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Find non convergent Cauchy sequence
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Let $B$ be the set of all bounded real sequences. Prove that $(B,rho)$, where $rho(x,y):=sup_ndfracx_n-y_nn$ is a non complete metric space.
I could show that this is a metric space and I also know that with the metric $d(x,y):=sup_n|x_n-y_n|$ this space is complete. How do I prove that it is not complete with $rho$?
functional-analysis convergence metric-spaces cauchy-sequences complete-spaces
asked Aug 6 at 15:49
mathstackuser
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up vote
2
down vote
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Let $B$ be the set of all bounded real sequences. Prove that $(B,rho)$, where $rho(x,y):=sup_ndfracx_n-y_nn$ is a non complete metric space.
I could show that this is a metric space and I also know that with the metric $d(x,y):=sup_n|x_n-y_n|$ this space is complete. How do I prove that it is not complete with $rho$?
functional-analysis convergence metric-spaces cauchy-sequences complete-spaces
asked Aug 6 at 15:49
mathstackuser
61011
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $B$ be the set of all bounded real sequences. Prove that $(B,rho)$, where $rho(x,y):=sup_ndfracx_n-y_nn$ is a non complete metric space.
I could show that this is a metric space and I also know that with the metric $d(x,y):=sup_n|x_n-y_n|$ this space is complete. How do I prove that it is not complete with $rho$?
functional-analysis convergence metric-spaces cauchy-sequences complete-spaces
asked Aug 6 at 15:49
mathstackuser
61011
Let $B$ be the set of all bounded real sequences. Prove that $(B,rho)$, where $rho(x,y):=sup_ndfracx_n-y_nn$ is a non complete metric space.
I could show that this is a metric space and I also know that with the metric $d(x,y):=sup_n|x_n-y_n|$ this space is complete. How do I prove that it is not complete with $rho$?
functional-analysis convergence metric-spaces cauchy-sequences complete-spaces
asked Aug 6 at 15:49
mathstackuser
61011
asked Aug 6 at 15:49
mathstackuser
61011
asked Aug 6 at 15:49
mathstackuser
61011
asked Aug 6 at 15:49
mathstackuser
61011
61011
add a comment |Â
add a comment |Â
1 Answer
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3
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The intuition you should have here is that the metric $rho$ reduces the information you have about the tails of the sequences you're looking at so we should try to make something bad happen there.
For $k leq n$, let $x_k^(n) = k^1/2$ and otherwise let $x_k^(n) = 0$. So then, for fixed $n$, $x^(n) in B$. It is straightforward to show that $(x^(n))_n geq 1$ is a Cauchy sequence for $rho$ (in fact, this is why I chose something like $k^1/2$). I claim that $x^(n)$ does not converge in $(B, rho)$ as $n to infty$.
To see this, first notice that if $rho(x^(n), x) to 0$ then in particular, for each fixed $k$, $x_k^(n) to x_k$. As a result, if $x^(n) to x$ in $(B, rho)$ then we must have $x = (1, 2^1/2, 3^1/2, 4^1/2, dots)$. But this is not a bounded sequence and hence doesn't lie in $B$, giving us a contradiction.
answered Aug 6 at 16:14
Rhys Steele
5,6101828
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The intuition you should have here is that the metric $rho$ reduces the information you have about the tails of the sequences you're looking at so we should try to make something bad happen there.
For $k leq n$, let $x_k^(n) = k^1/2$ and otherwise let $x_k^(n) = 0$. So then, for fixed $n$, $x^(n) in B$. It is straightforward to show that $(x^(n))_n geq 1$ is a Cauchy sequence for $rho$ (in fact, this is why I chose something like $k^1/2$). I claim that $x^(n)$ does not converge in $(B, rho)$ as $n to infty$.
To see this, first notice that if $rho(x^(n), x) to 0$ then in particular, for each fixed $k$, $x_k^(n) to x_k$. As a result, if $x^(n) to x$ in $(B, rho)$ then we must have $x = (1, 2^1/2, 3^1/2, 4^1/2, dots)$. But this is not a bounded sequence and hence doesn't lie in $B$, giving us a contradiction.
answered Aug 6 at 16:14
Rhys Steele
5,6101828
add a comment |Â
up vote
3
down vote
accepted
The intuition you should have here is that the metric $rho$ reduces the information you have about the tails of the sequences you're looking at so we should try to make something bad happen there.
For $k leq n$, let $x_k^(n) = k^1/2$ and otherwise let $x_k^(n) = 0$. So then, for fixed $n$, $x^(n) in B$. It is straightforward to show that $(x^(n))_n geq 1$ is a Cauchy sequence for $rho$ (in fact, this is why I chose something like $k^1/2$). I claim that $x^(n)$ does not converge in $(B, rho)$ as $n to infty$.
To see this, first notice that if $rho(x^(n), x) to 0$ then in particular, for each fixed $k$, $x_k^(n) to x_k$. As a result, if $x^(n) to x$ in $(B, rho)$ then we must have $x = (1, 2^1/2, 3^1/2, 4^1/2, dots)$. But this is not a bounded sequence and hence doesn't lie in $B$, giving us a contradiction.
answered Aug 6 at 16:14
Rhys Steele
5,6101828
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The intuition you should have here is that the metric $rho$ reduces the information you have about the tails of the sequences you're looking at so we should try to make something bad happen there.
For $k leq n$, let $x_k^(n) = k^1/2$ and otherwise let $x_k^(n) = 0$. So then, for fixed $n$, $x^(n) in B$. It is straightforward to show that $(x^(n))_n geq 1$ is a Cauchy sequence for $rho$ (in fact, this is why I chose something like $k^1/2$). I claim that $x^(n)$ does not converge in $(B, rho)$ as $n to infty$.
To see this, first notice that if $rho(x^(n), x) to 0$ then in particular, for each fixed $k$, $x_k^(n) to x_k$. As a result, if $x^(n) to x$ in $(B, rho)$ then we must have $x = (1, 2^1/2, 3^1/2, 4^1/2, dots)$. But this is not a bounded sequence and hence doesn't lie in $B$, giving us a contradiction.
answered Aug 6 at 16:14
Rhys Steele
5,6101828
The intuition you should have here is that the metric $rho$ reduces the information you have about the tails of the sequences you're looking at so we should try to make something bad happen there.
For $k leq n$, let $x_k^(n) = k^1/2$ and otherwise let $x_k^(n) = 0$. So then, for fixed $n$, $x^(n) in B$. It is straightforward to show that $(x^(n))_n geq 1$ is a Cauchy sequence for $rho$ (in fact, this is why I chose something like $k^1/2$). I claim that $x^(n)$ does not converge in $(B, rho)$ as $n to infty$.
To see this, first notice that if $rho(x^(n), x) to 0$ then in particular, for each fixed $k$, $x_k^(n) to x_k$. As a result, if $x^(n) to x$ in $(B, rho)$ then we must have $x = (1, 2^1/2, 3^1/2, 4^1/2, dots)$. But this is not a bounded sequence and hence doesn't lie in $B$, giving us a contradiction.
answered Aug 6 at 16:14
Rhys Steele
5,6101828
answered Aug 6 at 16:14
Rhys Steele
5,6101828
answered Aug 6 at 16:14
Rhys Steele
5,6101828
answered Aug 6 at 16:14
Rhys Steele
5,6101828
5,6101828
add a comment |Â
add a comment |Â
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