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.
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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.
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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 $A^\circ$.
Best Answer
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.