Proving that the non-negative real line is complete [duplicate]
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This question already has an answer here:
Showing that if a subset of a complete metric space is closed, it is also complete
5 answers
Heads up: I am very new to abstract algebra and proofs.
Take the non-negative real line $X = 0 cup mathbb{R}^+=[0, +infty)$. We know that $Xsubset mathbb{R}$ and that $mathbb{R}$ is complete. Define now the Euclidean metric $d(x,y) = lvert x - y rvert$. Then $(X,d)$ spans a metric space. I wish to show that this metric space is complete.
So for an arbitrary sequence $(x_n)$ we need $d(x_n, x) < varepsilon$, where $xin X$, that is the sequence has to converge to $x$ which is in the set we are working with. Then $(x_n)$ is Cauchy, and since the sequence was arbitrary $X$ is complete.
This is what I know we require, however, I am unsure if I am on the right track.
My attempt:
Consider a sequence that is Cauchy, $(x_m)$. Then for every $varepsilon$, there is an $N=N(varepsilon)$ such that $$d(x_m, x_r) = lvert x_m - x_rrvert < varepsilon .$$ As $mto infty$ $x_mto x$. How do I prove that $xin X$ so that the arbitrary Cauchy sequence converges to a point in $X$ (thereby completing the proof)?
Any help would be appreciated.
metric-spaces complete-spaces
edited Dec 7 '18 at 9:45
José Carlos Santos
159k22126231
asked Dec 7 '18 at 9:41
Sindre Bakke ØyenSindre Bakke Øyen
132
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marked as duplicate by Jean-Claude Arbaut, Community♦ Dec 7 '18 at 10:09
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.
add a comment |
$begingroup$
This question already has an answer here:
Showing that if a subset of a complete metric space is closed, it is also complete
5 answers
Heads up: I am very new to abstract algebra and proofs.
Take the non-negative real line $X = 0 cup mathbb{R}^+=[0, +infty)$. We know that $Xsubset mathbb{R}$ and that $mathbb{R}$ is complete. Define now the Euclidean metric $d(x,y) = lvert x - y rvert$. Then $(X,d)$ spans a metric space. I wish to show that this metric space is complete.
So for an arbitrary sequence $(x_n)$ we need $d(x_n, x) < varepsilon$, where $xin X$, that is the sequence has to converge to $x$ which is in the set we are working with. Then $(x_n)$ is Cauchy, and since the sequence was arbitrary $X$ is complete.
This is what I know we require, however, I am unsure if I am on the right track.
My attempt:
Consider a sequence that is Cauchy, $(x_m)$. Then for every $varepsilon$, there is an $N=N(varepsilon)$ such that $$d(x_m, x_r) = lvert x_m - x_rrvert < varepsilon .$$ As $mto infty$ $x_mto x$. How do I prove that $xin X$ so that the arbitrary Cauchy sequence converges to a point in $X$ (thereby completing the proof)?
Any help would be appreciated.
metric-spaces complete-spaces
edited Dec 7 '18 at 9:45
José Carlos Santos
159k22126231
asked Dec 7 '18 at 9:41
Sindre Bakke ØyenSindre Bakke Øyen
132
$endgroup$
marked as duplicate by Jean-Claude Arbaut, Community♦ Dec 7 '18 at 10:09
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.
add a comment |
$begingroup$
This question already has an answer here:
Showing that if a subset of a complete metric space is closed, it is also complete
5 answers
Heads up: I am very new to abstract algebra and proofs.
Take the non-negative real line $X = 0 cup mathbb{R}^+=[0, +infty)$. We know that $Xsubset mathbb{R}$ and that $mathbb{R}$ is complete. Define now the Euclidean metric $d(x,y) = lvert x - y rvert$. Then $(X,d)$ spans a metric space. I wish to show that this metric space is complete.
So for an arbitrary sequence $(x_n)$ we need $d(x_n, x) < varepsilon$, where $xin X$, that is the sequence has to converge to $x$ which is in the set we are working with. Then $(x_n)$ is Cauchy, and since the sequence was arbitrary $X$ is complete.
This is what I know we require, however, I am unsure if I am on the right track.
My attempt:
Consider a sequence that is Cauchy, $(x_m)$. Then for every $varepsilon$, there is an $N=N(varepsilon)$ such that $$d(x_m, x_r) = lvert x_m - x_rrvert < varepsilon .$$ As $mto infty$ $x_mto x$. How do I prove that $xin X$ so that the arbitrary Cauchy sequence converges to a point in $X$ (thereby completing the proof)?
Any help would be appreciated.
metric-spaces complete-spaces
edited Dec 7 '18 at 9:45
José Carlos Santos
159k22126231
asked Dec 7 '18 at 9:41
Sindre Bakke ØyenSindre Bakke Øyen
132
$endgroup$
This question already has an answer here:
Showing that if a subset of a complete metric space is closed, it is also complete
5 answers
Heads up: I am very new to abstract algebra and proofs.
Take the non-negative real line $X = 0 cup mathbb{R}^+=[0, +infty)$. We know that $Xsubset mathbb{R}$ and that $mathbb{R}$ is complete. Define now the Euclidean metric $d(x,y) = lvert x - y rvert$. Then $(X,d)$ spans a metric space. I wish to show that this metric space is complete.
So for an arbitrary sequence $(x_n)$ we need $d(x_n, x) < varepsilon$, where $xin X$, that is the sequence has to converge to $x$ which is in the set we are working with. Then $(x_n)$ is Cauchy, and since the sequence was arbitrary $X$ is complete.
This is what I know we require, however, I am unsure if I am on the right track.
My attempt:
Consider a sequence that is Cauchy, $(x_m)$. Then for every $varepsilon$, there is an $N=N(varepsilon)$ such that $$d(x_m, x_r) = lvert x_m - x_rrvert < varepsilon .$$ As $mto infty$ $x_mto x$. How do I prove that $xin X$ so that the arbitrary Cauchy sequence converges to a point in $X$ (thereby completing the proof)?
Any help would be appreciated.
This question already has an answer here:
Showing that if a subset of a complete metric space is closed, it is also complete
5 answers
metric-spaces complete-spaces
metric-spaces complete-spaces
edited Dec 7 '18 at 9:45
José Carlos Santos
159k22126231
asked Dec 7 '18 at 9:41
Sindre Bakke ØyenSindre Bakke Øyen
132
edited Dec 7 '18 at 9:45
José Carlos Santos
159k22126231
asked Dec 7 '18 at 9:41
Sindre Bakke ØyenSindre Bakke Øyen
132
edited Dec 7 '18 at 9:45
José Carlos Santos
159k22126231
edited Dec 7 '18 at 9:45
José Carlos Santos
159k22126231
edited Dec 7 '18 at 9:45
José Carlos Santos
159k22126231
159k22126231
asked Dec 7 '18 at 9:41
Sindre Bakke ØyenSindre Bakke Øyen
132
asked Dec 7 '18 at 9:41
Sindre Bakke ØyenSindre Bakke Øyen
132
asked Dec 7 '18 at 9:41
Sindre Bakke ØyenSindre Bakke Øyen
132
132
marked as duplicate by Jean-Claude Arbaut, Community♦ Dec 7 '18 at 10:09
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 Jean-Claude Arbaut, Community♦ Dec 7 '18 at 10:09
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.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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Since $[0,+infty)$ is a closed subset of $mathbb R$, if $(x_n)_{ninmathbb N}$ is a sequence of elements of $[0,+infty)$ which converges to a real number $x$, then $xin[0,+infty)$.
answered Dec 7 '18 at 9:43
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
If we would instead consider $X=[a, b] subset mathbb{R}$. Would this also be a closed subset of $mathbb{R}$? And in that case is $X$ also complete?
$endgroup$
– Sindre Bakke Øyen
Dec 7 '18 at 12:00
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Yes, you are right.
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– José Carlos Santos
Dec 7 '18 at 12:02
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $[0,+infty)$ is a closed subset of $mathbb R$, if $(x_n)_{ninmathbb N}$ is a sequence of elements of $[0,+infty)$ which converges to a real number $x$, then $xin[0,+infty)$.
answered Dec 7 '18 at 9:43
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
If we would instead consider $X=[a, b] subset mathbb{R}$. Would this also be a closed subset of $mathbb{R}$? And in that case is $X$ also complete?
$endgroup$
– Sindre Bakke Øyen
Dec 7 '18 at 12:00
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:02
add a comment |
$begingroup$
Since $[0,+infty)$ is a closed subset of $mathbb R$, if $(x_n)_{ninmathbb N}$ is a sequence of elements of $[0,+infty)$ which converges to a real number $x$, then $xin[0,+infty)$.
answered Dec 7 '18 at 9:43
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
$begingroup$
If we would instead consider $X=[a, b] subset mathbb{R}$. Would this also be a closed subset of $mathbb{R}$? And in that case is $X$ also complete?
$endgroup$
– Sindre Bakke Øyen
Dec 7 '18 at 12:00
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:02
add a comment |
$begingroup$
Since $[0,+infty)$ is a closed subset of $mathbb R$, if $(x_n)_{ninmathbb N}$ is a sequence of elements of $[0,+infty)$ which converges to a real number $x$, then $xin[0,+infty)$.
answered Dec 7 '18 at 9:43
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
Since $[0,+infty)$ is a closed subset of $mathbb R$, if $(x_n)_{ninmathbb N}$ is a sequence of elements of $[0,+infty)$ which converges to a real number $x$, then $xin[0,+infty)$.
answered Dec 7 '18 at 9:43
José Carlos SantosJosé Carlos Santos
159k22126231
answered Dec 7 '18 at 9:43
José Carlos SantosJosé Carlos Santos
159k22126231
answered Dec 7 '18 at 9:43
José Carlos SantosJosé Carlos Santos
159k22126231
answered Dec 7 '18 at 9:43
José Carlos SantosJosé Carlos Santos
159k22126231
159k22126231
$begingroup$
If we would instead consider $X=[a, b] subset mathbb{R}$. Would this also be a closed subset of $mathbb{R}$? And in that case is $X$ also complete?
$endgroup$
– Sindre Bakke Øyen
Dec 7 '18 at 12:00
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:02
add a comment |
$begingroup$
If we would instead consider $X=[a, b] subset mathbb{R}$. Would this also be a closed subset of $mathbb{R}$? And in that case is $X$ also complete?
$endgroup$
– Sindre Bakke Øyen
Dec 7 '18 at 12:00
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:02
$begingroup$
If we would instead consider $X=[a, b] subset mathbb{R}$. Would this also be a closed subset of $mathbb{R}$? And in that case is $X$ also complete?
$endgroup$
– Sindre Bakke Øyen
Dec 7 '18 at 12:00
$begingroup$
If we would instead consider $X=[a, b] subset mathbb{R}$. Would this also be a closed subset of $mathbb{R}$? And in that case is $X$ also complete?
$endgroup$
– Sindre Bakke Øyen
Dec 7 '18 at 12:00
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:02
$begingroup$
Yes, you are right.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:02
add a comment |
