Dtjytyk

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$A$ closed, $B$ compact. $A$ and $B$ are disjoint. Prove that the distance between them is positive.

Prove that the distance between a disjoint compact set and closed set is nonzero.Prove $X$ is compact if every family of closed sets with the finite intersection property have non-empty intersectionProve that if $f$ is a continuous mapping of a compact metric space $X$ into $mathbb{R}^k$, then $f(X)$ is closed and bounded.Prove compact subsets of metric spaces are closedDistance of compact sets- Proof without subsequence convergence and continuityClosed subsets of a compact metric space with distance zero from each otherProve that if $A subset C$ is bounded and closed, then $A$ is compact. (Proof guidance)Prove that if $A,B$ are closed and disjoint in a metric space, then they are contained in disjoint neighborhoodsClosed subsets of compact sets are compact (original proof)Is this proof that a continuous image of a compact metric space is compact correct?

1

$begingroup$

Let $M$ be a metric space, $Asubset M, Bsubset M, Acap B = emptyset, A text{closed}, B text{compact}.$

Define distance between sets as $D(A,B) = inf d(a,b)$ where $d$ is the metric on $M$.

I'm asked to prove that $D(A,B) > 0$ which is equivalent to proving that $D(A,B) neq 0.$

My attempt:

Assume $D(A,B) = 0.$

This means that $exists (a,b) s.t. d(a,b) < epsilon forall epsilon > 0.$ Let $a,b$ denote these specific $a,b$.

Take a sequence in $A$ including $a$ as last element. This sequence must converge to $b$ because of the above statement. This means there exists a sequence in $A$ converging to an element not in $A$. This is not possible since $A$ is closed.

Thus $D(A,B) neq 0.$

Is this proof correct? If it is not, how can I modify it / rewrite it. If it is, why do I need to know that $B$ is compact?

general-topology proof-verification convergence compactness

share|cite|improve this question

asked 2 days ago

HeuristicsHeuristics

536212

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You didn't justify why a sequence in $A$ could converge to a point in $B$. Your points $a$ and $b$ vary with $epsilon$.
  $endgroup$
  – Umberto P.
  2 days ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $M$ be a metric space, $Asubset M, Bsubset M, Acap B = emptyset, A text{closed}, B text{compact}.$

Define distance between sets as $D(A,B) = inf d(a,b)$ where $d$ is the metric on $M$.

I'm asked to prove that $D(A,B) > 0$ which is equivalent to proving that $D(A,B) neq 0.$

My attempt:

Assume $D(A,B) = 0.$

This means that $exists (a,b) s.t. d(a,b) < epsilon forall epsilon > 0.$ Let $a,b$ denote these specific $a,b$.

Take a sequence in $A$ including $a$ as last element. This sequence must converge to $b$ because of the above statement. This means there exists a sequence in $A$ converging to an element not in $A$. This is not possible since $A$ is closed.

Thus $D(A,B) neq 0.$

Is this proof correct? If it is not, how can I modify it / rewrite it. If it is, why do I need to know that $B$ is compact?

general-topology proof-verification convergence compactness

share|cite|improve this question

asked 2 days ago

HeuristicsHeuristics

536212

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You didn't justify why a sequence in $A$ could converge to a point in $B$. Your points $a$ and $b$ vary with $epsilon$.
  $endgroup$
  – Umberto P.
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $M$ be a metric space, $Asubset M, Bsubset M, Acap B = emptyset, A text{closed}, B text{compact}.$

Define distance between sets as $D(A,B) = inf d(a,b)$ where $d$ is the metric on $M$.

I'm asked to prove that $D(A,B) > 0$ which is equivalent to proving that $D(A,B) neq 0.$

My attempt:

Assume $D(A,B) = 0.$

This means that $exists (a,b) s.t. d(a,b) < epsilon forall epsilon > 0.$ Let $a,b$ denote these specific $a,b$.

Take a sequence in $A$ including $a$ as last element. This sequence must converge to $b$ because of the above statement. This means there exists a sequence in $A$ converging to an element not in $A$. This is not possible since $A$ is closed.

Thus $D(A,B) neq 0.$

Is this proof correct? If it is not, how can I modify it / rewrite it. If it is, why do I need to know that $B$ is compact?

general-topology proof-verification convergence compactness

share|cite|improve this question

asked 2 days ago

HeuristicsHeuristics

536212

$endgroup$

share|cite|improve this question

asked 2 days ago

HeuristicsHeuristics

536212

share|cite|improve this question

asked 2 days ago

HeuristicsHeuristics

536212

asked 2 days ago

HeuristicsHeuristics

536212

asked 2 days ago

HeuristicsHeuristics

536212

2 Answers
2

active

oldest

votes

6

$begingroup$

What follows is valid in a metric space, but out of habit I write $|a-b|$ in place of $d(a,b)$.

Every point $x in B$ belongs to the complement of $A$ which is open. Thus, for every $x in B$ there exists $r_x > 0$ with the property that $B(x,r_x) cap A = emptyset$. In particular $|a - x| ge r_x$ for all $x$.

The family ${ B(x,r_x/2) mid x in B}$ covers $B$, and since $B$ is compact there exists a finite subcover which we can label by ${B(x_k,r_k/2)}_{k=1}^n$.

Define $r = min{r_1,ldots,r_n}$ and note $r > 0$.

Select a point $b in B$ and $a in A$.

There exists an index $k$ with the property that $b in B(x_k,r_k/2)$, so that $|b - x_k| < r_k/2$.

For the same index $k$ you have also that $|a - x_k| ge r_k$.

The triangle inequality gives you
$$|a - b| ge |a - x_k| - |b - x_k| > frac{r_k}2 ge frac r2.$$

Since $a,b$ were arbitrary it follows that $d(A,B) ge dfrac r2$.

share|cite|improve this answer

answered 2 days ago

Umberto P.Umberto P.

39.8k13267

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, but where is my proof wrong?
  $endgroup$
  – Heuristics
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Heuristics a sequence has no last element. You need more work on the sequence argument. It can be made to work. E.g. use compactness.
  $endgroup$
  – Henno Brandsma
  2 days ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

For any set $A subseteq X$, the function $f_A:X to [0,infty), x to d(x,A)$ defined on a metric space $X$, is continuous. Also $f_A(p)=0$ iff $p in overline{A}$.

Now apply this for $A$, the closed set, and restrict the domain to the compact set $B$, say. Then $f_A |_{B}$ assumes a minimum $m in [0,infty)$ which cannot be $0$ as $f_p(A)=0$ for $p in A$ iff $p in overline{A} cap B = A cap B = emptyset$. It follows that for all $x in B$, $d(x,A) ge m$, so that $D(A,B) =m >0$.

share|cite|improve this answer

answered 2 days ago

Henno BrandsmaHenno Brandsma

112k348120

$endgroup$

add a comment | 

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6

$begingroup$

What follows is valid in a metric space, but out of habit I write $|a-b|$ in place of $d(a,b)$.

Every point $x in B$ belongs to the complement of $A$ which is open. Thus, for every $x in B$ there exists $r_x > 0$ with the property that $B(x,r_x) cap A = emptyset$. In particular $|a - x| ge r_x$ for all $x$.

The family ${ B(x,r_x/2) mid x in B}$ covers $B$, and since $B$ is compact there exists a finite subcover which we can label by ${B(x_k,r_k/2)}_{k=1}^n$.

Define $r = min{r_1,ldots,r_n}$ and note $r > 0$.

Select a point $b in B$ and $a in A$.

There exists an index $k$ with the property that $b in B(x_k,r_k/2)$, so that $|b - x_k| < r_k/2$.

For the same index $k$ you have also that $|a - x_k| ge r_k$.

The triangle inequality gives you
$$|a - b| ge |a - x_k| - |b - x_k| > frac{r_k}2 ge frac r2.$$

Since $a,b$ were arbitrary it follows that $d(A,B) ge dfrac r2$.

share|cite|improve this answer

answered 2 days ago

Umberto P.Umberto P.

39.8k13267

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, but where is my proof wrong?
  $endgroup$
  – Heuristics
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Heuristics a sequence has no last element. You need more work on the sequence argument. It can be made to work. E.g. use compactness.
  $endgroup$
  – Henno Brandsma
  2 days ago
  
  
  

  
  
  
  

add a comment | 

6

$begingroup$

What follows is valid in a metric space, but out of habit I write $|a-b|$ in place of $d(a,b)$.

Every point $x in B$ belongs to the complement of $A$ which is open. Thus, for every $x in B$ there exists $r_x > 0$ with the property that $B(x,r_x) cap A = emptyset$. In particular $|a - x| ge r_x$ for all $x$.

The family ${ B(x,r_x/2) mid x in B}$ covers $B$, and since $B$ is compact there exists a finite subcover which we can label by ${B(x_k,r_k/2)}_{k=1}^n$.

Define $r = min{r_1,ldots,r_n}$ and note $r > 0$.

Select a point $b in B$ and $a in A$.

There exists an index $k$ with the property that $b in B(x_k,r_k/2)$, so that $|b - x_k| < r_k/2$.

For the same index $k$ you have also that $|a - x_k| ge r_k$.

The triangle inequality gives you
$$|a - b| ge |a - x_k| - |b - x_k| > frac{r_k}2 ge frac r2.$$

Since $a,b$ were arbitrary it follows that $d(A,B) ge dfrac r2$.

share|cite|improve this answer

answered 2 days ago

Umberto P.Umberto P.

39.8k13267

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, but where is my proof wrong?
  $endgroup$
  – Heuristics
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Heuristics a sequence has no last element. You need more work on the sequence argument. It can be made to work. E.g. use compactness.
  $endgroup$
  – Henno Brandsma
  2 days ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

What follows is valid in a metric space, but out of habit I write $|a-b|$ in place of $d(a,b)$.

Every point $x in B$ belongs to the complement of $A$ which is open. Thus, for every $x in B$ there exists $r_x > 0$ with the property that $B(x,r_x) cap A = emptyset$. In particular $|a - x| ge r_x$ for all $x$.

The family ${ B(x,r_x/2) mid x in B}$ covers $B$, and since $B$ is compact there exists a finite subcover which we can label by ${B(x_k,r_k/2)}_{k=1}^n$.

Define $r = min{r_1,ldots,r_n}$ and note $r > 0$.

Select a point $b in B$ and $a in A$.

There exists an index $k$ with the property that $b in B(x_k,r_k/2)$, so that $|b - x_k| < r_k/2$.

For the same index $k$ you have also that $|a - x_k| ge r_k$.

The triangle inequality gives you
$$|a - b| ge |a - x_k| - |b - x_k| > frac{r_k}2 ge frac r2.$$

Since $a,b$ were arbitrary it follows that $d(A,B) ge dfrac r2$.

share|cite|improve this answer

answered 2 days ago

Umberto P.Umberto P.

39.8k13267

$endgroup$

share|cite|improve this answer

answered 2 days ago

Umberto P.Umberto P.

39.8k13267

answered 2 days ago

Umberto P.Umberto P.

39.8k13267

answered 2 days ago

Umberto P.Umberto P.

39.8k13267

1

$begingroup$

For any set $A subseteq X$, the function $f_A:X to [0,infty), x to d(x,A)$ defined on a metric space $X$, is continuous. Also $f_A(p)=0$ iff $p in overline{A}$.

Now apply this for $A$, the closed set, and restrict the domain to the compact set $B$, say. Then $f_A |_{B}$ assumes a minimum $m in [0,infty)$ which cannot be $0$ as $f_p(A)=0$ for $p in A$ iff $p in overline{A} cap B = A cap B = emptyset$. It follows that for all $x in B$, $d(x,A) ge m$, so that $D(A,B) =m >0$.

share|cite|improve this answer

answered 2 days ago

Henno BrandsmaHenno Brandsma

112k348120

$endgroup$

add a comment | 

1

$begingroup$

For any set $A subseteq X$, the function $f_A:X to [0,infty), x to d(x,A)$ defined on a metric space $X$, is continuous. Also $f_A(p)=0$ iff $p in overline{A}$.

Now apply this for $A$, the closed set, and restrict the domain to the compact set $B$, say. Then $f_A |_{B}$ assumes a minimum $m in [0,infty)$ which cannot be $0$ as $f_p(A)=0$ for $p in A$ iff $p in overline{A} cap B = A cap B = emptyset$. It follows that for all $x in B$, $d(x,A) ge m$, so that $D(A,B) =m >0$.

share|cite|improve this answer

answered 2 days ago

Henno BrandsmaHenno Brandsma

112k348120

$endgroup$

add a comment | 

$begingroup$

For any set $A subseteq X$, the function $f_A:X to [0,infty), x to d(x,A)$ defined on a metric space $X$, is continuous. Also $f_A(p)=0$ iff $p in overline{A}$.

Now apply this for $A$, the closed set, and restrict the domain to the compact set $B$, say. Then $f_A |_{B}$ assumes a minimum $m in [0,infty)$ which cannot be $0$ as $f_p(A)=0$ for $p in A$ iff $p in overline{A} cap B = A cap B = emptyset$. It follows that for all $x in B$, $d(x,A) ge m$, so that $D(A,B) =m >0$.

share|cite|improve this answer

answered 2 days ago

Henno BrandsmaHenno Brandsma

112k348120

$endgroup$

share|cite|improve this answer

answered 2 days ago

Henno BrandsmaHenno Brandsma

112k348120

answered 2 days ago

Henno BrandsmaHenno Brandsma

112k348120

answered 2 days ago

Henno BrandsmaHenno Brandsma

112k348120