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If $S subseteq X$ is closed, is $f(S,r)$ necessarily closed?

In a metric space, is the dilation of a closed set closed?In a normed vector space, if $S⊆X$ is closed, is $f(S,r)$ necessarily closed?Determine if the set $C ={ frac{1}{n}|n in mathbb{N} }$ is open or closed in the euclidian metricShow that $D ={ x + y mid x in (0,1) ,y in [1,2) }$ is open or closedDoes “uniformly isolated” imply closed?Show that $F subseteq X$ is closed iff $F cap K$ is closed for every compact set $Ksubseteq X$A non-empty closed proper set in a metric spaceDoes it follow that ${ (x,y) in Xtimes X mid d(x,y) < m } subseteq U$?Closed ball is a closed setTopology - Infinite intersection of closed and connected sets.Does $partial B(x_0, r) subseteq {x in X : d(x_0,x) = r }$ hold in an arbitrary metric space?$epsilon$ fattening of an open set needn't be open

3

$begingroup$

Let $X$ denote a metric space.

Whenever $S subseteq X$ and $r in mathbb{R}_{geq 0}$, write $f(S,r)$ for the following set.

$${x in X mid exists s in S : d(x,s) leq r}$$

Question. Given a closed set $S subseteq X$ and any $r in mathbb{R}_{geq 0}$, does it follow that $f(S,r)$ is closed?

If not, a counterexample $(X,S,r)$ would be appreciated.

metric-spaces examples-counterexamples

share|cite|improve this question

edited Aug 5 '14 at 12:40

goblin

asked Aug 3 '14 at 7:28

goblingoblin

37.1k1159193

$endgroup$

add a comment | 

3

$begingroup$

Let $X$ denote a metric space.

Whenever $S subseteq X$ and $r in mathbb{R}_{geq 0}$, write $f(S,r)$ for the following set.

$${x in X mid exists s in S : d(x,s) leq r}$$

Question. Given a closed set $S subseteq X$ and any $r in mathbb{R}_{geq 0}$, does it follow that $f(S,r)$ is closed?

If not, a counterexample $(X,S,r)$ would be appreciated.

metric-spaces examples-counterexamples

share|cite|improve this question

edited Aug 5 '14 at 12:40

goblin

asked Aug 3 '14 at 7:28

goblingoblin

37.1k1159193

$endgroup$

add a comment | 

$begingroup$

Let $X$ denote a metric space.

Whenever $S subseteq X$ and $r in mathbb{R}_{geq 0}$, write $f(S,r)$ for the following set.

$${x in X mid exists s in S : d(x,s) leq r}$$

Question. Given a closed set $S subseteq X$ and any $r in mathbb{R}_{geq 0}$, does it follow that $f(S,r)$ is closed?

If not, a counterexample $(X,S,r)$ would be appreciated.

metric-spaces examples-counterexamples

share|cite|improve this question

edited Aug 5 '14 at 12:40

goblin

asked Aug 3 '14 at 7:28

goblingoblin

37.1k1159193

$endgroup$

share|cite|improve this question

edited Aug 5 '14 at 12:40

goblin

asked Aug 3 '14 at 7:28

goblingoblin

37.1k1159193

share|cite|improve this question

edited Aug 5 '14 at 12:40

goblin

asked Aug 3 '14 at 7:28

goblingoblin

37.1k1159193

asked Aug 3 '14 at 7:28

goblingoblin

37.1k1159193

asked Aug 3 '14 at 7:28

goblingoblin

37.1k1159193

2 Answers
2

active

oldest

votes

3

$begingroup$

The answer is no.

Let $S={1-frac{1}{n}}_{n=1}^infty$. Let $X$ be the set $Scup [1.5,2]$ with its subspace metric. Then $f(S,1)=Scup [1.5,2)$ is not closed in $X$ since it is not of the form $Ccap X$ for some closed set $C$ of the reals.

We indeed see that $S$ is not a compact subset of $X$, in align with William's result.

share|cite|improve this answer

answered Aug 3 '14 at 10:47

Robert WolfeRobert Wolfe

5,98422763

$endgroup$

add a comment | 

3

$begingroup$

If in addition $C$ is compact, then the following proof shows $f(C,r)$ is closed for each positive real number $r$.

Suppose $z$ is a limit point of $f(C,r)$. For all $n$, find some $x_n in f(C,r)$ and $c_n in C$ such that $d(x_n, c_n) leq r$ and $d(x_n, z) leq frac{1}{n}$. Hence by the triangle inequality, $d(z, c_n) leq r + frac{1}{n}$. By sequential compactness, there is a subsequence of ${c_n : n in omega}$ which converges. To avoid notation, let just suppose the subsequence was just ${c_n : n in omega}$ and it coverges to the point $c$. Since $C$ is closed, $c in C$. Let $epsilon > 0$. Choose $N$ large such that for all $m > N$, $d(c, c_m) < frac{epsilon}{2}$ and $frac{1}{N} < frac{epsilon}{2}$. Then $d(z,c) leq r + d(z, c_m) + d(c_m, c) = r + frac{1}{m} + frac{epsilon}{2} leq r + frac{1}{N} + frac{epsilon}{2} leq r + frac{epsilon}{2} + frac{epsilon}{2} = r + epsilon$. Since $epsilon$ is arbitrary, $d(z,c) leq r$. Hence $z in f(C,r)$.

This seems to show that if $C$ is a compact set, then $f(C,r)$ is closed. At the moment, I am not sure whether or not being just closed is enough.

share|cite|improve this answer

answered Aug 3 '14 at 8:02

WilliamWilliam

17.3k22256

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice. So if $X = mathbb{R}^n$, then whenever $C subseteq X$ is compact, so too is $f(C,r),$ for all $r geq 0$.
  $endgroup$
  – goblin
  Aug 3 '14 at 9:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @goblin Yes. Note that I only showed $f(C,r)$ is closed. However if you are in $mathbb{R}^n$, compactness is equivalent to being closed and bounded. Hence in this case the set is indeed compact.
  $endgroup$
  – William
  Aug 3 '14 at 9:37
  

  
  
  
  

add a comment | 

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3

$begingroup$

The answer is no.

Let $S={1-frac{1}{n}}_{n=1}^infty$. Let $X$ be the set $Scup [1.5,2]$ with its subspace metric. Then $f(S,1)=Scup [1.5,2)$ is not closed in $X$ since it is not of the form $Ccap X$ for some closed set $C$ of the reals.

We indeed see that $S$ is not a compact subset of $X$, in align with William's result.

share|cite|improve this answer

answered Aug 3 '14 at 10:47

Robert WolfeRobert Wolfe

5,98422763

$endgroup$

add a comment | 

3

$begingroup$

The answer is no.

Let $S={1-frac{1}{n}}_{n=1}^infty$. Let $X$ be the set $Scup [1.5,2]$ with its subspace metric. Then $f(S,1)=Scup [1.5,2)$ is not closed in $X$ since it is not of the form $Ccap X$ for some closed set $C$ of the reals.

We indeed see that $S$ is not a compact subset of $X$, in align with William's result.

share|cite|improve this answer

answered Aug 3 '14 at 10:47

Robert WolfeRobert Wolfe

5,98422763

$endgroup$

add a comment | 

$begingroup$

The answer is no.

Let $S={1-frac{1}{n}}_{n=1}^infty$. Let $X$ be the set $Scup [1.5,2]$ with its subspace metric. Then $f(S,1)=Scup [1.5,2)$ is not closed in $X$ since it is not of the form $Ccap X$ for some closed set $C$ of the reals.

We indeed see that $S$ is not a compact subset of $X$, in align with William's result.

share|cite|improve this answer

answered Aug 3 '14 at 10:47

Robert WolfeRobert Wolfe

5,98422763

$endgroup$

share|cite|improve this answer

answered Aug 3 '14 at 10:47

Robert WolfeRobert Wolfe

5,98422763

answered Aug 3 '14 at 10:47

Robert WolfeRobert Wolfe

5,98422763

answered Aug 3 '14 at 10:47

Robert WolfeRobert Wolfe

5,98422763

3

$begingroup$

If in addition $C$ is compact, then the following proof shows $f(C,r)$ is closed for each positive real number $r$.

Suppose $z$ is a limit point of $f(C,r)$. For all $n$, find some $x_n in f(C,r)$ and $c_n in C$ such that $d(x_n, c_n) leq r$ and $d(x_n, z) leq frac{1}{n}$. Hence by the triangle inequality, $d(z, c_n) leq r + frac{1}{n}$. By sequential compactness, there is a subsequence of ${c_n : n in omega}$ which converges. To avoid notation, let just suppose the subsequence was just ${c_n : n in omega}$ and it coverges to the point $c$. Since $C$ is closed, $c in C$. Let $epsilon > 0$. Choose $N$ large such that for all $m > N$, $d(c, c_m) < frac{epsilon}{2}$ and $frac{1}{N} < frac{epsilon}{2}$. Then $d(z,c) leq r + d(z, c_m) + d(c_m, c) = r + frac{1}{m} + frac{epsilon}{2} leq r + frac{1}{N} + frac{epsilon}{2} leq r + frac{epsilon}{2} + frac{epsilon}{2} = r + epsilon$. Since $epsilon$ is arbitrary, $d(z,c) leq r$. Hence $z in f(C,r)$.

This seems to show that if $C$ is a compact set, then $f(C,r)$ is closed. At the moment, I am not sure whether or not being just closed is enough.

share|cite|improve this answer

answered Aug 3 '14 at 8:02

WilliamWilliam

17.3k22256

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice. So if $X = mathbb{R}^n$, then whenever $C subseteq X$ is compact, so too is $f(C,r),$ for all $r geq 0$.
  $endgroup$
  – goblin
  Aug 3 '14 at 9:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @goblin Yes. Note that I only showed $f(C,r)$ is closed. However if you are in $mathbb{R}^n$, compactness is equivalent to being closed and bounded. Hence in this case the set is indeed compact.
  $endgroup$
  – William
  Aug 3 '14 at 9:37
  

  
  
  
  

add a comment | 

3

$begingroup$

If in addition $C$ is compact, then the following proof shows $f(C,r)$ is closed for each positive real number $r$.

Suppose $z$ is a limit point of $f(C,r)$. For all $n$, find some $x_n in f(C,r)$ and $c_n in C$ such that $d(x_n, c_n) leq r$ and $d(x_n, z) leq frac{1}{n}$. Hence by the triangle inequality, $d(z, c_n) leq r + frac{1}{n}$. By sequential compactness, there is a subsequence of ${c_n : n in omega}$ which converges. To avoid notation, let just suppose the subsequence was just ${c_n : n in omega}$ and it coverges to the point $c$. Since $C$ is closed, $c in C$. Let $epsilon > 0$. Choose $N$ large such that for all $m > N$, $d(c, c_m) < frac{epsilon}{2}$ and $frac{1}{N} < frac{epsilon}{2}$. Then $d(z,c) leq r + d(z, c_m) + d(c_m, c) = r + frac{1}{m} + frac{epsilon}{2} leq r + frac{1}{N} + frac{epsilon}{2} leq r + frac{epsilon}{2} + frac{epsilon}{2} = r + epsilon$. Since $epsilon$ is arbitrary, $d(z,c) leq r$. Hence $z in f(C,r)$.

This seems to show that if $C$ is a compact set, then $f(C,r)$ is closed. At the moment, I am not sure whether or not being just closed is enough.

share|cite|improve this answer

answered Aug 3 '14 at 8:02

WilliamWilliam

17.3k22256

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice. So if $X = mathbb{R}^n$, then whenever $C subseteq X$ is compact, so too is $f(C,r),$ for all $r geq 0$.
  $endgroup$
  – goblin
  Aug 3 '14 at 9:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @goblin Yes. Note that I only showed $f(C,r)$ is closed. However if you are in $mathbb{R}^n$, compactness is equivalent to being closed and bounded. Hence in this case the set is indeed compact.
  $endgroup$
  – William
  Aug 3 '14 at 9:37
  

  
  
  
  

add a comment | 

$begingroup$

If in addition $C$ is compact, then the following proof shows $f(C,r)$ is closed for each positive real number $r$.

Suppose $z$ is a limit point of $f(C,r)$. For all $n$, find some $x_n in f(C,r)$ and $c_n in C$ such that $d(x_n, c_n) leq r$ and $d(x_n, z) leq frac{1}{n}$. Hence by the triangle inequality, $d(z, c_n) leq r + frac{1}{n}$. By sequential compactness, there is a subsequence of ${c_n : n in omega}$ which converges. To avoid notation, let just suppose the subsequence was just ${c_n : n in omega}$ and it coverges to the point $c$. Since $C$ is closed, $c in C$. Let $epsilon > 0$. Choose $N$ large such that for all $m > N$, $d(c, c_m) < frac{epsilon}{2}$ and $frac{1}{N} < frac{epsilon}{2}$. Then $d(z,c) leq r + d(z, c_m) + d(c_m, c) = r + frac{1}{m} + frac{epsilon}{2} leq r + frac{1}{N} + frac{epsilon}{2} leq r + frac{epsilon}{2} + frac{epsilon}{2} = r + epsilon$. Since $epsilon$ is arbitrary, $d(z,c) leq r$. Hence $z in f(C,r)$.

This seems to show that if $C$ is a compact set, then $f(C,r)$ is closed. At the moment, I am not sure whether or not being just closed is enough.

share|cite|improve this answer

answered Aug 3 '14 at 8:02

WilliamWilliam

17.3k22256

$endgroup$

share|cite|improve this answer

answered Aug 3 '14 at 8:02

WilliamWilliam

17.3k22256

answered Aug 3 '14 at 8:02

WilliamWilliam

17.3k22256

answered Aug 3 '14 at 8:02

WilliamWilliam

17.3k22256