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If $(X,d)$ is a metric space, then we say that a closed subset $A$ of $X$ is a z-set if for each number $k\gt 0$ there is a continuous map $f_k$ from $X$ into $X-A$ such that $d(x,f_k(x))\lt k$.

I am wondering if an AR which is a z-set in the Hilbert cube is a deformation retract of it?

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  • $\begingroup$ This is false even for finite-dimensional cubes: just take any non-contractible subset of the boundary as your $A$ (like $A=$ a 2-point set). $\endgroup$
    – Misha
    Commented Aug 17, 2012 at 6:27
  • $\begingroup$ I am sorry, i edit the question $\endgroup$
    – Pedro Perez
    Commented Aug 17, 2012 at 6:37
  • $\begingroup$ (I may be confused but in my opinion:) Every retract $X$ of any Hilbert cube $H$ is its deformation retract, i.e. $X$ is a deformation retract of $H$. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Feb 28, 2014 at 4:03

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