Is every contractible space a cone? It is easy to show that for any topological space $X$, the cone $CX$ is contractible. I am interested in the converse. If $Y$ is a contractible space, is $Y$ homeomorphic to $CX$ for some topological space $X$?
I was told the answer is no. However, I haven't been able to find a counterexample. I have two questions:


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*Is there a nice non-constructive way to see that counter-examples "should"/must exist?

*Does anyone have a counter-example?

 A: A fun counterexample: written in 1-dimensional strokes, the letter $\pi$.
A: A whole family of counter examples comes from finite spaces. Clearly, no finite nonempty space can be the cone of anything. But contractible finite spaces exist (there is even an entire relevant book, "Algebraic topology of finite topological spaces and applications", Barmak).
One way to quickly convince yourself that the answer must be 'no' is that cones are strongly related to $[0,1]$ (by construction!). So, any cone will have to admit some $\mathbb R$-like properties. However, contractibility is a very refined topological property and it is highly unlikely that just because a space is contractible that it will be strongly related to the real numbers. Maybe this counts as a non-constructive way to see that counterexamples "should" exist. 
For an explicit non-finite example, consider http://en.wikipedia.org/wiki/House_with_two_rooms
A: In Hatcher's algebraic topology the exercise of first chapter gives a example that there is a space which is contractible but not homotopic to a point.(i.e. the space can be contracted to one point but in the process you must move the point) And we all know that a cone can always be deformation retracted to a point(thanks for pointing out my fault). So I think that counterexample shows everything.
