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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.

R_D
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  • For a general topological space, it is not hard to construct a counte example,consider a cylander(a glass whose bottome removed), put a solid ball on it, call it $X$, now consider the the union of cylander with the boundry of the ball it is closed in $X$and contractible, call it $A$, then $A^\circ$ would be the cylander which is not contractible. – ali Dec 14 '21 at 13:42

1 Answers1

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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.

freakish
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