Determine all the Z-sets of topologist’s sine curve (how to start) 
Let $Y$ be a metric space and  $A\subseteq Y$  a closed subset. We say  $A$ is a Z-set in $Y$ if for every  $\epsilon>0$ there exist a continous mapping $f_{\epsilon}:Y\longrightarrow Y\setminus A $ such that $ d\left(f_{e}(y),y\right)<\epsilon$  for all $y\in Y$.
We are giving the exercise to find all Z-sets in $Y$ in regards to the clausure of the
Topologist's sine curve
i,e. $Y=\left\{ \left(x,\sin\left(\frac{1}{x}\right)\right):0<x\leq1\right\} \cup\left\{ \left(0,y\right):-1\leq y\leq1\right\} $

However I've been thinking about this exercise without improve (for example just trying to take $A=(0,0)$ one know $\sin\left(\frac{1}{x}\right)$ is continous except in $x=0$  but I can't see a way to make an $\epsilon$-mapping $f_\epsilon:Y\longrightarrow Y \setminus A$.
Also we haven't talk about other examples different from the hilbert Cube.

edit 1: So far I got that one Z set is of the form $A=\left \{ (0,y):y\in \left [ -1,1 \right ]\right \}$ but I can't prove using an $\epsilon$-mapping.

 A: We begin with some notation:
$p_x = (x,\sin\left(\frac{1}{x}\right))$ for $x \in (0,1]$
$S(M) = \{ p_x : x \in M \}$ for $M \subset (0,1]$
$L = \{ (0,y) :-1\le y \le 1 \}$
Since $Y$ is compact, all metrics on $Y$ are uniformly equivalent and we may work with $d((x,y),(x',y')) = \lvert x - x' \rvert + \lvert y - y' \rvert$.


*

*$Z  = L \cup \{ p_1 \}$ is a $Z$-set in $Y$. Let $\varepsilon > 0$.  Choose an integer $n \ge 0$ such that $b = (\pi/2 +2n\pi)^{-1} < \varepsilon$. Let $a = (3\pi/2 +2n\pi)^{-1}$. Then $S([a,b])$ is an arc connecting $p_a = (a,-1)$ with $p_b = (b,1)$. For each $y \in [-1,1]$ let $\phi(y) \in [a,b]$ be the unique point such that $(\phi(y),y) \in S([a,b])$. The function $\phi$ is continuous. Next choose $c \in [b,1)$ such that the diameter of $S([c,1])$ is $< \varepsilon$.Now define $$f_\varepsilon(x,y) = \begin{cases}
(x,y) & x \in [a,c] \\\
p_c & x \in [c,1] \\
(\phi(y),y) & x \in [0,a]
\end{cases}$$
This is a continuous function such that $f_\varepsilon(x,y) \in Y \setminus Z$ and $d(f_\varepsilon(x,y),(x,y)) < \varepsilon$. The latter is true for $x \in [c,1]$ by the choice of $c$ and for $x \in [0,a]$ because $d(f_\varepsilon(x,y),(x,y)) = 
 d((\phi(y),y),(x,y)) = \lvert \phi(y) - x \rvert \le b - 0 = b < \varepsilon$. In this construction the geometric intuition is that there is an obvious retraction $Y \to S([a,c])$ which moves points outside of $S([a,c])$ to a position in a distance $< \varepsilon$. Drawing a picture is helpful. Each point of $L \cup S((0,a])$ is moved horizontally to the right until it reaches the arc $S([a,b])$. Each point of the arc $S([c,1])$ is moved along this arc until it reaches $p_c$.

*No $\{p_x \}$ with $0 < x < 1$ is a $Z$-set in $Y$. The set $Y_x = Y \setminus \{p_x \}$ has the two connected components $C_1 = L  \cup S((0,x)) , C_2 = S((x,1])$. Let $\varepsilon = \min(x,1-x)$. Let $f : Y \to Y_x$ be any map. Since $Y$ is connected, we must have $f(Y) \subset C_i$ for some $i$. If $i=1$, then $d(f(p_1),p_1) \ge 1-x$ and if $i=2$, then $d(f(0,0),(0,0)) \ge x$. In both cases we have found a point $\xi$ such that $f(\xi)$ and $\xi$ do not have distance $<  \varepsilon$.

*This shows that $A \subset Y$ is a $Z$-set if and only if $A  \subset Z$. This is due the fact that subsets of $Z$-sets are again $Z$-sets. For the "only if" part assume that $A$ is a $Z$-set not contained in $Z$. Then it must contain a point $p_x$ with $x \in (0,1)$ so that $\{p_x \}$ should be a $Z$-set which contradicts 2.
