Compactness of Hausdorff metric
Multi tool use
$begingroup$
Let $(X,d)$ be a compact metric space. Let $(K,h)$ be the space of non-empty compact subsets of $X$ with the Hausdorff metric. Show that $K$ is compact.
First of all, I've found 2 related questions on stackexchange, but the answers only hint that the limit of a Cauchy sequence ${A_n}_{ngeq1}$ in $K$ is $A={xin X:$ there exists a sequence ${a_n}_{ngeq 1}$ with $a_nin A_n$ such that it converges to $x}$. I've also found lecture slides from Harvard about the same problem but all of the sources think it's easy to show ${A_n}$ converges to $A$, which is not obviously to me at all. How can we bound the Hausdorff distance between $A_n$ and $A$ based on our definition of $A$?
real-analysis general-topology
asked Jan 17 at 19:00
Fluffy SkyeFluffy Skye
1148
$endgroup$
add a comment |
$begingroup$
Let $(X,d)$ be a compact metric space. Let $(K,h)$ be the space of non-empty compact subsets of $X$ with the Hausdorff metric. Show that $K$ is compact.
First of all, I've found 2 related questions on stackexchange, but the answers only hint that the limit of a Cauchy sequence ${A_n}_{ngeq1}$ in $K$ is $A={xin X:$ there exists a sequence ${a_n}_{ngeq 1}$ with $a_nin A_n$ such that it converges to $x}$. I've also found lecture slides from Harvard about the same problem but all of the sources think it's easy to show ${A_n}$ converges to $A$, which is not obviously to me at all. How can we bound the Hausdorff distance between $A_n$ and $A$ based on our definition of $A$?
real-analysis general-topology
asked Jan 17 at 19:00
Fluffy SkyeFluffy Skye
1148
$endgroup$
$begingroup$
Another way is to show that the Hausdorff metric induces the Vietoris topology on the hyperspace of non-empty compact subsets. In that case, compactness of $K$ is a simple consequence of Alexander's subbase lemma.
$endgroup$
– Henno Brandsma
Jan 17 at 19:04
$begingroup$
@Herno Brandsma I just learnt about compactness. I know nothing about the things you mentioned :(
$endgroup$
– Fluffy Skye
Jan 17 at 19:05
$begingroup$
Duplicate here: math.stackexchange.com/questions/181158/…
$endgroup$
– T. Fo
Jan 17 at 19:40
add a comment |
$begingroup$
Let $(X,d)$ be a compact metric space. Let $(K,h)$ be the space of non-empty compact subsets of $X$ with the Hausdorff metric. Show that $K$ is compact.
First of all, I've found 2 related questions on stackexchange, but the answers only hint that the limit of a Cauchy sequence ${A_n}_{ngeq1}$ in $K$ is $A={xin X:$ there exists a sequence ${a_n}_{ngeq 1}$ with $a_nin A_n$ such that it converges to $x}$. I've also found lecture slides from Harvard about the same problem but all of the sources think it's easy to show ${A_n}$ converges to $A$, which is not obviously to me at all. How can we bound the Hausdorff distance between $A_n$ and $A$ based on our definition of $A$?
real-analysis general-topology
asked Jan 17 at 19:00
Fluffy SkyeFluffy Skye
1148
$endgroup$
Let $(X,d)$ be a compact metric space. Let $(K,h)$ be the space of non-empty compact subsets of $X$ with the Hausdorff metric. Show that $K$ is compact.
First of all, I've found 2 related questions on stackexchange, but the answers only hint that the limit of a Cauchy sequence ${A_n}_{ngeq1}$ in $K$ is $A={xin X:$ there exists a sequence ${a_n}_{ngeq 1}$ with $a_nin A_n$ such that it converges to $x}$. I've also found lecture slides from Harvard about the same problem but all of the sources think it's easy to show ${A_n}$ converges to $A$, which is not obviously to me at all. How can we bound the Hausdorff distance between $A_n$ and $A$ based on our definition of $A$?
real-analysis general-topology
real-analysis general-topology
asked Jan 17 at 19:00
Fluffy SkyeFluffy Skye
1148
asked Jan 17 at 19:00
Fluffy SkyeFluffy Skye
1148
asked Jan 17 at 19:00
Fluffy SkyeFluffy Skye
1148
asked Jan 17 at 19:00
Fluffy SkyeFluffy Skye
1148
asked Jan 17 at 19:00
Fluffy SkyeFluffy Skye
1148
1148
$begingroup$
Another way is to show that the Hausdorff metric induces the Vietoris topology on the hyperspace of non-empty compact subsets. In that case, compactness of $K$ is a simple consequence of Alexander's subbase lemma.
$endgroup$
– Henno Brandsma
Jan 17 at 19:04
$begingroup$
@Herno Brandsma I just learnt about compactness. I know nothing about the things you mentioned :(
$endgroup$
– Fluffy Skye
Jan 17 at 19:05
$begingroup$
Duplicate here: math.stackexchange.com/questions/181158/…
$endgroup$
– T. Fo
Jan 17 at 19:40
add a comment |
$begingroup$
Another way is to show that the Hausdorff metric induces the Vietoris topology on the hyperspace of non-empty compact subsets. In that case, compactness of $K$ is a simple consequence of Alexander's subbase lemma.
$endgroup$
– Henno Brandsma
Jan 17 at 19:04
$begingroup$
@Herno Brandsma I just learnt about compactness. I know nothing about the things you mentioned :(
$endgroup$
– Fluffy Skye
Jan 17 at 19:05
$begingroup$
Duplicate here: math.stackexchange.com/questions/181158/…
$endgroup$
– T. Fo
Jan 17 at 19:40
$begingroup$
Another way is to show that the Hausdorff metric induces the Vietoris topology on the hyperspace of non-empty compact subsets. In that case, compactness of $K$ is a simple consequence of Alexander's subbase lemma.
$endgroup$
– Henno Brandsma
Jan 17 at 19:04
$begingroup$
Another way is to show that the Hausdorff metric induces the Vietoris topology on the hyperspace of non-empty compact subsets. In that case, compactness of $K$ is a simple consequence of Alexander's subbase lemma.
$endgroup$
– Henno Brandsma
Jan 17 at 19:04
$begingroup$
@Herno Brandsma I just learnt about compactness. I know nothing about the things you mentioned :(
$endgroup$
– Fluffy Skye
Jan 17 at 19:05
$begingroup$
@Herno Brandsma I just learnt about compactness. I know nothing about the things you mentioned :(
$endgroup$
– Fluffy Skye
Jan 17 at 19:05
$begingroup$
Duplicate here: math.stackexchange.com/questions/181158/…
$endgroup$
– T. Fo
Jan 17 at 19:40
$begingroup$
Duplicate here: math.stackexchange.com/questions/181158/…
$endgroup$
– T. Fo
Jan 17 at 19:40
add a comment |
1 Answer
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You have to show that $K_n$ converges in the Hausdorff sense to $A$, and also that $A$ is compact.
To show that $A $ is compact, it is sufficent to show that $A $ is closed. By sequential caracterisation, it not too hard to make it.
To show the other point, i will give some notation.
Let for $B$ included in $X$, $B_epsilon$ the union of balls of radius $epsilon $ centered in a point of $B $ (i.e. the points at à distance $leq epsilon $ of $B$).
Then the Hausdorff distance between 2 sets $B$ and $C$ is the infinimum of the epsilon such that $B_epsilon$ contains $C$ and $C_epsilon$ contains $B$.
You can show that the distance between $B$ and $C$ is the same as the distance between the closure of $B$ and the closure of $C$
$(K_n)$ is Cauchy. So for $n$ and $m$ greater than a $n_0$, $K_n$ and $K_m$ are at a Hausdorff distance $leq epsilon$. So it is the same for $K_n$ and the closure of the union of $K_{n'}$ for n' greater than $n_0$. This last set contains $A $, so you can include $A $ in ${K_n}_epsilon $.
The other inclusion, let $x$ in $K_n$ and build à sequence $(x_{n'}) $ of elements in $K_{n'}$ at à distance $leq epsilon$ of $x$. Extract a convergent subsequence, then the limit will be in $A$ and at a distance $leq epsilon$ of $x$. This concludes for the inclusion $K_n$ in $A_epsilon$.
Then, to conclude, you need to show that a sequence of compacts has Cauchy subsequence.
Hint to show it :
first show it if $X$ is a finite set with the discrete metric.
then, cover $X$ by à finite number of balls of radius < 1 ; by using this and the preceding point (look at the indices of the balls which intersects with $K_n$), you can extract a subsequence of $(K_n)$ on which the terms are espaced of a hausdorff distance <1. Then continue with $1/2^n$ and do a diagonal argument.
answered Jan 17 at 20:16
DLeMeurDLeMeur
3168
$endgroup$
$begingroup$
For the second inclusion. How do you extract a convergent subsequence? Generally it's not true that a closed all in a metric space is compact.
$endgroup$
– Fluffy Skye
Jan 17 at 22:58
$begingroup$
You lost me when you introduced Kn' for proving the first inclusion. What is the same for Kn and union of Kn'?
$endgroup$
– Fluffy Skye
Jan 17 at 23:05
$begingroup$
But it is true that a closed space included in a compact space is compact. (That is why I said it is sufficent to show A closed to show A compact here). Kn and the (closure of) the union of Kn' are still at a Hausdorff distance $leq epsilon $.
$endgroup$
– DLeMeur
Jan 17 at 23:20
add a comment |
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$begingroup$
You have to show that $K_n$ converges in the Hausdorff sense to $A$, and also that $A$ is compact.
To show that $A $ is compact, it is sufficent to show that $A $ is closed. By sequential caracterisation, it not too hard to make it.
To show the other point, i will give some notation.
Let for $B$ included in $X$, $B_epsilon$ the union of balls of radius $epsilon $ centered in a point of $B $ (i.e. the points at à distance $leq epsilon $ of $B$).
Then the Hausdorff distance between 2 sets $B$ and $C$ is the infinimum of the epsilon such that $B_epsilon$ contains $C$ and $C_epsilon$ contains $B$.
You can show that the distance between $B$ and $C$ is the same as the distance between the closure of $B$ and the closure of $C$
$(K_n)$ is Cauchy. So for $n$ and $m$ greater than a $n_0$, $K_n$ and $K_m$ are at a Hausdorff distance $leq epsilon$. So it is the same for $K_n$ and the closure of the union of $K_{n'}$ for n' greater than $n_0$. This last set contains $A $, so you can include $A $ in ${K_n}_epsilon $.
The other inclusion, let $x$ in $K_n$ and build à sequence $(x_{n'}) $ of elements in $K_{n'}$ at à distance $leq epsilon$ of $x$. Extract a convergent subsequence, then the limit will be in $A$ and at a distance $leq epsilon$ of $x$. This concludes for the inclusion $K_n$ in $A_epsilon$.
Then, to conclude, you need to show that a sequence of compacts has Cauchy subsequence.
Hint to show it :
first show it if $X$ is a finite set with the discrete metric.
then, cover $X$ by à finite number of balls of radius < 1 ; by using this and the preceding point (look at the indices of the balls which intersects with $K_n$), you can extract a subsequence of $(K_n)$ on which the terms are espaced of a hausdorff distance <1. Then continue with $1/2^n$ and do a diagonal argument.
answered Jan 17 at 20:16
DLeMeurDLeMeur
3168
$endgroup$
$begingroup$
For the second inclusion. How do you extract a convergent subsequence? Generally it's not true that a closed all in a metric space is compact.
$endgroup$
– Fluffy Skye
Jan 17 at 22:58
$begingroup$
You lost me when you introduced Kn' for proving the first inclusion. What is the same for Kn and union of Kn'?
$endgroup$
– Fluffy Skye
Jan 17 at 23:05
$begingroup$
But it is true that a closed space included in a compact space is compact. (That is why I said it is sufficent to show A closed to show A compact here). Kn and the (closure of) the union of Kn' are still at a Hausdorff distance $leq epsilon $.
$endgroup$
– DLeMeur
Jan 17 at 23:20
add a comment |
$begingroup$
You have to show that $K_n$ converges in the Hausdorff sense to $A$, and also that $A$ is compact.
To show that $A $ is compact, it is sufficent to show that $A $ is closed. By sequential caracterisation, it not too hard to make it.
To show the other point, i will give some notation.
Let for $B$ included in $X$, $B_epsilon$ the union of balls of radius $epsilon $ centered in a point of $B $ (i.e. the points at à distance $leq epsilon $ of $B$).
Then the Hausdorff distance between 2 sets $B$ and $C$ is the infinimum of the epsilon such that $B_epsilon$ contains $C$ and $C_epsilon$ contains $B$.
You can show that the distance between $B$ and $C$ is the same as the distance between the closure of $B$ and the closure of $C$
$(K_n)$ is Cauchy. So for $n$ and $m$ greater than a $n_0$, $K_n$ and $K_m$ are at a Hausdorff distance $leq epsilon$. So it is the same for $K_n$ and the closure of the union of $K_{n'}$ for n' greater than $n_0$. This last set contains $A $, so you can include $A $ in ${K_n}_epsilon $.
The other inclusion, let $x$ in $K_n$ and build à sequence $(x_{n'}) $ of elements in $K_{n'}$ at à distance $leq epsilon$ of $x$. Extract a convergent subsequence, then the limit will be in $A$ and at a distance $leq epsilon$ of $x$. This concludes for the inclusion $K_n$ in $A_epsilon$.
Then, to conclude, you need to show that a sequence of compacts has Cauchy subsequence.
Hint to show it :
first show it if $X$ is a finite set with the discrete metric.
then, cover $X$ by à finite number of balls of radius < 1 ; by using this and the preceding point (look at the indices of the balls which intersects with $K_n$), you can extract a subsequence of $(K_n)$ on which the terms are espaced of a hausdorff distance <1. Then continue with $1/2^n$ and do a diagonal argument.
answered Jan 17 at 20:16
DLeMeurDLeMeur
3168
$endgroup$
$begingroup$
For the second inclusion. How do you extract a convergent subsequence? Generally it's not true that a closed all in a metric space is compact.
$endgroup$
– Fluffy Skye
Jan 17 at 22:58
$begingroup$
You lost me when you introduced Kn' for proving the first inclusion. What is the same for Kn and union of Kn'?
$endgroup$
– Fluffy Skye
Jan 17 at 23:05
$begingroup$
But it is true that a closed space included in a compact space is compact. (That is why I said it is sufficent to show A closed to show A compact here). Kn and the (closure of) the union of Kn' are still at a Hausdorff distance $leq epsilon $.
$endgroup$
– DLeMeur
Jan 17 at 23:20
add a comment |
$begingroup$
You have to show that $K_n$ converges in the Hausdorff sense to $A$, and also that $A$ is compact.
To show that $A $ is compact, it is sufficent to show that $A $ is closed. By sequential caracterisation, it not too hard to make it.
To show the other point, i will give some notation.
Let for $B$ included in $X$, $B_epsilon$ the union of balls of radius $epsilon $ centered in a point of $B $ (i.e. the points at à distance $leq epsilon $ of $B$).
Then the Hausdorff distance between 2 sets $B$ and $C$ is the infinimum of the epsilon such that $B_epsilon$ contains $C$ and $C_epsilon$ contains $B$.
You can show that the distance between $B$ and $C$ is the same as the distance between the closure of $B$ and the closure of $C$
$(K_n)$ is Cauchy. So for $n$ and $m$ greater than a $n_0$, $K_n$ and $K_m$ are at a Hausdorff distance $leq epsilon$. So it is the same for $K_n$ and the closure of the union of $K_{n'}$ for n' greater than $n_0$. This last set contains $A $, so you can include $A $ in ${K_n}_epsilon $.
The other inclusion, let $x$ in $K_n$ and build à sequence $(x_{n'}) $ of elements in $K_{n'}$ at à distance $leq epsilon$ of $x$. Extract a convergent subsequence, then the limit will be in $A$ and at a distance $leq epsilon$ of $x$. This concludes for the inclusion $K_n$ in $A_epsilon$.
Then, to conclude, you need to show that a sequence of compacts has Cauchy subsequence.
Hint to show it :
first show it if $X$ is a finite set with the discrete metric.
then, cover $X$ by à finite number of balls of radius < 1 ; by using this and the preceding point (look at the indices of the balls which intersects with $K_n$), you can extract a subsequence of $(K_n)$ on which the terms are espaced of a hausdorff distance <1. Then continue with $1/2^n$ and do a diagonal argument.
answered Jan 17 at 20:16
DLeMeurDLeMeur
3168
$endgroup$
You have to show that $K_n$ converges in the Hausdorff sense to $A$, and also that $A$ is compact.
To show that $A $ is compact, it is sufficent to show that $A $ is closed. By sequential caracterisation, it not too hard to make it.
To show the other point, i will give some notation.
Let for $B$ included in $X$, $B_epsilon$ the union of balls of radius $epsilon $ centered in a point of $B $ (i.e. the points at à distance $leq epsilon $ of $B$).
Then the Hausdorff distance between 2 sets $B$ and $C$ is the infinimum of the epsilon such that $B_epsilon$ contains $C$ and $C_epsilon$ contains $B$.
You can show that the distance between $B$ and $C$ is the same as the distance between the closure of $B$ and the closure of $C$
$(K_n)$ is Cauchy. So for $n$ and $m$ greater than a $n_0$, $K_n$ and $K_m$ are at a Hausdorff distance $leq epsilon$. So it is the same for $K_n$ and the closure of the union of $K_{n'}$ for n' greater than $n_0$. This last set contains $A $, so you can include $A $ in ${K_n}_epsilon $.
The other inclusion, let $x$ in $K_n$ and build à sequence $(x_{n'}) $ of elements in $K_{n'}$ at à distance $leq epsilon$ of $x$. Extract a convergent subsequence, then the limit will be in $A$ and at a distance $leq epsilon$ of $x$. This concludes for the inclusion $K_n$ in $A_epsilon$.
Then, to conclude, you need to show that a sequence of compacts has Cauchy subsequence.
Hint to show it :
first show it if $X$ is a finite set with the discrete metric.
then, cover $X$ by à finite number of balls of radius < 1 ; by using this and the preceding point (look at the indices of the balls which intersects with $K_n$), you can extract a subsequence of $(K_n)$ on which the terms are espaced of a hausdorff distance <1. Then continue with $1/2^n$ and do a diagonal argument.
answered Jan 17 at 20:16
DLeMeurDLeMeur
3168
edited Jan 17 at 20:28
answered Jan 17 at 20:16
DLeMeurDLeMeur
3168
answered Jan 17 at 20:16
DLeMeurDLeMeur
3168
answered Jan 17 at 20:16
DLeMeurDLeMeur
3168
3168
$begingroup$
For the second inclusion. How do you extract a convergent subsequence? Generally it's not true that a closed all in a metric space is compact.
$endgroup$
– Fluffy Skye
Jan 17 at 22:58
$begingroup$
You lost me when you introduced Kn' for proving the first inclusion. What is the same for Kn and union of Kn'?
$endgroup$
– Fluffy Skye
Jan 17 at 23:05
$begingroup$
But it is true that a closed space included in a compact space is compact. (That is why I said it is sufficent to show A closed to show A compact here). Kn and the (closure of) the union of Kn' are still at a Hausdorff distance $leq epsilon $.
$endgroup$
– DLeMeur
Jan 17 at 23:20
add a comment |
$begingroup$
For the second inclusion. How do you extract a convergent subsequence? Generally it's not true that a closed all in a metric space is compact.
$endgroup$
– Fluffy Skye
Jan 17 at 22:58
$begingroup$
You lost me when you introduced Kn' for proving the first inclusion. What is the same for Kn and union of Kn'?
$endgroup$
– Fluffy Skye
Jan 17 at 23:05
$begingroup$
But it is true that a closed space included in a compact space is compact. (That is why I said it is sufficent to show A closed to show A compact here). Kn and the (closure of) the union of Kn' are still at a Hausdorff distance $leq epsilon $.
$endgroup$
– DLeMeur
Jan 17 at 23:20
$begingroup$
For the second inclusion. How do you extract a convergent subsequence? Generally it's not true that a closed all in a metric space is compact.
$endgroup$
– Fluffy Skye
Jan 17 at 22:58
$begingroup$
For the second inclusion. How do you extract a convergent subsequence? Generally it's not true that a closed all in a metric space is compact.
$endgroup$
– Fluffy Skye
Jan 17 at 22:58
$begingroup$
You lost me when you introduced Kn' for proving the first inclusion. What is the same for Kn and union of Kn'?
$endgroup$
– Fluffy Skye
Jan 17 at 23:05
$begingroup$
You lost me when you introduced Kn' for proving the first inclusion. What is the same for Kn and union of Kn'?
$endgroup$
– Fluffy Skye
Jan 17 at 23:05
$begingroup$
But it is true that a closed space included in a compact space is compact. (That is why I said it is sufficent to show A closed to show A compact here). Kn and the (closure of) the union of Kn' are still at a Hausdorff distance $leq epsilon $.
$endgroup$
– DLeMeur
Jan 17 at 23:20
$begingroup$
But it is true that a closed space included in a compact space is compact. (That is why I said it is sufficent to show A closed to show A compact here). Kn and the (closure of) the union of Kn' are still at a Hausdorff distance $leq epsilon $.
$endgroup$
– DLeMeur
Jan 17 at 23:20
add a comment |
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LMeSoF,7,4Ou qANDR7,07KQ,j w,Cs6efl2d0KyQ zS4GGUaUxn8OZZIq,atfBXWqC6SsJq
$begingroup$
Another way is to show that the Hausdorff metric induces the Vietoris topology on the hyperspace of non-empty compact subsets. In that case, compactness of $K$ is a simple consequence of Alexander's subbase lemma.
$endgroup$
– Henno Brandsma
Jan 17 at 19:04
$begingroup$
@Herno Brandsma I just learnt about compactness. I know nothing about the things you mentioned :(
$endgroup$
– Fluffy Skye
Jan 17 at 19:05
$begingroup$
Duplicate here: math.stackexchange.com/questions/181158/…
$endgroup$
– T. Fo
Jan 17 at 19:40