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Every compact metric space is complete, proof verification [duplicate]
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This question already has an answer here:
Every compact metric space is complete
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I want to give a proof, that every compact metric space is complete.
Proof:
Let $(X,d)$ be a compact metric space. Let $(x_n)_ninmathbbN$ be a Cauchy sequence.
The theorem of Bolzano-Weierstraß yields, that $(x_n)$ has a convergent subsequence $(x_n_k)_n_kinmathbbN$.
Note $x:=lim_n_ktoinfty x_n_k$.
I want to show, that $d(x_n, x)to 0$.
Let $epsilon >0$ be arbitrary.
Since $d(x_n_k, x)to 0$ exists for $epsilon/2 >0$ a $N'inmathbbN$ such that $d(x_n_k, x)<epsilon/2$ for every $n_kgeq N'$.
Likewise since $(x_n)$ is a Cauchy sequence, we find for $epsilon/2>0$ a $N''inmathbbN$ such that for every $n,mgeq N''$ holds, that $d(x_n,x_m)<epsilon/2$.
Take $N:=maxN', N''$ and we conclude the proof:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
Therefor $lim_ntoinfty x_nto x$ converges and $(X,d)$ is complete.
I would appreciate your thoughts on my proof.
Thanks in advance.
proof-verification metric-spaces
asked Jul 20 at 9:14
Cornman
2,47721127
marked as duplicate by uniquesolution, Delta-u, amWhy, user133281, Parcly Taxel Jul 20 at 17:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
 |Â
show 2 more comments
up vote
0
down vote
favorite
This question already has an answer here:
Every compact metric space is complete
6 answers
I want to give a proof, that every compact metric space is complete.
Proof:
Let $(X,d)$ be a compact metric space. Let $(x_n)_ninmathbbN$ be a Cauchy sequence.
The theorem of Bolzano-Weierstraß yields, that $(x_n)$ has a convergent subsequence $(x_n_k)_n_kinmathbbN$.
Note $x:=lim_n_ktoinfty x_n_k$.
I want to show, that $d(x_n, x)to 0$.
Let $epsilon >0$ be arbitrary.
Since $d(x_n_k, x)to 0$ exists for $epsilon/2 >0$ a $N'inmathbbN$ such that $d(x_n_k, x)<epsilon/2$ for every $n_kgeq N'$.
Likewise since $(x_n)$ is a Cauchy sequence, we find for $epsilon/2>0$ a $N''inmathbbN$ such that for every $n,mgeq N''$ holds, that $d(x_n,x_m)<epsilon/2$.
Take $N:=maxN', N''$ and we conclude the proof:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
Therefor $lim_ntoinfty x_nto x$ converges and $(X,d)$ is complete.
I would appreciate your thoughts on my proof.
Thanks in advance.
proof-verification metric-spaces
asked Jul 20 at 9:14
Cornman
2,47721127
marked as duplicate by uniquesolution, Delta-u, amWhy, user133281, Parcly Taxel Jul 20 at 17:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
What is Bolzano Weierstrass theorem say in a metric space?
– Kavi Rama Murthy
Jul 20 at 9:18
1
@KaviRamaMurthy: Every sequence in a compact metric space has a convergent subsequence.
– Mathematician 42
Jul 20 at 9:19
@Mathematician42 $B-W$ is a theorem applied on $mathbb R^n$. Not a general metric space as far as I know.
– Vera
Jul 20 at 9:19
In my lecture notes it got called like that. It is the theorem Mathematican 42 stated.
– Cornman
Jul 20 at 9:20
It's simply called "sequentially compact" I guess, but you could call the equivalence of compactness and sequentially compactness in metric spaces a Bolzano-Weierstrass type theorem.
– Mathematician 42
Jul 20 at 9:20
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Every compact metric space is complete
6 answers
I want to give a proof, that every compact metric space is complete.
Proof:
Let $(X,d)$ be a compact metric space. Let $(x_n)_ninmathbbN$ be a Cauchy sequence.
The theorem of Bolzano-Weierstraß yields, that $(x_n)$ has a convergent subsequence $(x_n_k)_n_kinmathbbN$.
Note $x:=lim_n_ktoinfty x_n_k$.
I want to show, that $d(x_n, x)to 0$.
Let $epsilon >0$ be arbitrary.
Since $d(x_n_k, x)to 0$ exists for $epsilon/2 >0$ a $N'inmathbbN$ such that $d(x_n_k, x)<epsilon/2$ for every $n_kgeq N'$.
Likewise since $(x_n)$ is a Cauchy sequence, we find for $epsilon/2>0$ a $N''inmathbbN$ such that for every $n,mgeq N''$ holds, that $d(x_n,x_m)<epsilon/2$.
Take $N:=maxN', N''$ and we conclude the proof:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
Therefor $lim_ntoinfty x_nto x$ converges and $(X,d)$ is complete.
I would appreciate your thoughts on my proof.
Thanks in advance.
proof-verification metric-spaces
asked Jul 20 at 9:14
Cornman
2,47721127
This question already has an answer here:
Every compact metric space is complete
6 answers
I want to give a proof, that every compact metric space is complete.
Proof:
Let $(X,d)$ be a compact metric space. Let $(x_n)_ninmathbbN$ be a Cauchy sequence.
The theorem of Bolzano-Weierstraß yields, that $(x_n)$ has a convergent subsequence $(x_n_k)_n_kinmathbbN$.
Note $x:=lim_n_ktoinfty x_n_k$.
I want to show, that $d(x_n, x)to 0$.
Let $epsilon >0$ be arbitrary.
Since $d(x_n_k, x)to 0$ exists for $epsilon/2 >0$ a $N'inmathbbN$ such that $d(x_n_k, x)<epsilon/2$ for every $n_kgeq N'$.
Likewise since $(x_n)$ is a Cauchy sequence, we find for $epsilon/2>0$ a $N''inmathbbN$ such that for every $n,mgeq N''$ holds, that $d(x_n,x_m)<epsilon/2$.
Take $N:=maxN', N''$ and we conclude the proof:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
Therefor $lim_ntoinfty x_nto x$ converges and $(X,d)$ is complete.
I would appreciate your thoughts on my proof.
Thanks in advance.
This question already has an answer here:
Every compact metric space is complete
6 answers
proof-verification metric-spaces
asked Jul 20 at 9:14
Cornman
2,47721127
asked Jul 20 at 9:14
Cornman
2,47721127
asked Jul 20 at 9:14
Cornman
2,47721127
asked Jul 20 at 9:14
Cornman
2,47721127
2,47721127
marked as duplicate by uniquesolution, Delta-u, amWhy, user133281, Parcly Taxel Jul 20 at 17:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by uniquesolution, Delta-u, amWhy, user133281, Parcly Taxel Jul 20 at 17:01
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
What is Bolzano Weierstrass theorem say in a metric space?
– Kavi Rama Murthy
Jul 20 at 9:18
1
@KaviRamaMurthy: Every sequence in a compact metric space has a convergent subsequence.
– Mathematician 42
Jul 20 at 9:19
@Mathematician42 $B-W$ is a theorem applied on $mathbb R^n$. Not a general metric space as far as I know.
– Vera
Jul 20 at 9:19
In my lecture notes it got called like that. It is the theorem Mathematican 42 stated.
– Cornman
Jul 20 at 9:20
It's simply called "sequentially compact" I guess, but you could call the equivalence of compactness and sequentially compactness in metric spaces a Bolzano-Weierstrass type theorem.
– Mathematician 42
Jul 20 at 9:20
 |Â
show 2 more comments
What is Bolzano Weierstrass theorem say in a metric space?
– Kavi Rama Murthy
Jul 20 at 9:18
1
@KaviRamaMurthy: Every sequence in a compact metric space has a convergent subsequence.
– Mathematician 42
Jul 20 at 9:19
@Mathematician42 $B-W$ is a theorem applied on $mathbb R^n$. Not a general metric space as far as I know.
– Vera
Jul 20 at 9:19
In my lecture notes it got called like that. It is the theorem Mathematican 42 stated.
– Cornman
Jul 20 at 9:20
It's simply called "sequentially compact" I guess, but you could call the equivalence of compactness and sequentially compactness in metric spaces a Bolzano-Weierstrass type theorem.
– Mathematician 42
Jul 20 at 9:20
What is Bolzano Weierstrass theorem say in a metric space?
– Kavi Rama Murthy
Jul 20 at 9:18
What is Bolzano Weierstrass theorem say in a metric space?
– Kavi Rama Murthy
Jul 20 at 9:18
1
1
@KaviRamaMurthy: Every sequence in a compact metric space has a convergent subsequence.
– Mathematician 42
Jul 20 at 9:19
@KaviRamaMurthy: Every sequence in a compact metric space has a convergent subsequence.
– Mathematician 42
Jul 20 at 9:19
@Mathematician42 $B-W$ is a theorem applied on $mathbb R^n$. Not a general metric space as far as I know.
– Vera
Jul 20 at 9:19
@Mathematician42 $B-W$ is a theorem applied on $mathbb R^n$. Not a general metric space as far as I know.
– Vera
Jul 20 at 9:19
In my lecture notes it got called like that. It is the theorem Mathematican 42 stated.
– Cornman
Jul 20 at 9:20
In my lecture notes it got called like that. It is the theorem Mathematican 42 stated.
– Cornman
Jul 20 at 9:20
It's simply called "sequentially compact" I guess, but you could call the equivalence of compactness and sequentially compactness in metric spaces a Bolzano-Weierstrass type theorem.
– Mathematician 42
Jul 20 at 9:20
It's simply called "sequentially compact" I guess, but you could call the equivalence of compactness and sequentially compactness in metric spaces a Bolzano-Weierstrass type theorem.
– Mathematician 42
Jul 20 at 9:20
 |Â
show 2 more comments
1 Answer
1
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up vote
1
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You wrote:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
You get a correct proof if you wrote: if $nge N$ and $n_k ge N$ then:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
answered Jul 20 at 9:17
Fred
37.3k1237
The starting point of the proof of OP is wrong. B-W theorem is only for Euclidean spaces.
– Kavi Rama Murthy
Jul 20 at 9:20
Since $X$ is compact, every sequence in $X$ has a convergent subsequence.
– Fred
Jul 20 at 9:23
But my choice of $N$ is correct, and everything that leads up to it? I just had to mention what you stated?
– Cornman
Jul 20 at 9:24
1
Yes, your choice of $N$ is correct. Just mention what I have stated.
– Fred
Jul 20 at 9:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You wrote:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
You get a correct proof if you wrote: if $nge N$ and $n_k ge N$ then:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
answered Jul 20 at 9:17
Fred
37.3k1237
The starting point of the proof of OP is wrong. B-W theorem is only for Euclidean spaces.
– Kavi Rama Murthy
Jul 20 at 9:20
Since $X$ is compact, every sequence in $X$ has a convergent subsequence.
– Fred
Jul 20 at 9:23
But my choice of $N$ is correct, and everything that leads up to it? I just had to mention what you stated?
– Cornman
Jul 20 at 9:24
1
Yes, your choice of $N$ is correct. Just mention what I have stated.
– Fred
Jul 20 at 9:26
add a comment |Â
up vote
1
down vote
accepted
You wrote:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
You get a correct proof if you wrote: if $nge N$ and $n_k ge N$ then:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
answered Jul 20 at 9:17
Fred
37.3k1237
The starting point of the proof of OP is wrong. B-W theorem is only for Euclidean spaces.
– Kavi Rama Murthy
Jul 20 at 9:20
Since $X$ is compact, every sequence in $X$ has a convergent subsequence.
– Fred
Jul 20 at 9:23
But my choice of $N$ is correct, and everything that leads up to it? I just had to mention what you stated?
– Cornman
Jul 20 at 9:24
1
Yes, your choice of $N$ is correct. Just mention what I have stated.
– Fred
Jul 20 at 9:26
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You wrote:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
You get a correct proof if you wrote: if $nge N$ and $n_k ge N$ then:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
answered Jul 20 at 9:17
Fred
37.3k1237
You wrote:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
You get a correct proof if you wrote: if $nge N$ and $n_k ge N$ then:
$d(x_n, x)leq d(x_n, x_n_k)+d(x_n_k,x)<epsilon/2+epsilon/2=epsilon$.
answered Jul 20 at 9:17
Fred
37.3k1237
answered Jul 20 at 9:17
Fred
37.3k1237
answered Jul 20 at 9:17
Fred
37.3k1237
answered Jul 20 at 9:17
Fred
37.3k1237
37.3k1237
The starting point of the proof of OP is wrong. B-W theorem is only for Euclidean spaces.
– Kavi Rama Murthy
Jul 20 at 9:20
Since $X$ is compact, every sequence in $X$ has a convergent subsequence.
– Fred
Jul 20 at 9:23
But my choice of $N$ is correct, and everything that leads up to it? I just had to mention what you stated?
– Cornman
Jul 20 at 9:24
1
Yes, your choice of $N$ is correct. Just mention what I have stated.
– Fred
Jul 20 at 9:26
add a comment |Â
The starting point of the proof of OP is wrong. B-W theorem is only for Euclidean spaces.
– Kavi Rama Murthy
Jul 20 at 9:20
Since $X$ is compact, every sequence in $X$ has a convergent subsequence.
– Fred
Jul 20 at 9:23
But my choice of $N$ is correct, and everything that leads up to it? I just had to mention what you stated?
– Cornman
Jul 20 at 9:24
1
Yes, your choice of $N$ is correct. Just mention what I have stated.
– Fred
Jul 20 at 9:26
The starting point of the proof of OP is wrong. B-W theorem is only for Euclidean spaces.
– Kavi Rama Murthy
Jul 20 at 9:20
The starting point of the proof of OP is wrong. B-W theorem is only for Euclidean spaces.
– Kavi Rama Murthy
Jul 20 at 9:20
Since $X$ is compact, every sequence in $X$ has a convergent subsequence.
– Fred
Jul 20 at 9:23
Since $X$ is compact, every sequence in $X$ has a convergent subsequence.
– Fred
Jul 20 at 9:23
But my choice of $N$ is correct, and everything that leads up to it? I just had to mention what you stated?
– Cornman
Jul 20 at 9:24
But my choice of $N$ is correct, and everything that leads up to it? I just had to mention what you stated?
– Cornman
Jul 20 at 9:24
1
1
Yes, your choice of $N$ is correct. Just mention what I have stated.
– Fred
Jul 20 at 9:26
Yes, your choice of $N$ is correct. Just mention what I have stated.
– Fred
Jul 20 at 9:26
add a comment |Â
What is Bolzano Weierstrass theorem say in a metric space?
– Kavi Rama Murthy
Jul 20 at 9:18
1
@KaviRamaMurthy: Every sequence in a compact metric space has a convergent subsequence.
– Mathematician 42
Jul 20 at 9:19
@Mathematician42 $B-W$ is a theorem applied on $mathbb R^n$. Not a general metric space as far as I know.
– Vera
Jul 20 at 9:19
In my lecture notes it got called like that. It is the theorem Mathematican 42 stated.
– Cornman
Jul 20 at 9:20
It's simply called "sequentially compact" I guess, but you could call the equivalence of compactness and sequentially compactness in metric spaces a Bolzano-Weierstrass type theorem.
– Mathematician 42
Jul 20 at 9:20