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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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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). – Misha Aug 17 '12 at 6:27
I am sorry, i edit the question – Pedro Perez Aug 17 '12 at 6:37
(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 $H$. – Włodzimierz Holsztyński Feb 28 '14 at 4:03

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