR.

Then, I have tried to find an example of a $B \subseteq \mathbb{Q}$ such that $\mathbb{Q} \setminus B$ is not a Retract, and then, for Lemma, this $B$ would not be a $\sigma$z-set and then check if this set $B$ have empty interior in $Q$ and is closed. I could not find it.

$\mathbf{Note 2.}$ This is my first post. So excuse me if I broke any rule or something similar in the forum when writing the question. Thanks <3.

References.

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

[2] http://www.scielo.org.co/pdf/rein/v31n2/v31n2a05.pdf

$\def\cube{\mathbf Q}$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.