Mixing
Real numbers equipped with the metric $ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space
I have to show that the real numbers equipped with the metric
$ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space.
Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that.
Can anybody help me with this.
Thanks for helping me.
functional-analysis metric-spaces
edited Dec 9 '14 at 13:37
Martin Sleziak
44.6k8115271
asked May 31 '12 at 22:49
srijansrijan
6,35563979
|
show 4 more comments
I have to show that the real numbers equipped with the metric
$ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space.
Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that.
Can anybody help me with this.
Thanks for helping me.
functional-analysis metric-spaces
edited Dec 9 '14 at 13:37
Martin Sleziak
44.6k8115271
asked May 31 '12 at 22:49
srijansrijan
6,35563979
5
The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
– Chris Eagle
May 31 '12 at 22:54
4
Try the sequence $x_n = n$. Draw a picture.
– copper.hat
May 31 '12 at 23:00
1
You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
– copper.hat
May 31 '12 at 23:10
1
$d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
– copper.hat
May 31 '12 at 23:26
2
I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
– copper.hat
May 31 '12 at 23:37
|
show 4 more comments
I have to show that the real numbers equipped with the metric
$ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space.
Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that.
Can anybody help me with this.
Thanks for helping me.
functional-analysis metric-spaces
edited Dec 9 '14 at 13:37
Martin Sleziak
44.6k8115271
asked May 31 '12 at 22:49
srijansrijan
6,35563979
I have to show that the real numbers equipped with the metric
$ d (x,y) = | arctan(x) - arctan(y)| $ is an incomplete metric space.
Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that.
Can anybody help me with this.
Thanks for helping me.
functional-analysis metric-spaces
functional-analysis metric-spaces
edited Dec 9 '14 at 13:37
Martin Sleziak
44.6k8115271
asked May 31 '12 at 22:49
srijansrijan
6,35563979
edited Dec 9 '14 at 13:37
Martin Sleziak
44.6k8115271
asked May 31 '12 at 22:49
srijansrijan
6,35563979
edited Dec 9 '14 at 13:37
Martin Sleziak
44.6k8115271
edited Dec 9 '14 at 13:37
Martin Sleziak
44.6k8115271
edited Dec 9 '14 at 13:37
Martin Sleziak
44.6k8115271
44.6k8115271
asked May 31 '12 at 22:49
srijansrijan
6,35563979
asked May 31 '12 at 22:49
srijansrijan
6,35563979
asked May 31 '12 at 22:49
srijansrijan
6,35563979
6,35563979
5
The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
– Chris Eagle
May 31 '12 at 22:54
4
Try the sequence $x_n = n$. Draw a picture.
– copper.hat
May 31 '12 at 23:00
1
You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
– copper.hat
May 31 '12 at 23:10
1
$d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
– copper.hat
May 31 '12 at 23:26
2
I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
– copper.hat
May 31 '12 at 23:37
|
show 4 more comments
5
The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
– Chris Eagle
May 31 '12 at 22:54
4
Try the sequence $x_n = n$. Draw a picture.
– copper.hat
May 31 '12 at 23:00
1
You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
– copper.hat
May 31 '12 at 23:10
1
$d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
– copper.hat
May 31 '12 at 23:26
2
I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
– copper.hat
May 31 '12 at 23:37
5
5
The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
– Chris Eagle
May 31 '12 at 22:54
The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
– Chris Eagle
May 31 '12 at 22:54
4
4
Try the sequence $x_n = n$. Draw a picture.
– copper.hat
May 31 '12 at 23:00
Try the sequence $x_n = n$. Draw a picture.
– copper.hat
May 31 '12 at 23:00
1
1
You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
– copper.hat
May 31 '12 at 23:10
You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
– copper.hat
May 31 '12 at 23:10
1
1
$d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
– copper.hat
May 31 '12 at 23:26
$d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
– copper.hat
May 31 '12 at 23:26
2
2
I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
– copper.hat
May 31 '12 at 23:37
I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
– copper.hat
May 31 '12 at 23:37
|
show 4 more comments
3 Answers
3
active
oldest
votes
Sorry for reviving such an old problem...
Anyways, what is important here is that $text{arctan}$ is a bijection from $mathbb{R}$ to $( -pi/2, pi/2 )$, and it is an isometry if we give $left(-pi/2, pi/2right)$ the metric it carries as a subspace of $mathbb{R}$ with the usual metric. If $f: X to Y$ is a surjective isometry then the Cauchy sequences in $Y$ are the images of Cauchy sequences under $f$, and the convergent sequences in $Y$ are the images of convergent sequences under $f$, so $Y$ is complete if and only if $X$ is. Thus $(mathbb{R}, d)$ is complete if and only if $(-pi/2, pi/2)$ with the metric $text{dist}(x, y) = |x - y|$ is complete. But $(-pi/2, pi/2)$ is not complete since ${pi/2 - 1/n}_{n=1}^infty$ is Cauchy and does not converge in $(-pi/2, pi/2)$.
edited Nov 4 '18 at 23:11
Adam
32
answered Dec 9 '14 at 12:50
user149792user149792
4,06842467
add a comment |
Consider $x_n = n$. Let $varepsilon > 0$ and choose $displaystyle N > tanbigg(fracpi2 - varepsilonbigg)$. If $m, n > N$ then $displaystyle {arctan m, arctan m} subseteq bigg(arctan N, fracpi2bigg)$. Thus $$d(x_m, x_n) = vert arctan m - arctan n vert leq bigg vert fracpi2 - arctan N biggvert < bigg vert fracpi2 - fracpi2 + varepsilon biggvert= varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n to infty$, $arctan x_n to pi/2$. But $(x_n)$ does not converge to any element in $mathbb R$ since there is no $x in mathbb R$ such that $arctan x = pi/2$.
edited Nov 21 '18 at 19:05
Yadati Kiran
1,692619
answered Feb 3 '13 at 10:48
Robert CardonaRobert Cardona
5,25523498
1
I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
– Jonathan
Sep 29 '17 at 17:11
add a comment |
Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $mathbb{R}$, but will be under this metric.
answered May 31 '12 at 22:58
Cameron BuieCameron Buie
85.1k771155
thanks i took sequence $x_n = n$ as suggested by copper.hat
– srijan
May 31 '12 at 23:33
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
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active
oldest
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votes
Sorry for reviving such an old problem...
Anyways, what is important here is that $text{arctan}$ is a bijection from $mathbb{R}$ to $( -pi/2, pi/2 )$, and it is an isometry if we give $left(-pi/2, pi/2right)$ the metric it carries as a subspace of $mathbb{R}$ with the usual metric. If $f: X to Y$ is a surjective isometry then the Cauchy sequences in $Y$ are the images of Cauchy sequences under $f$, and the convergent sequences in $Y$ are the images of convergent sequences under $f$, so $Y$ is complete if and only if $X$ is. Thus $(mathbb{R}, d)$ is complete if and only if $(-pi/2, pi/2)$ with the metric $text{dist}(x, y) = |x - y|$ is complete. But $(-pi/2, pi/2)$ is not complete since ${pi/2 - 1/n}_{n=1}^infty$ is Cauchy and does not converge in $(-pi/2, pi/2)$.
edited Nov 4 '18 at 23:11
Adam
32
answered Dec 9 '14 at 12:50
user149792user149792
4,06842467
add a comment |
Sorry for reviving such an old problem...
Anyways, what is important here is that $text{arctan}$ is a bijection from $mathbb{R}$ to $( -pi/2, pi/2 )$, and it is an isometry if we give $left(-pi/2, pi/2right)$ the metric it carries as a subspace of $mathbb{R}$ with the usual metric. If $f: X to Y$ is a surjective isometry then the Cauchy sequences in $Y$ are the images of Cauchy sequences under $f$, and the convergent sequences in $Y$ are the images of convergent sequences under $f$, so $Y$ is complete if and only if $X$ is. Thus $(mathbb{R}, d)$ is complete if and only if $(-pi/2, pi/2)$ with the metric $text{dist}(x, y) = |x - y|$ is complete. But $(-pi/2, pi/2)$ is not complete since ${pi/2 - 1/n}_{n=1}^infty$ is Cauchy and does not converge in $(-pi/2, pi/2)$.
edited Nov 4 '18 at 23:11
Adam
32
answered Dec 9 '14 at 12:50
user149792user149792
4,06842467
add a comment |
Sorry for reviving such an old problem...
Anyways, what is important here is that $text{arctan}$ is a bijection from $mathbb{R}$ to $( -pi/2, pi/2 )$, and it is an isometry if we give $left(-pi/2, pi/2right)$ the metric it carries as a subspace of $mathbb{R}$ with the usual metric. If $f: X to Y$ is a surjective isometry then the Cauchy sequences in $Y$ are the images of Cauchy sequences under $f$, and the convergent sequences in $Y$ are the images of convergent sequences under $f$, so $Y$ is complete if and only if $X$ is. Thus $(mathbb{R}, d)$ is complete if and only if $(-pi/2, pi/2)$ with the metric $text{dist}(x, y) = |x - y|$ is complete. But $(-pi/2, pi/2)$ is not complete since ${pi/2 - 1/n}_{n=1}^infty$ is Cauchy and does not converge in $(-pi/2, pi/2)$.
edited Nov 4 '18 at 23:11
Adam
32
answered Dec 9 '14 at 12:50
user149792user149792
4,06842467
Sorry for reviving such an old problem...
Anyways, what is important here is that $text{arctan}$ is a bijection from $mathbb{R}$ to $( -pi/2, pi/2 )$, and it is an isometry if we give $left(-pi/2, pi/2right)$ the metric it carries as a subspace of $mathbb{R}$ with the usual metric. If $f: X to Y$ is a surjective isometry then the Cauchy sequences in $Y$ are the images of Cauchy sequences under $f$, and the convergent sequences in $Y$ are the images of convergent sequences under $f$, so $Y$ is complete if and only if $X$ is. Thus $(mathbb{R}, d)$ is complete if and only if $(-pi/2, pi/2)$ with the metric $text{dist}(x, y) = |x - y|$ is complete. But $(-pi/2, pi/2)$ is not complete since ${pi/2 - 1/n}_{n=1}^infty$ is Cauchy and does not converge in $(-pi/2, pi/2)$.
edited Nov 4 '18 at 23:11
Adam
32
answered Dec 9 '14 at 12:50
user149792user149792
4,06842467
edited Nov 4 '18 at 23:11
Adam
32
edited Nov 4 '18 at 23:11
Adam
32
edited Nov 4 '18 at 23:11
Adam
32
32
answered Dec 9 '14 at 12:50
user149792user149792
4,06842467
answered Dec 9 '14 at 12:50
user149792user149792
4,06842467
answered Dec 9 '14 at 12:50
user149792user149792
4,06842467
4,06842467
add a comment |
add a comment |
Consider $x_n = n$. Let $varepsilon > 0$ and choose $displaystyle N > tanbigg(fracpi2 - varepsilonbigg)$. If $m, n > N$ then $displaystyle {arctan m, arctan m} subseteq bigg(arctan N, fracpi2bigg)$. Thus $$d(x_m, x_n) = vert arctan m - arctan n vert leq bigg vert fracpi2 - arctan N biggvert < bigg vert fracpi2 - fracpi2 + varepsilon biggvert= varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n to infty$, $arctan x_n to pi/2$. But $(x_n)$ does not converge to any element in $mathbb R$ since there is no $x in mathbb R$ such that $arctan x = pi/2$.
edited Nov 21 '18 at 19:05
Yadati Kiran
1,692619
answered Feb 3 '13 at 10:48
Robert CardonaRobert Cardona
5,25523498
1
I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
– Jonathan
Sep 29 '17 at 17:11
add a comment |
Consider $x_n = n$. Let $varepsilon > 0$ and choose $displaystyle N > tanbigg(fracpi2 - varepsilonbigg)$. If $m, n > N$ then $displaystyle {arctan m, arctan m} subseteq bigg(arctan N, fracpi2bigg)$. Thus $$d(x_m, x_n) = vert arctan m - arctan n vert leq bigg vert fracpi2 - arctan N biggvert < bigg vert fracpi2 - fracpi2 + varepsilon biggvert= varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n to infty$, $arctan x_n to pi/2$. But $(x_n)$ does not converge to any element in $mathbb R$ since there is no $x in mathbb R$ such that $arctan x = pi/2$.
edited Nov 21 '18 at 19:05
Yadati Kiran
1,692619
answered Feb 3 '13 at 10:48
Robert CardonaRobert Cardona
5,25523498
1
I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
– Jonathan
Sep 29 '17 at 17:11
add a comment |
Consider $x_n = n$. Let $varepsilon > 0$ and choose $displaystyle N > tanbigg(fracpi2 - varepsilonbigg)$. If $m, n > N$ then $displaystyle {arctan m, arctan m} subseteq bigg(arctan N, fracpi2bigg)$. Thus $$d(x_m, x_n) = vert arctan m - arctan n vert leq bigg vert fracpi2 - arctan N biggvert < bigg vert fracpi2 - fracpi2 + varepsilon biggvert= varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n to infty$, $arctan x_n to pi/2$. But $(x_n)$ does not converge to any element in $mathbb R$ since there is no $x in mathbb R$ such that $arctan x = pi/2$.
edited Nov 21 '18 at 19:05
Yadati Kiran
1,692619
answered Feb 3 '13 at 10:48
Robert CardonaRobert Cardona
5,25523498
Consider $x_n = n$. Let $varepsilon > 0$ and choose $displaystyle N > tanbigg(fracpi2 - varepsilonbigg)$. If $m, n > N$ then $displaystyle {arctan m, arctan m} subseteq bigg(arctan N, fracpi2bigg)$. Thus $$d(x_m, x_n) = vert arctan m - arctan n vert leq bigg vert fracpi2 - arctan N biggvert < bigg vert fracpi2 - fracpi2 + varepsilon biggvert= varepsilon$$ Thus $(x_n)$ is a Cauchy sequence. Observe that as $n to infty$, $arctan x_n to pi/2$. But $(x_n)$ does not converge to any element in $mathbb R$ since there is no $x in mathbb R$ such that $arctan x = pi/2$.
edited Nov 21 '18 at 19:05
Yadati Kiran
1,692619
answered Feb 3 '13 at 10:48
Robert CardonaRobert Cardona
5,25523498
edited Nov 21 '18 at 19:05
Yadati Kiran
1,692619
edited Nov 21 '18 at 19:05
Yadati Kiran
1,692619
edited Nov 21 '18 at 19:05
Yadati Kiran
1,692619
1,692619
answered Feb 3 '13 at 10:48
Robert CardonaRobert Cardona
5,25523498
answered Feb 3 '13 at 10:48
Robert CardonaRobert Cardona
5,25523498
answered Feb 3 '13 at 10:48
Robert CardonaRobert Cardona
5,25523498
5,25523498
1
I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
– Jonathan
Sep 29 '17 at 17:11
add a comment |
1
I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
– Jonathan
Sep 29 '17 at 17:11
1
1
I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
– Jonathan
Sep 29 '17 at 17:11
I think you meant : Observe that as $n to infty$, $arctan(x_n) to pi/2$.
– Jonathan
Sep 29 '17 at 17:11
add a comment |
Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $mathbb{R}$, but will be under this metric.
answered May 31 '12 at 22:58
Cameron BuieCameron Buie
85.1k771155
thanks i took sequence $x_n = n$ as suggested by copper.hat
– srijan
May 31 '12 at 23:33
add a comment |
Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $mathbb{R}$, but will be under this metric.
answered May 31 '12 at 22:58
Cameron BuieCameron Buie
85.1k771155
thanks i took sequence $x_n = n$ as suggested by copper.hat
– srijan
May 31 '12 at 23:33
add a comment |
Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $mathbb{R}$, but will be under this metric.
answered May 31 '12 at 22:58
Cameron BuieCameron Buie
85.1k771155
Consider a sequence that grows without bound. Such a sequence isn't Cauchy in the usual metric on $mathbb{R}$, but will be under this metric.
answered May 31 '12 at 22:58
Cameron BuieCameron Buie
85.1k771155
answered May 31 '12 at 22:58
Cameron BuieCameron Buie
85.1k771155
answered May 31 '12 at 22:58
Cameron BuieCameron Buie
85.1k771155
answered May 31 '12 at 22:58
Cameron BuieCameron Buie
85.1k771155
85.1k771155
thanks i took sequence $x_n = n$ as suggested by copper.hat
– srijan
May 31 '12 at 23:33
add a comment |
thanks i took sequence $x_n = n$ as suggested by copper.hat
– srijan
May 31 '12 at 23:33
thanks i took sequence $x_n = n$ as suggested by copper.hat
– srijan
May 31 '12 at 23:33
thanks i took sequence $x_n = n$ as suggested by copper.hat
– srijan
May 31 '12 at 23:33
add a comment |
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5
The simplest way to do this is to notice that $x mapsto arctan(x)$ is an isometry for your space to the open interval $(-pi/2,pi/2)$ with its usual metric.
– Chris Eagle
May 31 '12 at 22:54
4
Try the sequence $x_n = n$. Draw a picture.
– copper.hat
May 31 '12 at 23:00
1
You can show easily that it does not converge to any number, ie, for any $y$, show $d(n,y)$ does not converge to zero.
– copper.hat
May 31 '12 at 23:10
1
$d(n,y) = | arctan(n) - arctan(y)|$. So $d(n,y) to (frac{pi}{2}-y)$. $y$ is a fixed number, so $arctan(y) < frac{pi}{2}$.
– copper.hat
May 31 '12 at 23:26
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I am showing that $x_n$ does not converge to any fixed y. $arctan y$ is a fixed number, it doesn't tend towards anything but itself.
– copper.hat
May 31 '12 at 23:37