Mixing
Compactness in $mathbb R^n $
$begingroup$
I am not sure if I'm missing something subtle or I'm just totally wrong.
Is the following true?
$Ksubseteq mathbb R^n$ is compact (usual topology with 2-norm)
if and only if
$Ksubseteq mathbb R^n$ is compact under a norm on $mathbb R^n$?
I would think equivalence of norms would play a role here.
general-topology
asked Jan 23 at 18:20
chhrochhro
1,310311
$endgroup$
add a comment |
$begingroup$
I am not sure if I'm missing something subtle or I'm just totally wrong.
Is the following true?
$Ksubseteq mathbb R^n$ is compact (usual topology with 2-norm)
if and only if
$Ksubseteq mathbb R^n$ is compact under a norm on $mathbb R^n$?
I would think equivalence of norms would play a role here.
general-topology
asked Jan 23 at 18:20
chhrochhro
1,310311
$endgroup$
$begingroup$
The statement is true. As you said the norms are equivalent which means that the open sets with respect to all norms are the same. Therefore by the definition of compactness (any open cover has a finite sub-cover) we have that $K$ is compact with respect to all norms.
$endgroup$
– Yanko
Jan 23 at 18:24
$begingroup$
@Yanko, I see. When you put it that way, I think it's much clearer. I was thinking of applying Heine-Borel, and that's why I'm hung up on the details.
$endgroup$
– chhro
Jan 23 at 18:25
add a comment |
$begingroup$
I am not sure if I'm missing something subtle or I'm just totally wrong.
Is the following true?
$Ksubseteq mathbb R^n$ is compact (usual topology with 2-norm)
if and only if
$Ksubseteq mathbb R^n$ is compact under a norm on $mathbb R^n$?
I would think equivalence of norms would play a role here.
general-topology
asked Jan 23 at 18:20
chhrochhro
1,310311
$endgroup$
I am not sure if I'm missing something subtle or I'm just totally wrong.
Is the following true?
$Ksubseteq mathbb R^n$ is compact (usual topology with 2-norm)
if and only if
$Ksubseteq mathbb R^n$ is compact under a norm on $mathbb R^n$?
I would think equivalence of norms would play a role here.
general-topology
general-topology
asked Jan 23 at 18:20
chhrochhro
1,310311
asked Jan 23 at 18:20
chhrochhro
1,310311
asked Jan 23 at 18:20
chhrochhro
1,310311
asked Jan 23 at 18:20
chhrochhro
1,310311
asked Jan 23 at 18:20
chhrochhro
1,310311
1,310311
$begingroup$
The statement is true. As you said the norms are equivalent which means that the open sets with respect to all norms are the same. Therefore by the definition of compactness (any open cover has a finite sub-cover) we have that $K$ is compact with respect to all norms.
$endgroup$
– Yanko
Jan 23 at 18:24
$begingroup$
@Yanko, I see. When you put it that way, I think it's much clearer. I was thinking of applying Heine-Borel, and that's why I'm hung up on the details.
$endgroup$
– chhro
Jan 23 at 18:25
add a comment |
$begingroup$
The statement is true. As you said the norms are equivalent which means that the open sets with respect to all norms are the same. Therefore by the definition of compactness (any open cover has a finite sub-cover) we have that $K$ is compact with respect to all norms.
$endgroup$
– Yanko
Jan 23 at 18:24
$begingroup$
@Yanko, I see. When you put it that way, I think it's much clearer. I was thinking of applying Heine-Borel, and that's why I'm hung up on the details.
$endgroup$
– chhro
Jan 23 at 18:25
$begingroup$
The statement is true. As you said the norms are equivalent which means that the open sets with respect to all norms are the same. Therefore by the definition of compactness (any open cover has a finite sub-cover) we have that $K$ is compact with respect to all norms.
$endgroup$
– Yanko
Jan 23 at 18:24
$begingroup$
The statement is true. As you said the norms are equivalent which means that the open sets with respect to all norms are the same. Therefore by the definition of compactness (any open cover has a finite sub-cover) we have that $K$ is compact with respect to all norms.
$endgroup$
– Yanko
Jan 23 at 18:24
$begingroup$
@Yanko, I see. When you put it that way, I think it's much clearer. I was thinking of applying Heine-Borel, and that's why I'm hung up on the details.
$endgroup$
– chhro
Jan 23 at 18:25
$begingroup$
@Yanko, I see. When you put it that way, I think it's much clearer. I was thinking of applying Heine-Borel, and that's why I'm hung up on the details.
$endgroup$
– chhro
Jan 23 at 18:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is true. As you said this is due to the fact that in finite dimension all norms are equivalent.
So here if $K$ is compact it just means :
$$exists M, forall x in K, |x|_2 leq M text{ and } K text{ is closed }$$
Now take any norm $N$ on $mathbb{R}^n$. Since all norms are equivalent then there is such that : $forall x in mathbb{R}^n, | x |_2 leq C N(x)$. Hence $K$ is also bounded for the norm $N$.
You can check that "closeness" still holds on $K$ with $N$.
answered Jan 23 at 18:25
ThinkingThinking
1,22716
$endgroup$
$begingroup$
Great! I was more sure of the closed condition being true, and was more wary of the boundedness condition. Turned out to be fine as you said!
$endgroup$
– chhro
Jan 23 at 18:28
add a comment |
$begingroup$
All norms on $mathbb{R}^n$ are equivalent which means they yield the same topology and in particular the same compact subsets.
answered Jan 23 at 20:16
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
This is true. As you said this is due to the fact that in finite dimension all norms are equivalent.
So here if $K$ is compact it just means :
$$exists M, forall x in K, |x|_2 leq M text{ and } K text{ is closed }$$
Now take any norm $N$ on $mathbb{R}^n$. Since all norms are equivalent then there is such that : $forall x in mathbb{R}^n, | x |_2 leq C N(x)$. Hence $K$ is also bounded for the norm $N$.
You can check that "closeness" still holds on $K$ with $N$.
answered Jan 23 at 18:25
ThinkingThinking
1,22716
$endgroup$
$begingroup$
Great! I was more sure of the closed condition being true, and was more wary of the boundedness condition. Turned out to be fine as you said!
$endgroup$
– chhro
Jan 23 at 18:28
add a comment |
$begingroup$
This is true. As you said this is due to the fact that in finite dimension all norms are equivalent.
So here if $K$ is compact it just means :
$$exists M, forall x in K, |x|_2 leq M text{ and } K text{ is closed }$$
Now take any norm $N$ on $mathbb{R}^n$. Since all norms are equivalent then there is such that : $forall x in mathbb{R}^n, | x |_2 leq C N(x)$. Hence $K$ is also bounded for the norm $N$.
You can check that "closeness" still holds on $K$ with $N$.
answered Jan 23 at 18:25
ThinkingThinking
1,22716
$endgroup$
$begingroup$
Great! I was more sure of the closed condition being true, and was more wary of the boundedness condition. Turned out to be fine as you said!
$endgroup$
– chhro
Jan 23 at 18:28
add a comment |
$begingroup$
This is true. As you said this is due to the fact that in finite dimension all norms are equivalent.
So here if $K$ is compact it just means :
$$exists M, forall x in K, |x|_2 leq M text{ and } K text{ is closed }$$
Now take any norm $N$ on $mathbb{R}^n$. Since all norms are equivalent then there is such that : $forall x in mathbb{R}^n, | x |_2 leq C N(x)$. Hence $K$ is also bounded for the norm $N$.
You can check that "closeness" still holds on $K$ with $N$.
answered Jan 23 at 18:25
ThinkingThinking
1,22716
$endgroup$
This is true. As you said this is due to the fact that in finite dimension all norms are equivalent.
So here if $K$ is compact it just means :
$$exists M, forall x in K, |x|_2 leq M text{ and } K text{ is closed }$$
Now take any norm $N$ on $mathbb{R}^n$. Since all norms are equivalent then there is such that : $forall x in mathbb{R}^n, | x |_2 leq C N(x)$. Hence $K$ is also bounded for the norm $N$.
You can check that "closeness" still holds on $K$ with $N$.
answered Jan 23 at 18:25
ThinkingThinking
1,22716
answered Jan 23 at 18:25
ThinkingThinking
1,22716
answered Jan 23 at 18:25
ThinkingThinking
1,22716
answered Jan 23 at 18:25
ThinkingThinking
1,22716
1,22716
$begingroup$
Great! I was more sure of the closed condition being true, and was more wary of the boundedness condition. Turned out to be fine as you said!
$endgroup$
– chhro
Jan 23 at 18:28
add a comment |
$begingroup$
Great! I was more sure of the closed condition being true, and was more wary of the boundedness condition. Turned out to be fine as you said!
$endgroup$
– chhro
Jan 23 at 18:28
$begingroup$
Great! I was more sure of the closed condition being true, and was more wary of the boundedness condition. Turned out to be fine as you said!
$endgroup$
– chhro
Jan 23 at 18:28
$begingroup$
Great! I was more sure of the closed condition being true, and was more wary of the boundedness condition. Turned out to be fine as you said!
$endgroup$
– chhro
Jan 23 at 18:28
add a comment |
$begingroup$
All norms on $mathbb{R}^n$ are equivalent which means they yield the same topology and in particular the same compact subsets.
answered Jan 23 at 20:16
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
$begingroup$
All norms on $mathbb{R}^n$ are equivalent which means they yield the same topology and in particular the same compact subsets.
answered Jan 23 at 20:16
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
add a comment |
$begingroup$
All norms on $mathbb{R}^n$ are equivalent which means they yield the same topology and in particular the same compact subsets.
answered Jan 23 at 20:16
Henno BrandsmaHenno Brandsma
112k348121
$endgroup$
All norms on $mathbb{R}^n$ are equivalent which means they yield the same topology and in particular the same compact subsets.
answered Jan 23 at 20:16
Henno BrandsmaHenno Brandsma
112k348121
answered Jan 23 at 20:16
Henno BrandsmaHenno Brandsma
112k348121
answered Jan 23 at 20:16
Henno BrandsmaHenno Brandsma
112k348121
answered Jan 23 at 20:16
Henno BrandsmaHenno Brandsma
112k348121
112k348121
add a comment |
add a comment |
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$begingroup$
The statement is true. As you said the norms are equivalent which means that the open sets with respect to all norms are the same. Therefore by the definition of compactness (any open cover has a finite sub-cover) we have that $K$ is compact with respect to all norms.
$endgroup$
– Yanko
Jan 23 at 18:24
$begingroup$
@Yanko, I see. When you put it that way, I think it's much clearer. I was thinking of applying Heine-Borel, and that's why I'm hung up on the details.
$endgroup$
– chhro
Jan 23 at 18:25