Condition for equality of “distance to the boundary” and “distance to the set”
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Let $(X,d)$ be a metric space, $A subseteq X$ and $x notin A$. What are the condition which make true the statement
$d(x,A) = d(x,partial A)$
If A is compact (not empty) I can think of a way to find a point $y in A$ such that $d(x,y)=d(x,A)$. And if $X$ is a normed space that should be enough to prove $y in partial A$. Is there something beside this?
metric-spaces
asked Jan 3 at 8:08
Markus SteinerMarkus Steiner
1036
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add a comment |
$begingroup$
Let $(X,d)$ be a metric space, $A subseteq X$ and $x notin A$. What are the condition which make true the statement
$d(x,A) = d(x,partial A)$
If A is compact (not empty) I can think of a way to find a point $y in A$ such that $d(x,y)=d(x,A)$. And if $X$ is a normed space that should be enough to prove $y in partial A$. Is there something beside this?
metric-spaces
asked Jan 3 at 8:08
Markus SteinerMarkus Steiner
1036
$endgroup$
$begingroup$
One condition: if $Aneqvarnothing$ then $A$ should not be clopen. So this will not work in e.g. discrete metric space.
$endgroup$
– drhab
Jan 3 at 8:27
$begingroup$
If $X$ is a normed space, then $A$ need only be weakly compact. All closed balls are weakly closed, so their intersection with weakly compact sets are weakly compact. If you consider the nested family of sets $B[x; d(x, A) + 1/n] cap C$, they should have a non-empty intersection which lies on the boundary.
$endgroup$
– Theo Bendit
Jan 3 at 22:57
add a comment |
$begingroup$
Let $(X,d)$ be a metric space, $A subseteq X$ and $x notin A$. What are the condition which make true the statement
$d(x,A) = d(x,partial A)$
If A is compact (not empty) I can think of a way to find a point $y in A$ such that $d(x,y)=d(x,A)$. And if $X$ is a normed space that should be enough to prove $y in partial A$. Is there something beside this?
metric-spaces
asked Jan 3 at 8:08
Markus SteinerMarkus Steiner
1036
$endgroup$
Let $(X,d)$ be a metric space, $A subseteq X$ and $x notin A$. What are the condition which make true the statement
$d(x,A) = d(x,partial A)$
If A is compact (not empty) I can think of a way to find a point $y in A$ such that $d(x,y)=d(x,A)$. And if $X$ is a normed space that should be enough to prove $y in partial A$. Is there something beside this?
metric-spaces
metric-spaces
asked Jan 3 at 8:08
Markus SteinerMarkus Steiner
1036
asked Jan 3 at 8:08
Markus SteinerMarkus Steiner
1036
asked Jan 3 at 8:08
Markus SteinerMarkus Steiner
1036
asked Jan 3 at 8:08
Markus SteinerMarkus Steiner
1036
asked Jan 3 at 8:08
Markus SteinerMarkus Steiner
1036
1036
$begingroup$
One condition: if $Aneqvarnothing$ then $A$ should not be clopen. So this will not work in e.g. discrete metric space.
$endgroup$
– drhab
Jan 3 at 8:27
$begingroup$
If $X$ is a normed space, then $A$ need only be weakly compact. All closed balls are weakly closed, so their intersection with weakly compact sets are weakly compact. If you consider the nested family of sets $B[x; d(x, A) + 1/n] cap C$, they should have a non-empty intersection which lies on the boundary.
$endgroup$
– Theo Bendit
Jan 3 at 22:57
add a comment |
$begingroup$
One condition: if $Aneqvarnothing$ then $A$ should not be clopen. So this will not work in e.g. discrete metric space.
$endgroup$
– drhab
Jan 3 at 8:27
$begingroup$
If $X$ is a normed space, then $A$ need only be weakly compact. All closed balls are weakly closed, so their intersection with weakly compact sets are weakly compact. If you consider the nested family of sets $B[x; d(x, A) + 1/n] cap C$, they should have a non-empty intersection which lies on the boundary.
$endgroup$
– Theo Bendit
Jan 3 at 22:57
$begingroup$
One condition: if $Aneqvarnothing$ then $A$ should not be clopen. So this will not work in e.g. discrete metric space.
$endgroup$
– drhab
Jan 3 at 8:27
$begingroup$
One condition: if $Aneqvarnothing$ then $A$ should not be clopen. So this will not work in e.g. discrete metric space.
$endgroup$
– drhab
Jan 3 at 8:27
$begingroup$
If $X$ is a normed space, then $A$ need only be weakly compact. All closed balls are weakly closed, so their intersection with weakly compact sets are weakly compact. If you consider the nested family of sets $B[x; d(x, A) + 1/n] cap C$, they should have a non-empty intersection which lies on the boundary.
$endgroup$
– Theo Bendit
Jan 3 at 22:57
$begingroup$
If $X$ is a normed space, then $A$ need only be weakly compact. All closed balls are weakly closed, so their intersection with weakly compact sets are weakly compact. If you consider the nested family of sets $B[x; d(x, A) + 1/n] cap C$, they should have a non-empty intersection which lies on the boundary.
$endgroup$
– Theo Bendit
Jan 3 at 22:57
add a comment |
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$begingroup$
One condition: if $Aneqvarnothing$ then $A$ should not be clopen. So this will not work in e.g. discrete metric space.
$endgroup$
– drhab
Jan 3 at 8:27
$begingroup$
If $X$ is a normed space, then $A$ need only be weakly compact. All closed balls are weakly closed, so their intersection with weakly compact sets are weakly compact. If you consider the nested family of sets $B[x; d(x, A) + 1/n] cap C$, they should have a non-empty intersection which lies on the boundary.
$endgroup$
– Theo Bendit
Jan 3 at 22:57