A question about metric space
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Let $$X = left{ (x_n)_{n=1}^{infty} mid x_n in mathbb{R}, quad 0 leq x_n leq 1 quad forall n in mathbb{N} right}.$$
The set $X$ equipped with metric $d(x,y) = sup { mid x_n - y_n mid : n in mathbb{N} }$.
I know that:
The metric space $Y$ is compact if and only if the every sequence in $Y$ has a convergent sub-sequence.
But I could not prove, why $X$ is not compact by defined metric?
Please help me.
real-analysis functional-analysis analysis
edited Nov 22 at 14:57
Davide Giraudo
124k16150259
asked Nov 22 at 14:33
joe
874
add a comment |
up vote
-1
down vote
favorite
Let $$X = left{ (x_n)_{n=1}^{infty} mid x_n in mathbb{R}, quad 0 leq x_n leq 1 quad forall n in mathbb{N} right}.$$
The set $X$ equipped with metric $d(x,y) = sup { mid x_n - y_n mid : n in mathbb{N} }$.
I know that:
The metric space $Y$ is compact if and only if the every sequence in $Y$ has a convergent sub-sequence.
But I could not prove, why $X$ is not compact by defined metric?
Please help me.
real-analysis functional-analysis analysis
edited Nov 22 at 14:57
Davide Giraudo
124k16150259
asked Nov 22 at 14:33
joe
874
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:51
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $$X = left{ (x_n)_{n=1}^{infty} mid x_n in mathbb{R}, quad 0 leq x_n leq 1 quad forall n in mathbb{N} right}.$$
The set $X$ equipped with metric $d(x,y) = sup { mid x_n - y_n mid : n in mathbb{N} }$.
I know that:
The metric space $Y$ is compact if and only if the every sequence in $Y$ has a convergent sub-sequence.
But I could not prove, why $X$ is not compact by defined metric?
Please help me.
real-analysis functional-analysis analysis
edited Nov 22 at 14:57
Davide Giraudo
124k16150259
asked Nov 22 at 14:33
joe
874
Let $$X = left{ (x_n)_{n=1}^{infty} mid x_n in mathbb{R}, quad 0 leq x_n leq 1 quad forall n in mathbb{N} right}.$$
The set $X$ equipped with metric $d(x,y) = sup { mid x_n - y_n mid : n in mathbb{N} }$.
I know that:
The metric space $Y$ is compact if and only if the every sequence in $Y$ has a convergent sub-sequence.
But I could not prove, why $X$ is not compact by defined metric?
Please help me.
real-analysis functional-analysis analysis
real-analysis functional-analysis analysis
edited Nov 22 at 14:57
Davide Giraudo
124k16150259
asked Nov 22 at 14:33
joe
874
edited Nov 22 at 14:57
Davide Giraudo
124k16150259
asked Nov 22 at 14:33
joe
874
edited Nov 22 at 14:57
Davide Giraudo
124k16150259
edited Nov 22 at 14:57
Davide Giraudo
124k16150259
edited Nov 22 at 14:57
Davide Giraudo
124k16150259
124k16150259
asked Nov 22 at 14:33
joe
874
asked Nov 22 at 14:33
joe
874
asked Nov 22 at 14:33
joe
874
874
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:51
add a comment |
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:51
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:51
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:51
add a comment |
2 Answers
2
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up vote
3
down vote
You can prove the space is not compact by finding a sequence without a convergent subsequence.
Hint:
Like a lot of times, you should try your search for a counterexample by looking at simple things.
Think about very simple vectors in $X$. Very simple ones. With a whole lot of zeroes. Like, as many zeroes as you can get (i.e., with as few nonzero elements as possible). Also, to make them even simpler, have the nonzero element be something simple. Like a constant value. Some simple constant value, so it'll be easy to calculate with it.
answered Nov 22 at 14:37
5xum
89.4k393161
add a comment |
up vote
0
down vote
That is not compact .
Counterexample: $x_n:-n^{th}$ coordinate is 1 else $0$.
answered Nov 22 at 14:50
Shubham
1,5921519
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You can prove the space is not compact by finding a sequence without a convergent subsequence.
Hint:
Like a lot of times, you should try your search for a counterexample by looking at simple things.
Think about very simple vectors in $X$. Very simple ones. With a whole lot of zeroes. Like, as many zeroes as you can get (i.e., with as few nonzero elements as possible). Also, to make them even simpler, have the nonzero element be something simple. Like a constant value. Some simple constant value, so it'll be easy to calculate with it.
answered Nov 22 at 14:37
5xum
89.4k393161
add a comment |
up vote
3
down vote
You can prove the space is not compact by finding a sequence without a convergent subsequence.
Hint:
Like a lot of times, you should try your search for a counterexample by looking at simple things.
Think about very simple vectors in $X$. Very simple ones. With a whole lot of zeroes. Like, as many zeroes as you can get (i.e., with as few nonzero elements as possible). Also, to make them even simpler, have the nonzero element be something simple. Like a constant value. Some simple constant value, so it'll be easy to calculate with it.
answered Nov 22 at 14:37
5xum
89.4k393161
add a comment |
up vote
3
down vote
up vote
3
down vote
You can prove the space is not compact by finding a sequence without a convergent subsequence.
Hint:
Like a lot of times, you should try your search for a counterexample by looking at simple things.
Think about very simple vectors in $X$. Very simple ones. With a whole lot of zeroes. Like, as many zeroes as you can get (i.e., with as few nonzero elements as possible). Also, to make them even simpler, have the nonzero element be something simple. Like a constant value. Some simple constant value, so it'll be easy to calculate with it.
answered Nov 22 at 14:37
5xum
89.4k393161
You can prove the space is not compact by finding a sequence without a convergent subsequence.
Hint:
Like a lot of times, you should try your search for a counterexample by looking at simple things.
Think about very simple vectors in $X$. Very simple ones. With a whole lot of zeroes. Like, as many zeroes as you can get (i.e., with as few nonzero elements as possible). Also, to make them even simpler, have the nonzero element be something simple. Like a constant value. Some simple constant value, so it'll be easy to calculate with it.
answered Nov 22 at 14:37
5xum
89.4k393161
answered Nov 22 at 14:37
5xum
89.4k393161
answered Nov 22 at 14:37
5xum
89.4k393161
answered Nov 22 at 14:37
5xum
89.4k393161
89.4k393161
add a comment |
add a comment |
up vote
0
down vote
That is not compact .
Counterexample: $x_n:-n^{th}$ coordinate is 1 else $0$.
answered Nov 22 at 14:50
Shubham
1,5921519
add a comment |
up vote
0
down vote
That is not compact .
Counterexample: $x_n:-n^{th}$ coordinate is 1 else $0$.
answered Nov 22 at 14:50
Shubham
1,5921519
add a comment |
up vote
0
down vote
up vote
0
down vote
That is not compact .
Counterexample: $x_n:-n^{th}$ coordinate is 1 else $0$.
answered Nov 22 at 14:50
Shubham
1,5921519
That is not compact .
Counterexample: $x_n:-n^{th}$ coordinate is 1 else $0$.
answered Nov 22 at 14:50
Shubham
1,5921519
answered Nov 22 at 14:50
Shubham
1,5921519
answered Nov 22 at 14:50
Shubham
1,5921519
answered Nov 22 at 14:50
Shubham
1,5921519
1,5921519
add a comment |
add a comment |
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You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:51