Is the interior of a closed, contractible set contractible if it is path connected?

My question is closely related to my other question. Since I got the answer for what I asked I’ve decided to ask my modified question separately.

If $A$ is a closed, contractible subset of a topological space $X$ and if $A^\circ$ is path connected then is $A^\circ$ contractible?

If we relax the condition of being closed then the answer provided in my other question works as a counter example.

Solution 1:

Consider the following:

$$X=\bigg\{(x,y,z)\in\mathbb{R}^3\ \bigg|\ 1\leq \sqrt{x^2+y^2} \leq 2\bigg\}$$ $$Y=\bigg\{(x,y,0)\in\mathbb{R}^3\ \bigg| \sqrt{x^2+y^2} \leq 1\bigg\}$$ $$A=X\cup Y$$

i.e. $A$ is the union of an infinite thick cylinder with a flat disk inside it.

Then $A$ is closed (union of two closed subsets) and contractible (first deform $A$ onto $Y$ via $t\cdot (x,y,z)\mapsto (x,y,tz)$ , and then $Y$ onto the origin), but

$$A^\circ=X^\circ=\bigg\{(x,y,z)\in\mathbb{R}^3\ \bigg|\ 1< \sqrt{x^2+y^2} < 2\bigg\}$$

is path connected and not contractible. This can be even generalized when $A$ is a closure of an open set by making $Y=\{(x,y,z)\in\mathbb{R}^3\ |\ |z|\leq\sqrt{x^2+y^2} \leq 1\}$.

Generally "being contractible" does not behave well under closure and interior operators.