# Is the interior of a contractible space contractible if it is path connected?

Let $$A$$ be a contractible subset of a topological space $$X$$. A set is said to be contractible if it has the same homotopy type as a point. Alternately, a space is contractible if the identity map of the space is homotopic to a constant map.

The first thing that I was wondering is if $$A$$ is contractible then what could we say about $$A^\circ$$ and $$\overline A$$; are they also contractible? The answer to both is a clear no.

1. Let $$X=\mathbb R^2$$. Take $$A$$ to be the union of two closed discs, touching at a single point. Then $$A^\circ$$ is the union of the interiors of each disc and this space is not even path connected.

2. Let $$X=\mathbb R^2$$. Take $$A$$ to be the circle $$(S^1)$$ minus a point. This is homeomorphic to $$\mathbb R$$ and is therefore contractible. The closure is $$S^1$$ which is not.

The first answer gave rise to the following question: If $$A$$ is contractible and $$A^\circ$$ is path connected can we say that $$A^\circ$$ is contractible?

So far I have neither been able to come up with a proof or a counter example. The only thing I can say is that we may assume without loss of generality that $$A$$ contracts to a point in $$A^\circ$$.

Take the standard open cone in $$\mathbb{R}^3$$ (i.e. it does not contain the tip of the cone), about each point on the cone include an (open) line segment normal to that point of length $$f(x)$$, were $$f$$ is some function that tends to 0 fast enough so that the resulting space is homeomorphic (in the obvious way) to the open cone crossed with $$\mathbb{R}$$. Finally, add the cone point.
The space we have described is contractible to the cone point, but its interior is homeomorphic to the open cone crossed with $$\mathbb{R}$$, hence, noncontractible.
• Wow! I didn't realise how simple the answer would be! Infact if $D$ is the open unit disc in $\mathbb R^2$ then we can simply take $A=(D^2-\{(0,0)\}\times R)\bigcup \{(0,0,0)\}$ and this would work right?