Proving $B={f:fin C_{[0,1]}text{ and }:d(f,0)leqslant 1}$ is not compact
$begingroup$
Let $(C_{[0,1]},d)$ be the metric space defined with the supremum metric.
Let $B={f:fin C_{[0,1]}text{ and}:d(f,0)leqslant 1}$ where $0$ denotes the constant function form $[0,1]$ into $mathbb{R}$.
Prove that $B$ is not compact.
Since we are working on a metric space I can use the compactness definition that every convergent sequence has a convergent subsequence so I am trying to find a function that acts as a counterexample to prove the space is not compact.
Question:
Is my strategy right? If so, how should I find the sequence of functions?
Thanks in advance!
general-topology metric-spaces
$endgroup$
add a comment |
$begingroup$
Let $(C_{[0,1]},d)$ be the metric space defined with the supremum metric.
Let $B={f:fin C_{[0,1]}text{ and}:d(f,0)leqslant 1}$ where $0$ denotes the constant function form $[0,1]$ into $mathbb{R}$.
Prove that $B$ is not compact.
Since we are working on a metric space I can use the compactness definition that every convergent sequence has a convergent subsequence so I am trying to find a function that acts as a counterexample to prove the space is not compact.
Question:
Is my strategy right? If so, how should I find the sequence of functions?
Thanks in advance!
general-topology metric-spaces
$endgroup$
add a comment |
$begingroup$
Let $(C_{[0,1]},d)$ be the metric space defined with the supremum metric.
Let $B={f:fin C_{[0,1]}text{ and}:d(f,0)leqslant 1}$ where $0$ denotes the constant function form $[0,1]$ into $mathbb{R}$.
Prove that $B$ is not compact.
Since we are working on a metric space I can use the compactness definition that every convergent sequence has a convergent subsequence so I am trying to find a function that acts as a counterexample to prove the space is not compact.
Question:
Is my strategy right? If so, how should I find the sequence of functions?
Thanks in advance!
general-topology metric-spaces
$endgroup$
Let $(C_{[0,1]},d)$ be the metric space defined with the supremum metric.
Let $B={f:fin C_{[0,1]}text{ and}:d(f,0)leqslant 1}$ where $0$ denotes the constant function form $[0,1]$ into $mathbb{R}$.
Prove that $B$ is not compact.
Since we are working on a metric space I can use the compactness definition that every convergent sequence has a convergent subsequence so I am trying to find a function that acts as a counterexample to prove the space is not compact.
Question:
Is my strategy right? If so, how should I find the sequence of functions?
Thanks in advance!
general-topology metric-spaces
general-topology metric-spaces
edited Dec 16 '18 at 22:10
Bernard
123k741116
123k741116
asked Dec 16 '18 at 21:55
Pedro GomesPedro Gomes
1,9252721
1,9252721
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1 Answer
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$begingroup$
You can indeed use a sequence, e.g. $f_n(x) = x^n$ is a good candidate.
$endgroup$
$begingroup$
Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 16 '18 at 21:59
$begingroup$
Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:00
$begingroup$
So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
$endgroup$
– Pedro Gomes
Dec 16 '18 at 22:21
$begingroup$
The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
$endgroup$
– Matematleta
Dec 16 '18 at 22:27
$begingroup$
@PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:27
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
You can indeed use a sequence, e.g. $f_n(x) = x^n$ is a good candidate.
$endgroup$
$begingroup$
Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 16 '18 at 21:59
$begingroup$
Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:00
$begingroup$
So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
$endgroup$
– Pedro Gomes
Dec 16 '18 at 22:21
$begingroup$
The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
$endgroup$
– Matematleta
Dec 16 '18 at 22:27
$begingroup$
@PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:27
add a comment |
$begingroup$
You can indeed use a sequence, e.g. $f_n(x) = x^n$ is a good candidate.
$endgroup$
$begingroup$
Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 16 '18 at 21:59
$begingroup$
Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:00
$begingroup$
So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
$endgroup$
– Pedro Gomes
Dec 16 '18 at 22:21
$begingroup$
The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
$endgroup$
– Matematleta
Dec 16 '18 at 22:27
$begingroup$
@PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:27
add a comment |
$begingroup$
You can indeed use a sequence, e.g. $f_n(x) = x^n$ is a good candidate.
$endgroup$
You can indeed use a sequence, e.g. $f_n(x) = x^n$ is a good candidate.
edited Dec 17 '18 at 12:30
answered Dec 16 '18 at 21:57
Henno BrandsmaHenno Brandsma
112k348121
112k348121
$begingroup$
Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 16 '18 at 21:59
$begingroup$
Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:00
$begingroup$
So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
$endgroup$
– Pedro Gomes
Dec 16 '18 at 22:21
$begingroup$
The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
$endgroup$
– Matematleta
Dec 16 '18 at 22:27
$begingroup$
@PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:27
add a comment |
$begingroup$
Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 16 '18 at 21:59
$begingroup$
Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:00
$begingroup$
So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
$endgroup$
– Pedro Gomes
Dec 16 '18 at 22:21
$begingroup$
The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
$endgroup$
– Matematleta
Dec 16 '18 at 22:27
$begingroup$
@PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:27
$begingroup$
Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 16 '18 at 21:59
$begingroup$
Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
$endgroup$
– Pedro Gomes
Dec 16 '18 at 21:59
$begingroup$
Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:00
$begingroup$
Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:00
$begingroup$
So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
$endgroup$
– Pedro Gomes
Dec 16 '18 at 22:21
$begingroup$
So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
$endgroup$
– Pedro Gomes
Dec 16 '18 at 22:21
$begingroup$
The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
$endgroup$
– Matematleta
Dec 16 '18 at 22:27
$begingroup$
The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
$endgroup$
– Matematleta
Dec 16 '18 at 22:27
$begingroup$
@PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:27
$begingroup$
@PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
$endgroup$
– Henno Brandsma
Dec 16 '18 at 22:27
add a comment |
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