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Distance from a point in a set to a subset of that set is Lipschitz

Distance of a point to a subset.If $x geq 0$, $y geq 0$, $x leq D$, and $0 leq 1/(1+x) - 1/(1+y) leq C$, then is there an upper bound on the value of $y - x$?A continuous selection from a countable family of uniformly Lipschitz functions is LipschitzEvery 1-Lipschitz function in the closed unit ball has a fixed pointDistance from a point to a set in a metric spaceLipschitz constant for Linear Time Varying dynamic systemDensity of Lipschitz functions in the set of uniformly continuous functionsDistance between points obtained with Lipschitz functions

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5

1

$begingroup$

The question I'm stuck on is the following:

Let $(X,d)$ be a metric space and let $Y$ be a subset of $X$. If $xin X$, define the distance $d(x,Y)$ as $inf(d(x,y):yin Y$. Show that the mapping from $X$ to $mathbbR:xrightarrow d(x,Y)$ is Lipschitz, i.e. that there exists a constant $C>0$ such that $|d(x,Y)-d(x',Y)|le Cd(x,x'), x,x'in X$.

I'm quite lost as to how to approach it because there is no upper bound for $d(x,x')$ and thus the left side can easily go off to infinity. How can I relate distance between two points and the difference in their distances to $Y$ in a way that one constant works for the entire set?

real-analysis metric-spaces lipschitz-functions

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asked 8 hours ago

cruijfcruijf

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• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $C=1$ works. A picture may help.
  $endgroup$
  – Jochen
  7 hours ago
  

  
  
  
  

add a comment
 | 

5

1

$begingroup$

The question I'm stuck on is the following:

Let $(X,d)$ be a metric space and let $Y$ be a subset of $X$. If $xin X$, define the distance $d(x,Y)$ as $inf(d(x,y):yin Y$. Show that the mapping from $X$ to $mathbbR:xrightarrow d(x,Y)$ is Lipschitz, i.e. that there exists a constant $C>0$ such that $|d(x,Y)-d(x',Y)|le Cd(x,x'), x,x'in X$.

I'm quite lost as to how to approach it because there is no upper bound for $d(x,x')$ and thus the left side can easily go off to infinity. How can I relate distance between two points and the difference in their distances to $Y$ in a way that one constant works for the entire set?

real-analysis metric-spaces lipschitz-functions

share|cite|improve this question

asked 8 hours ago

cruijfcruijf

2833 bronze badges

New contributor

cruijf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $C=1$ works. A picture may help.
  $endgroup$
  – Jochen
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

The question I'm stuck on is the following:

Let $(X,d)$ be a metric space and let $Y$ be a subset of $X$. If $xin X$, define the distance $d(x,Y)$ as $inf(d(x,y):yin Y$. Show that the mapping from $X$ to $mathbbR:xrightarrow d(x,Y)$ is Lipschitz, i.e. that there exists a constant $C>0$ such that $|d(x,Y)-d(x',Y)|le Cd(x,x'), x,x'in X$.

I'm quite lost as to how to approach it because there is no upper bound for $d(x,x')$ and thus the left side can easily go off to infinity. How can I relate distance between two points and the difference in their distances to $Y$ in a way that one constant works for the entire set?

real-analysis metric-spaces lipschitz-functions

share|cite|improve this question

asked 8 hours ago

cruijfcruijf

2833 bronze badges

New contributor

cruijf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 8 hours ago

cruijfcruijf

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cruijf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 8 hours ago

cruijfcruijf

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New contributor

cruijf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

cruijfcruijf

2833 bronze badges

New contributor

cruijf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

cruijfcruijf

2833 bronze badges

2 Answers
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3

$begingroup$

Let $x,y in X$

then $forall z in Y$ we have $d(x,Y) leq d(x,z) leq d(x,y)+d(y,z)$

Thus $$d(x,Y) leq d(x,y)+d(y,Y) Longrightarrow d(x,Y)-d(y,Y) leq d(x,y)$$

Now similarly

$forall z in Y$ we have $$d(y,Y) leq d(y,z) leq d(x,y)+d(x,z)Longrightarrow d(x,Y) leq d(x,y) +d(x,Y)$$

So $d(y,Y)-d(x,Y) leq d(x,y)$

Combining the above we have that $|d(y,Y)-d(x,Y) |leq d(x,y)$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Marios GretsasMarios Gretsas

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much!
  $endgroup$
  – cruijf
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome.
  $endgroup$
  – Marios Gretsas
  7 hours ago
  

  
  
  
  

add a comment
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1

$begingroup$

We have $d(x, x')+d(x', y) ge d(x, y)ge d(x, Y)$ for any $yin Y$.

So, taking infimum for $yin Y$, we get
$$d(x, x') +d(x', Y) ge d(x, Y)$$
So, $d(x, Y)-d(x', Y) le d(x, x')$.

By symmetry, we also get $d(x',Y) - d(x,Y)le d(x, x')$.

share|cite|improve this answer

answered 7 hours ago

BerciBerci

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3

$begingroup$

Let $x,y in X$

then $forall z in Y$ we have $d(x,Y) leq d(x,z) leq d(x,y)+d(y,z)$

Thus $$d(x,Y) leq d(x,y)+d(y,Y) Longrightarrow d(x,Y)-d(y,Y) leq d(x,y)$$

Now similarly

$forall z in Y$ we have $$d(y,Y) leq d(y,z) leq d(x,y)+d(x,z)Longrightarrow d(x,Y) leq d(x,y) +d(x,Y)$$

So $d(y,Y)-d(x,Y) leq d(x,y)$

Combining the above we have that $|d(y,Y)-d(x,Y) |leq d(x,y)$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Marios GretsasMarios Gretsas

9,95011 gold badge1515 silver badges4040 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much!
  $endgroup$
  – cruijf
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome.
  $endgroup$
  – Marios Gretsas
  7 hours ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

Let $x,y in X$

then $forall z in Y$ we have $d(x,Y) leq d(x,z) leq d(x,y)+d(y,z)$

Thus $$d(x,Y) leq d(x,y)+d(y,Y) Longrightarrow d(x,Y)-d(y,Y) leq d(x,y)$$

Now similarly

$forall z in Y$ we have $$d(y,Y) leq d(y,z) leq d(x,y)+d(x,z)Longrightarrow d(x,Y) leq d(x,y) +d(x,Y)$$

So $d(y,Y)-d(x,Y) leq d(x,y)$

Combining the above we have that $|d(y,Y)-d(x,Y) |leq d(x,y)$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Marios GretsasMarios Gretsas

9,95011 gold badge1515 silver badges4040 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much!
  $endgroup$
  – cruijf
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome.
  $endgroup$
  – Marios Gretsas
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Let $x,y in X$

then $forall z in Y$ we have $d(x,Y) leq d(x,z) leq d(x,y)+d(y,z)$

Thus $$d(x,Y) leq d(x,y)+d(y,Y) Longrightarrow d(x,Y)-d(y,Y) leq d(x,y)$$

Now similarly

$forall z in Y$ we have $$d(y,Y) leq d(y,z) leq d(x,y)+d(x,z)Longrightarrow d(x,Y) leq d(x,y) +d(x,Y)$$

So $d(y,Y)-d(x,Y) leq d(x,y)$

Combining the above we have that $|d(y,Y)-d(x,Y) |leq d(x,y)$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Marios GretsasMarios Gretsas

9,95011 gold badge1515 silver badges4040 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

Marios GretsasMarios Gretsas

9,95011 gold badge1515 silver badges4040 bronze badges

answered 7 hours ago

Marios GretsasMarios Gretsas

9,95011 gold badge1515 silver badges4040 bronze badges

answered 7 hours ago

Marios GretsasMarios Gretsas

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1

$begingroup$

We have $d(x, x')+d(x', y) ge d(x, y)ge d(x, Y)$ for any $yin Y$.

So, taking infimum for $yin Y$, we get
$$d(x, x') +d(x', Y) ge d(x, Y)$$
So, $d(x, Y)-d(x', Y) le d(x, x')$.

By symmetry, we also get $d(x',Y) - d(x,Y)le d(x, x')$.

share|cite|improve this answer

answered 7 hours ago

BerciBerci

65.7k22 gold badges4242 silver badges7676 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

We have $d(x, x')+d(x', y) ge d(x, y)ge d(x, Y)$ for any $yin Y$.

So, taking infimum for $yin Y$, we get
$$d(x, x') +d(x', Y) ge d(x, Y)$$
So, $d(x, Y)-d(x', Y) le d(x, x')$.

By symmetry, we also get $d(x',Y) - d(x,Y)le d(x, x')$.

share|cite|improve this answer

answered 7 hours ago

BerciBerci

65.7k22 gold badges4242 silver badges7676 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

We have $d(x, x')+d(x', y) ge d(x, y)ge d(x, Y)$ for any $yin Y$.

So, taking infimum for $yin Y$, we get
$$d(x, x') +d(x', Y) ge d(x, Y)$$
So, $d(x, Y)-d(x', Y) le d(x, x')$.

By symmetry, we also get $d(x',Y) - d(x,Y)le d(x, x')$.

share|cite|improve this answer

answered 7 hours ago

BerciBerci

65.7k22 gold badges4242 silver badges7676 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

BerciBerci

65.7k22 gold badges4242 silver badges7676 bronze badges

answered 7 hours ago

BerciBerci

65.7k22 gold badges4242 silver badges7676 bronze badges

answered 7 hours ago

BerciBerci

65.7k22 gold badges4242 silver badges7676 bronze badges