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Showing a metric space to be incomplete [duplicate]
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Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space
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I'm supposed to show that the set of real numbers, $R$, is an incomplete metric space with metric $d(x,y)=|text{tan}^{-1}(x)-text{tan}^{-1}(y)|$.
My issue: What we usually do in such problems is that we find a cauchy sequence which doesn't converge in same metric space. Now since all of the terms of the sequence $(x_n)_{ngeq1}$ are supposed to come from real numbers, shouldn't the limit itself be in real numbers too? Also, we were taught a theorem in class "Every Cauchy sequence of real numbers converges" which means that the limit will exist too. Then why would it be incomplete? Thanks!
real-analysis metric-spaces
asked Jan 28 at 9:49
Ankit KumarAnkit Kumar
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marked as duplicate by José Carlos Santos real-analysis
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Jan 28 at 9:52
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.
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$begingroup$
This question already has an answer here:
Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space
3 answers
I'm supposed to show that the set of real numbers, $R$, is an incomplete metric space with metric $d(x,y)=|text{tan}^{-1}(x)-text{tan}^{-1}(y)|$.
My issue: What we usually do in such problems is that we find a cauchy sequence which doesn't converge in same metric space. Now since all of the terms of the sequence $(x_n)_{ngeq1}$ are supposed to come from real numbers, shouldn't the limit itself be in real numbers too? Also, we were taught a theorem in class "Every Cauchy sequence of real numbers converges" which means that the limit will exist too. Then why would it be incomplete? Thanks!
real-analysis metric-spaces
asked Jan 28 at 9:49
Ankit KumarAnkit Kumar
1,542221
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marked as duplicate by José Carlos Santos real-analysis
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Jan 28 at 9:52
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.
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Have a look at math.stackexchange.com/questions/152243/…
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– Olba12
Jan 28 at 9:52
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$begingroup$
This question already has an answer here:
Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space
3 answers
I'm supposed to show that the set of real numbers, $R$, is an incomplete metric space with metric $d(x,y)=|text{tan}^{-1}(x)-text{tan}^{-1}(y)|$.
My issue: What we usually do in such problems is that we find a cauchy sequence which doesn't converge in same metric space. Now since all of the terms of the sequence $(x_n)_{ngeq1}$ are supposed to come from real numbers, shouldn't the limit itself be in real numbers too? Also, we were taught a theorem in class "Every Cauchy sequence of real numbers converges" which means that the limit will exist too. Then why would it be incomplete? Thanks!
real-analysis metric-spaces
asked Jan 28 at 9:49
Ankit KumarAnkit Kumar
1,542221
$endgroup$
This question already has an answer here:
Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space
3 answers
I'm supposed to show that the set of real numbers, $R$, is an incomplete metric space with metric $d(x,y)=|text{tan}^{-1}(x)-text{tan}^{-1}(y)|$.
My issue: What we usually do in such problems is that we find a cauchy sequence which doesn't converge in same metric space. Now since all of the terms of the sequence $(x_n)_{ngeq1}$ are supposed to come from real numbers, shouldn't the limit itself be in real numbers too? Also, we were taught a theorem in class "Every Cauchy sequence of real numbers converges" which means that the limit will exist too. Then why would it be incomplete? Thanks!
This question already has an answer here:
Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space
3 answers
real-analysis metric-spaces
real-analysis metric-spaces
asked Jan 28 at 9:49
Ankit KumarAnkit Kumar
1,542221
asked Jan 28 at 9:49
Ankit KumarAnkit Kumar
1,542221
asked Jan 28 at 9:49
Ankit KumarAnkit Kumar
1,542221
asked Jan 28 at 9:49
Ankit KumarAnkit Kumar
1,542221
asked Jan 28 at 9:49
Ankit KumarAnkit Kumar
1,542221
1,542221
marked as duplicate by José Carlos Santos real-analysis
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Jan 28 at 9:52
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 José Carlos Santos real-analysis
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Jan 28 at 9:52
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.
1
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Have a look at math.stackexchange.com/questions/152243/…
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– Olba12
Jan 28 at 9:52
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1
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Have a look at math.stackexchange.com/questions/152243/…
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– Olba12
Jan 28 at 9:52
1
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Have a look at math.stackexchange.com/questions/152243/…
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– Olba12
Jan 28 at 9:52
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Have a look at math.stackexchange.com/questions/152243/…
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– Olba12
Jan 28 at 9:52
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1 Answer
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Consider $u_n=n$, $u_n$ is a Cauchy sequence since $lim_ntan^{-1}(n)=pi/2$ and does not have a limit since there does not exists a number $x$ with $tan^{-1}(x)=pi/2$.
answered Jan 28 at 9:51
Tsemo AristideTsemo Aristide
59.9k11446
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Oh got it! So the convergence is determined by the metric right? Like "Every Cauchy sequence of real numbers converges" is probably true only for the usual metric. Am I correct?
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– Ankit Kumar
Jan 28 at 9:54
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1 Answer
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1 Answer
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$begingroup$
Consider $u_n=n$, $u_n$ is a Cauchy sequence since $lim_ntan^{-1}(n)=pi/2$ and does not have a limit since there does not exists a number $x$ with $tan^{-1}(x)=pi/2$.
answered Jan 28 at 9:51
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
$begingroup$
Oh got it! So the convergence is determined by the metric right? Like "Every Cauchy sequence of real numbers converges" is probably true only for the usual metric. Am I correct?
$endgroup$
– Ankit Kumar
Jan 28 at 9:54
add a comment |
$begingroup$
Consider $u_n=n$, $u_n$ is a Cauchy sequence since $lim_ntan^{-1}(n)=pi/2$ and does not have a limit since there does not exists a number $x$ with $tan^{-1}(x)=pi/2$.
answered Jan 28 at 9:51
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
$begingroup$
Oh got it! So the convergence is determined by the metric right? Like "Every Cauchy sequence of real numbers converges" is probably true only for the usual metric. Am I correct?
$endgroup$
– Ankit Kumar
Jan 28 at 9:54
add a comment |
$begingroup$
Consider $u_n=n$, $u_n$ is a Cauchy sequence since $lim_ntan^{-1}(n)=pi/2$ and does not have a limit since there does not exists a number $x$ with $tan^{-1}(x)=pi/2$.
answered Jan 28 at 9:51
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
Consider $u_n=n$, $u_n$ is a Cauchy sequence since $lim_ntan^{-1}(n)=pi/2$ and does not have a limit since there does not exists a number $x$ with $tan^{-1}(x)=pi/2$.
answered Jan 28 at 9:51
Tsemo AristideTsemo Aristide
59.9k11446
edited Jan 28 at 9:54
answered Jan 28 at 9:51
Tsemo AristideTsemo Aristide
59.9k11446
answered Jan 28 at 9:51
Tsemo AristideTsemo Aristide
59.9k11446
answered Jan 28 at 9:51
Tsemo AristideTsemo Aristide
59.9k11446
59.9k11446
$begingroup$
Oh got it! So the convergence is determined by the metric right? Like "Every Cauchy sequence of real numbers converges" is probably true only for the usual metric. Am I correct?
$endgroup$
– Ankit Kumar
Jan 28 at 9:54
add a comment |
$begingroup$
Oh got it! So the convergence is determined by the metric right? Like "Every Cauchy sequence of real numbers converges" is probably true only for the usual metric. Am I correct?
$endgroup$
– Ankit Kumar
Jan 28 at 9:54
$begingroup$
Oh got it! So the convergence is determined by the metric right? Like "Every Cauchy sequence of real numbers converges" is probably true only for the usual metric. Am I correct?
$endgroup$
– Ankit Kumar
Jan 28 at 9:54
$begingroup$
Oh got it! So the convergence is determined by the metric right? Like "Every Cauchy sequence of real numbers converges" is probably true only for the usual metric. Am I correct?
$endgroup$
– Ankit Kumar
Jan 28 at 9:54
add a comment |
1
$begingroup$
Have a look at math.stackexchange.com/questions/152243/…
$endgroup$
– Olba12
Jan 28 at 9:52