Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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 at 4:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.