Interior of a contractible set
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Let $O, F subset mathbb R^n$ for $n ge 2$. Suppose $O$ is open, $F$ is closed and $O subset F$ with closure $bar{O} = F$. If we further assume that both $O$ and $F$ are contractible to the same fixed point $p in O$. Can we conclude that $F^{circ} = O$?
general-topology
asked Dec 4 '18 at 8:16
user1101010user1101010
7751730
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add a comment |
$begingroup$
Let $O, F subset mathbb R^n$ for $n ge 2$. Suppose $O$ is open, $F$ is closed and $O subset F$ with closure $bar{O} = F$. If we further assume that both $O$ and $F$ are contractible to the same fixed point $p in O$. Can we conclude that $F^{circ} = O$?
general-topology
asked Dec 4 '18 at 8:16
user1101010user1101010
7751730
$endgroup$
add a comment |
$begingroup$
Let $O, F subset mathbb R^n$ for $n ge 2$. Suppose $O$ is open, $F$ is closed and $O subset F$ with closure $bar{O} = F$. If we further assume that both $O$ and $F$ are contractible to the same fixed point $p in O$. Can we conclude that $F^{circ} = O$?
general-topology
asked Dec 4 '18 at 8:16
user1101010user1101010
7751730
$endgroup$
Let $O, F subset mathbb R^n$ for $n ge 2$. Suppose $O$ is open, $F$ is closed and $O subset F$ with closure $bar{O} = F$. If we further assume that both $O$ and $F$ are contractible to the same fixed point $p in O$. Can we conclude that $F^{circ} = O$?
general-topology
general-topology
asked Dec 4 '18 at 8:16
user1101010user1101010
7751730
asked Dec 4 '18 at 8:16
user1101010user1101010
7751730
asked Dec 4 '18 at 8:16
user1101010user1101010
7751730
asked Dec 4 '18 at 8:16
user1101010user1101010
7751730
asked Dec 4 '18 at 8:16
user1101010user1101010
7751730
7751730
add a comment |
add a comment |
1 Answer
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$begingroup$
No. Let $F = D^n = { x in mathbb{R}^n mid lVert x rVert le 1 }$ and $O = mathring{F} setminus [1/2,1) times { (0,dots,0) }$. Both are contractible to $0$.
answered Dec 4 '18 at 14:59
Paul FrostPaul Frost
10k3932
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
No. Let $F = D^n = { x in mathbb{R}^n mid lVert x rVert le 1 }$ and $O = mathring{F} setminus [1/2,1) times { (0,dots,0) }$. Both are contractible to $0$.
answered Dec 4 '18 at 14:59
Paul FrostPaul Frost
10k3932
$endgroup$
add a comment |
$begingroup$
No. Let $F = D^n = { x in mathbb{R}^n mid lVert x rVert le 1 }$ and $O = mathring{F} setminus [1/2,1) times { (0,dots,0) }$. Both are contractible to $0$.
answered Dec 4 '18 at 14:59
Paul FrostPaul Frost
10k3932
$endgroup$
add a comment |
$begingroup$
No. Let $F = D^n = { x in mathbb{R}^n mid lVert x rVert le 1 }$ and $O = mathring{F} setminus [1/2,1) times { (0,dots,0) }$. Both are contractible to $0$.
answered Dec 4 '18 at 14:59
Paul FrostPaul Frost
10k3932
$endgroup$
No. Let $F = D^n = { x in mathbb{R}^n mid lVert x rVert le 1 }$ and $O = mathring{F} setminus [1/2,1) times { (0,dots,0) }$. Both are contractible to $0$.
answered Dec 4 '18 at 14:59
Paul FrostPaul Frost
10k3932
answered Dec 4 '18 at 14:59
Paul FrostPaul Frost
10k3932
answered Dec 4 '18 at 14:59
Paul FrostPaul Frost
10k3932
answered Dec 4 '18 at 14:59
Paul FrostPaul Frost
10k3932
10k3932
add a comment |
add a comment |
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