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Metric space and subsetsReverse Triangle Inequality ProofEquivalence relation on a continuumHow to prove easily that a generalized ellipse is $C^1$How to prove any closed set in $mathbbR$ is $G_delta$Definitions of supremumConvergence of a point and that of its projectionCompactess of the sum closed balls whose centers is a compact set.Distance to a closed setTo show that the function $f(x) = inf d(x,x_n) : n in Bbb N $ is uniformly continuous on X.Proving that if $f:D to mathbbR$ is continuous and compact, then $f(x_n)$ is a Cauchy sequence.Prove that $ f$ is uniformly continuousShow that if $f$ continuous then $f(overlineX)subset overlinef(X)$ for all $Xsubset R$A closed set in vector space of limited sequences $l^infty$Let $f$ be a real uniformly continuous function on the bounded set $E$ in $mathbbR^1$. Prove that $f$ is bounded on $E$Let $f: mathscrB(mathbbN;mathbbR)rightarrow mathbbR$ given by $f(x) = liminf x_n.$ Prove that $f$ is continuous.Proof that the generalized inverse of an increasing right-continuous function is also right-continuousIs $A=y in BbbR: y=lim_nto inftyf(x_n), textfor some sequence~ x_n to +infty$ compact for a continuous $f$?

5

5

$begingroup$

I want to prove that given a metric space $(M,d)$ and $F subset M$, then the function $f_F: M to Bbb R$ given by $f_F(x) = d(x,F) = infd(x,y) : y in M$ is continuous.

Take $x in M$. If $x in F$ it is obvious. I am convinced that it is continuous, since if you take a sequence $(x_n)_n in Bbb N$ which converges to $x in M$, then the distances from the $x_n$ to $F$ go close to the distance from $x$ to $F$, since the $x_n$ go close to $x$.

But I'm having much trouble writing it. My teacher said to "just forget about the $inf$ and work with sequences", but that doesn't help me in nothing at all.

Given $epsilon > 0$ and a sequence $(x_n)_x in Bbb N$ such that $x_n to x$, I must prove that $f_F(x_n) to f_F(x)$, that is, find $n_epsilon in Bbb N$ such that: $$n > n_epsilon implies |d(x_n,F) - d(x,F)| < epsilon $$

Surely if $y in F$, we have $d(x,y) geq d(x,F)$. Other than this, I don't know how to deal with these infs, and this absolute value. I suppose I'll have to use the triangle inequality. Can someone help me do it?

real-analysis metric-spaces

share|cite|improve this question

asked Sep 24 '14 at 18:15

Ivo TerekIvo Terek

46.7k954146

$endgroup$

add a comment | 

5

5

$begingroup$

I want to prove that given a metric space $(M,d)$ and $F subset M$, then the function $f_F: M to Bbb R$ given by $f_F(x) = d(x,F) = infd(x,y) : y in M$ is continuous.

Take $x in M$. If $x in F$ it is obvious. I am convinced that it is continuous, since if you take a sequence $(x_n)_n in Bbb N$ which converges to $x in M$, then the distances from the $x_n$ to $F$ go close to the distance from $x$ to $F$, since the $x_n$ go close to $x$.

But I'm having much trouble writing it. My teacher said to "just forget about the $inf$ and work with sequences", but that doesn't help me in nothing at all.

Given $epsilon > 0$ and a sequence $(x_n)_x in Bbb N$ such that $x_n to x$, I must prove that $f_F(x_n) to f_F(x)$, that is, find $n_epsilon in Bbb N$ such that: $$n > n_epsilon implies |d(x_n,F) - d(x,F)| < epsilon $$

Surely if $y in F$, we have $d(x,y) geq d(x,F)$. Other than this, I don't know how to deal with these infs, and this absolute value. I suppose I'll have to use the triangle inequality. Can someone help me do it?

real-analysis metric-spaces

share|cite|improve this question

asked Sep 24 '14 at 18:15

Ivo TerekIvo Terek

46.7k954146

$endgroup$

add a comment | 

$begingroup$

I want to prove that given a metric space $(M,d)$ and $F subset M$, then the function $f_F: M to Bbb R$ given by $f_F(x) = d(x,F) = infd(x,y) : y in M$ is continuous.

Take $x in M$. If $x in F$ it is obvious. I am convinced that it is continuous, since if you take a sequence $(x_n)_n in Bbb N$ which converges to $x in M$, then the distances from the $x_n$ to $F$ go close to the distance from $x$ to $F$, since the $x_n$ go close to $x$.

But I'm having much trouble writing it. My teacher said to "just forget about the $inf$ and work with sequences", but that doesn't help me in nothing at all.

Given $epsilon > 0$ and a sequence $(x_n)_x in Bbb N$ such that $x_n to x$, I must prove that $f_F(x_n) to f_F(x)$, that is, find $n_epsilon in Bbb N$ such that: $$n > n_epsilon implies |d(x_n,F) - d(x,F)| < epsilon $$

Surely if $y in F$, we have $d(x,y) geq d(x,F)$. Other than this, I don't know how to deal with these infs, and this absolute value. I suppose I'll have to use the triangle inequality. Can someone help me do it?

real-analysis metric-spaces

share|cite|improve this question

asked Sep 24 '14 at 18:15

Ivo TerekIvo Terek

46.7k954146

$endgroup$

share|cite|improve this question

asked Sep 24 '14 at 18:15

Ivo TerekIvo Terek

46.7k954146

share|cite|improve this question

asked Sep 24 '14 at 18:15

Ivo TerekIvo Terek

46.7k954146

asked Sep 24 '14 at 18:15

Ivo TerekIvo Terek

46.7k954146

asked Sep 24 '14 at 18:15

Ivo TerekIvo Terek

46.7k954146

2 Answers
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15

$begingroup$

From $d(x',y)leq d(x',x)+d(x,y)$, by taking infimum we get
$d(x',F)leq d(x',x)+d(x,F)$.
Similarly, $d(x,F)leq d(x',x)+d(x',F)$.
Consequently, $|d(x',F)-d(x,F)|<d(x',x)$.

share|cite|improve this answer

answered Sep 24 '14 at 22:18

Fabio LucchiniFabio Lucchini

9,50111426

$endgroup$

add a comment | 

0

$begingroup$

Here is a more worked example without "taking infimum" using the inverse triangle inequality:

We wish to show for all $x, y in M$, $lvert f_F(x) - f_X(y)rvert le d(x, y)$.

Case: $d(x, z) ge d(y, z)$.

Fix the $z$ such that both bounds are true. By the definitions of infimum, we have two bounds (let $epsilon > 0$):

beginalign
forall z in F quad &f_F(x) le d(x, z) \
exists z in F quad &f_F(x) + epsilon > d(x, z)
endalign

By the inverse triangle inequality,

beginalign
& d(x, z) - d(y,z) le d(x,y) \
implies & f_F(x) - d(y, z) le d(x,y) \
implies & f_F(x) - f_F(y) - epsilon le d(x,y) \
implies & f_F(x) - f_F(y) le d(x,y) + epsilon
endalign

Since this holds for all $epsilon > 0$, $f_F(x) - f_F(y) le d(x,y)$. The case of $d(x,z) < d(y,z)$ is similar, so we have proved $lvert f_F(x) - f_F(y)rvert le d(x, y)$. Now simply apply our definition of continuity using $delta = epsilon$ (actually $f_F$ is Lipschitz continuous with constant $1$).

share|cite|improve this answer

edited Mar 22 at 6:21

answered Mar 22 at 0:38

qwrqwr

6,68242755

$endgroup$

add a comment | 

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15

$begingroup$

From $d(x',y)leq d(x',x)+d(x,y)$, by taking infimum we get
$d(x',F)leq d(x',x)+d(x,F)$.
Similarly, $d(x,F)leq d(x',x)+d(x',F)$.
Consequently, $|d(x',F)-d(x,F)|<d(x',x)$.

share|cite|improve this answer

answered Sep 24 '14 at 22:18

Fabio LucchiniFabio Lucchini

9,50111426

$endgroup$

add a comment | 

15

$begingroup$

From $d(x',y)leq d(x',x)+d(x,y)$, by taking infimum we get
$d(x',F)leq d(x',x)+d(x,F)$.
Similarly, $d(x,F)leq d(x',x)+d(x',F)$.
Consequently, $|d(x',F)-d(x,F)|<d(x',x)$.

share|cite|improve this answer

answered Sep 24 '14 at 22:18

Fabio LucchiniFabio Lucchini

9,50111426

$endgroup$

add a comment | 

$begingroup$

From $d(x',y)leq d(x',x)+d(x,y)$, by taking infimum we get
$d(x',F)leq d(x',x)+d(x,F)$.
Similarly, $d(x,F)leq d(x',x)+d(x',F)$.
Consequently, $|d(x',F)-d(x,F)|<d(x',x)$.

share|cite|improve this answer

answered Sep 24 '14 at 22:18

Fabio LucchiniFabio Lucchini

9,50111426

$endgroup$

share|cite|improve this answer

answered Sep 24 '14 at 22:18

Fabio LucchiniFabio Lucchini

9,50111426

answered Sep 24 '14 at 22:18

Fabio LucchiniFabio Lucchini

9,50111426

answered Sep 24 '14 at 22:18

Fabio LucchiniFabio Lucchini

9,50111426

0

$begingroup$

Here is a more worked example without "taking infimum" using the inverse triangle inequality:

We wish to show for all $x, y in M$, $lvert f_F(x) - f_X(y)rvert le d(x, y)$.

Case: $d(x, z) ge d(y, z)$.

Fix the $z$ such that both bounds are true. By the definitions of infimum, we have two bounds (let $epsilon > 0$):

beginalign
forall z in F quad &f_F(x) le d(x, z) \
exists z in F quad &f_F(x) + epsilon > d(x, z)
endalign

By the inverse triangle inequality,

beginalign
& d(x, z) - d(y,z) le d(x,y) \
implies & f_F(x) - d(y, z) le d(x,y) \
implies & f_F(x) - f_F(y) - epsilon le d(x,y) \
implies & f_F(x) - f_F(y) le d(x,y) + epsilon
endalign

Since this holds for all $epsilon > 0$, $f_F(x) - f_F(y) le d(x,y)$. The case of $d(x,z) < d(y,z)$ is similar, so we have proved $lvert f_F(x) - f_F(y)rvert le d(x, y)$. Now simply apply our definition of continuity using $delta = epsilon$ (actually $f_F$ is Lipschitz continuous with constant $1$).

share|cite|improve this answer

edited Mar 22 at 6:21

answered Mar 22 at 0:38

qwrqwr

6,68242755

$endgroup$

add a comment | 

0

$begingroup$

Here is a more worked example without "taking infimum" using the inverse triangle inequality:

We wish to show for all $x, y in M$, $lvert f_F(x) - f_X(y)rvert le d(x, y)$.

Case: $d(x, z) ge d(y, z)$.

Fix the $z$ such that both bounds are true. By the definitions of infimum, we have two bounds (let $epsilon > 0$):

beginalign
forall z in F quad &f_F(x) le d(x, z) \
exists z in F quad &f_F(x) + epsilon > d(x, z)
endalign

By the inverse triangle inequality,

beginalign
& d(x, z) - d(y,z) le d(x,y) \
implies & f_F(x) - d(y, z) le d(x,y) \
implies & f_F(x) - f_F(y) - epsilon le d(x,y) \
implies & f_F(x) - f_F(y) le d(x,y) + epsilon
endalign

Since this holds for all $epsilon > 0$, $f_F(x) - f_F(y) le d(x,y)$. The case of $d(x,z) < d(y,z)$ is similar, so we have proved $lvert f_F(x) - f_F(y)rvert le d(x, y)$. Now simply apply our definition of continuity using $delta = epsilon$ (actually $f_F$ is Lipschitz continuous with constant $1$).

share|cite|improve this answer

edited Mar 22 at 6:21

answered Mar 22 at 0:38

qwrqwr

6,68242755

$endgroup$

add a comment | 

$begingroup$

Here is a more worked example without "taking infimum" using the inverse triangle inequality:

We wish to show for all $x, y in M$, $lvert f_F(x) - f_X(y)rvert le d(x, y)$.

Case: $d(x, z) ge d(y, z)$.

Fix the $z$ such that both bounds are true. By the definitions of infimum, we have two bounds (let $epsilon > 0$):

beginalign
forall z in F quad &f_F(x) le d(x, z) \
exists z in F quad &f_F(x) + epsilon > d(x, z)
endalign

By the inverse triangle inequality,

beginalign
& d(x, z) - d(y,z) le d(x,y) \
implies & f_F(x) - d(y, z) le d(x,y) \
implies & f_F(x) - f_F(y) - epsilon le d(x,y) \
implies & f_F(x) - f_F(y) le d(x,y) + epsilon
endalign

Since this holds for all $epsilon > 0$, $f_F(x) - f_F(y) le d(x,y)$. The case of $d(x,z) < d(y,z)$ is similar, so we have proved $lvert f_F(x) - f_F(y)rvert le d(x, y)$. Now simply apply our definition of continuity using $delta = epsilon$ (actually $f_F$ is Lipschitz continuous with constant $1$).

share|cite|improve this answer

edited Mar 22 at 6:21

answered Mar 22 at 0:38

qwrqwr

6,68242755

$endgroup$

share|cite|improve this answer

edited Mar 22 at 6:21

answered Mar 22 at 0:38

qwrqwr

6,68242755

answered Mar 22 at 0:38

qwrqwr

6,68242755

answered Mar 22 at 0:38

qwrqwr

6,68242755