Mixing
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Finding the closure of set [duplicate]
$begingroup$
This question already has an answer here:
Is the set defined by y=sin(1/x) open or closed?
3 answers
$S=left {(x,sin(frac{1}{x} )) ;0<xleq 1right }$
I know that $bar{S}=(left { 0right }times[-1 ,1])cup S$
How could I prove it ?
general-topology
asked Jan 14 at 9:56
Pedro AlvarèsPedro Alvarès
636
$endgroup$
marked as duplicate by Kavi Rama Murthy, Yanko, Cameron Buie, Misha Lavrov, max_zorn Jan 26 at 22:47
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:
Is the set defined by y=sin(1/x) open or closed?
3 answers
$S=left {(x,sin(frac{1}{x} )) ;0<xleq 1right }$
I know that $bar{S}=(left { 0right }times[-1 ,1])cup S$
How could I prove it ?
general-topology
asked Jan 14 at 9:56
Pedro AlvarèsPedro Alvarès
636
$endgroup$
marked as duplicate by Kavi Rama Murthy, Yanko, Cameron Buie, Misha Lavrov, max_zorn Jan 26 at 22:47
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.
3
$begingroup$
If you 'know' it you must have made some attempt. Can you show us what you have done so far?
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:59
1
$begingroup$
Show that $bar{S} subseteq (left { 0right }times[-1 ,1])cup S $ by reasoning on the image of $sin$, and then show that $ (left { 0right }times[-1 ,1])cup S subseteq bar{S} $ by defining sequences on $S$ whose limits are in $(left { 0right }times[-1 ,1])$.
$endgroup$
– RcnSc
Jan 14 at 10:18
$begingroup$
If $(0,x)in (left { 0 right }times [-1,1])$ , then x=sin(r) for some r, let $(frac{1}{r+2npi },sin(r+2npi ))$ be sequence of S, then it converges to (0,x), so the second inclusion you mentioned is done right ? Or is what I did wrong ?
$endgroup$
– Pedro Alvarès
Jan 14 at 10:47
add a comment |
$begingroup$
This question already has an answer here:
Is the set defined by y=sin(1/x) open or closed?
3 answers
$S=left {(x,sin(frac{1}{x} )) ;0<xleq 1right }$
I know that $bar{S}=(left { 0right }times[-1 ,1])cup S$
How could I prove it ?
general-topology
asked Jan 14 at 9:56
Pedro AlvarèsPedro Alvarès
636
$endgroup$
This question already has an answer here:
Is the set defined by y=sin(1/x) open or closed?
3 answers
$S=left {(x,sin(frac{1}{x} )) ;0<xleq 1right }$
I know that $bar{S}=(left { 0right }times[-1 ,1])cup S$
How could I prove it ?
This question already has an answer here:
Is the set defined by y=sin(1/x) open or closed?
3 answers
general-topology
general-topology
asked Jan 14 at 9:56
Pedro AlvarèsPedro Alvarès
636
asked Jan 14 at 9:56
Pedro AlvarèsPedro Alvarès
636
asked Jan 14 at 9:56
Pedro AlvarèsPedro Alvarès
636
asked Jan 14 at 9:56
Pedro AlvarèsPedro Alvarès
636
asked Jan 14 at 9:56
Pedro AlvarèsPedro Alvarès
636
636
marked as duplicate by Kavi Rama Murthy, Yanko, Cameron Buie, Misha Lavrov, max_zorn Jan 26 at 22:47
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 Kavi Rama Murthy, Yanko, Cameron Buie, Misha Lavrov, max_zorn Jan 26 at 22:47
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.
3
$begingroup$
If you 'know' it you must have made some attempt. Can you show us what you have done so far?
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:59
1
$begingroup$
Show that $bar{S} subseteq (left { 0right }times[-1 ,1])cup S $ by reasoning on the image of $sin$, and then show that $ (left { 0right }times[-1 ,1])cup S subseteq bar{S} $ by defining sequences on $S$ whose limits are in $(left { 0right }times[-1 ,1])$.
$endgroup$
– RcnSc
Jan 14 at 10:18
$begingroup$
If $(0,x)in (left { 0 right }times [-1,1])$ , then x=sin(r) for some r, let $(frac{1}{r+2npi },sin(r+2npi ))$ be sequence of S, then it converges to (0,x), so the second inclusion you mentioned is done right ? Or is what I did wrong ?
$endgroup$
– Pedro Alvarès
Jan 14 at 10:47
add a comment |
3
$begingroup$
If you 'know' it you must have made some attempt. Can you show us what you have done so far?
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:59
1
$begingroup$
Show that $bar{S} subseteq (left { 0right }times[-1 ,1])cup S $ by reasoning on the image of $sin$, and then show that $ (left { 0right }times[-1 ,1])cup S subseteq bar{S} $ by defining sequences on $S$ whose limits are in $(left { 0right }times[-1 ,1])$.
$endgroup$
– RcnSc
Jan 14 at 10:18
$begingroup$
If $(0,x)in (left { 0 right }times [-1,1])$ , then x=sin(r) for some r, let $(frac{1}{r+2npi },sin(r+2npi ))$ be sequence of S, then it converges to (0,x), so the second inclusion you mentioned is done right ? Or is what I did wrong ?
$endgroup$
– Pedro Alvarès
Jan 14 at 10:47
3
3
$begingroup$
If you 'know' it you must have made some attempt. Can you show us what you have done so far?
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:59
$begingroup$
If you 'know' it you must have made some attempt. Can you show us what you have done so far?
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:59
1
1
$begingroup$
Show that $bar{S} subseteq (left { 0right }times[-1 ,1])cup S $ by reasoning on the image of $sin$, and then show that $ (left { 0right }times[-1 ,1])cup S subseteq bar{S} $ by defining sequences on $S$ whose limits are in $(left { 0right }times[-1 ,1])$.
$endgroup$
– RcnSc
Jan 14 at 10:18
$begingroup$
Show that $bar{S} subseteq (left { 0right }times[-1 ,1])cup S $ by reasoning on the image of $sin$, and then show that $ (left { 0right }times[-1 ,1])cup S subseteq bar{S} $ by defining sequences on $S$ whose limits are in $(left { 0right }times[-1 ,1])$.
$endgroup$
– RcnSc
Jan 14 at 10:18
$begingroup$
If $(0,x)in (left { 0 right }times [-1,1])$ , then x=sin(r) for some r, let $(frac{1}{r+2npi },sin(r+2npi ))$ be sequence of S, then it converges to (0,x), so the second inclusion you mentioned is done right ? Or is what I did wrong ?
$endgroup$
– Pedro Alvarès
Jan 14 at 10:47
$begingroup$
If $(0,x)in (left { 0 right }times [-1,1])$ , then x=sin(r) for some r, let $(frac{1}{r+2npi },sin(r+2npi ))$ be sequence of S, then it converges to (0,x), so the second inclusion you mentioned is done right ? Or is what I did wrong ?
$endgroup$
– Pedro Alvarès
Jan 14 at 10:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose $(x_n, sin(frac{1}{x_n}))$ is a sequence in $S$ that converges (in the plane) to $(p,q)$. This means that $x_n to p$ and $sin(frac{1}{x_n}) to q$.
But $x_n in (0,1]$ so $p in overline{(0,1]}=[0,1]$. If $p >0$ we know by continuity of the sine and the $frac{1}{x}$ function that $q_ntosin( frac1{p})=q$ (as limits are unique) so that $(p,q) in S$. Otherwise $p=0$ and we can only say that $q in [-1,1]$ as all values $sin(frac{1}{x_n}) in [-1,1]$ and $[-1,1]$ is closed.
This shows that $S subseteq overline{S} subseteq Scup ({0}times [-1,1])$.
On the other hand, if $(0,q)$ is in ${0}times [-1,1]$, then pick $p>0$ such that $sin(p) = q$. Then $p_n=frac{1}{p+2npi}$, for $n in mathbb{N}$ is in $(0,1]$ for $nge 1$ and $p_n to 0$ and $sin(frac{1}{p_n}) = sin(p+2npi) = sin(p)=q$ for all those $n$, and $(p_n,q)=(p_n, sin(frac{1}{p_n})) in S$ and this sequence converges to $(0,q)$ and so $(0,q) in overline{S}$. So we have now shown:
$$Scup ({0}times[-1,1]) subseteq overline{S} subseteq S cup ({0}times[-1,1])$$
and we have equality.
edited Jan 15 at 0:17
thmusic
898
answered Jan 14 at 23:16
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Let $x$ be any point in $L = {0}×[-1,1]$.
Show for all open $U$ nhood $x$, $U cap S$ is not empty.
Thus $L$ subset $overline S$ and $L cup S$ subset $overline S$.
To show there are no other points $p$ in $overline S$ find an open ball about $p$ that misses $S$.
edited Jan 14 at 23:46
whiskeyo
1368
answered Jan 14 at 11:22
William ElliotWilliam Elliot
8,1522720
$endgroup$
$begingroup$
Shouldn't your chosen x be a couple ?
$endgroup$
– Pedro Alvarès
Jan 14 at 11:54
$begingroup$
@AnyBany Correction made.
$endgroup$
– William Elliot
Jan 14 at 21:38
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose $(x_n, sin(frac{1}{x_n}))$ is a sequence in $S$ that converges (in the plane) to $(p,q)$. This means that $x_n to p$ and $sin(frac{1}{x_n}) to q$.
But $x_n in (0,1]$ so $p in overline{(0,1]}=[0,1]$. If $p >0$ we know by continuity of the sine and the $frac{1}{x}$ function that $q_ntosin( frac1{p})=q$ (as limits are unique) so that $(p,q) in S$. Otherwise $p=0$ and we can only say that $q in [-1,1]$ as all values $sin(frac{1}{x_n}) in [-1,1]$ and $[-1,1]$ is closed.
This shows that $S subseteq overline{S} subseteq Scup ({0}times [-1,1])$.
On the other hand, if $(0,q)$ is in ${0}times [-1,1]$, then pick $p>0$ such that $sin(p) = q$. Then $p_n=frac{1}{p+2npi}$, for $n in mathbb{N}$ is in $(0,1]$ for $nge 1$ and $p_n to 0$ and $sin(frac{1}{p_n}) = sin(p+2npi) = sin(p)=q$ for all those $n$, and $(p_n,q)=(p_n, sin(frac{1}{p_n})) in S$ and this sequence converges to $(0,q)$ and so $(0,q) in overline{S}$. So we have now shown:
$$Scup ({0}times[-1,1]) subseteq overline{S} subseteq S cup ({0}times[-1,1])$$
and we have equality.
edited Jan 15 at 0:17
thmusic
898
answered Jan 14 at 23:16
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Suppose $(x_n, sin(frac{1}{x_n}))$ is a sequence in $S$ that converges (in the plane) to $(p,q)$. This means that $x_n to p$ and $sin(frac{1}{x_n}) to q$.
But $x_n in (0,1]$ so $p in overline{(0,1]}=[0,1]$. If $p >0$ we know by continuity of the sine and the $frac{1}{x}$ function that $q_ntosin( frac1{p})=q$ (as limits are unique) so that $(p,q) in S$. Otherwise $p=0$ and we can only say that $q in [-1,1]$ as all values $sin(frac{1}{x_n}) in [-1,1]$ and $[-1,1]$ is closed.
This shows that $S subseteq overline{S} subseteq Scup ({0}times [-1,1])$.
On the other hand, if $(0,q)$ is in ${0}times [-1,1]$, then pick $p>0$ such that $sin(p) = q$. Then $p_n=frac{1}{p+2npi}$, for $n in mathbb{N}$ is in $(0,1]$ for $nge 1$ and $p_n to 0$ and $sin(frac{1}{p_n}) = sin(p+2npi) = sin(p)=q$ for all those $n$, and $(p_n,q)=(p_n, sin(frac{1}{p_n})) in S$ and this sequence converges to $(0,q)$ and so $(0,q) in overline{S}$. So we have now shown:
$$Scup ({0}times[-1,1]) subseteq overline{S} subseteq S cup ({0}times[-1,1])$$
and we have equality.
edited Jan 15 at 0:17
thmusic
898
answered Jan 14 at 23:16
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
add a comment |
$begingroup$
Suppose $(x_n, sin(frac{1}{x_n}))$ is a sequence in $S$ that converges (in the plane) to $(p,q)$. This means that $x_n to p$ and $sin(frac{1}{x_n}) to q$.
But $x_n in (0,1]$ so $p in overline{(0,1]}=[0,1]$. If $p >0$ we know by continuity of the sine and the $frac{1}{x}$ function that $q_ntosin( frac1{p})=q$ (as limits are unique) so that $(p,q) in S$. Otherwise $p=0$ and we can only say that $q in [-1,1]$ as all values $sin(frac{1}{x_n}) in [-1,1]$ and $[-1,1]$ is closed.
This shows that $S subseteq overline{S} subseteq Scup ({0}times [-1,1])$.
On the other hand, if $(0,q)$ is in ${0}times [-1,1]$, then pick $p>0$ such that $sin(p) = q$. Then $p_n=frac{1}{p+2npi}$, for $n in mathbb{N}$ is in $(0,1]$ for $nge 1$ and $p_n to 0$ and $sin(frac{1}{p_n}) = sin(p+2npi) = sin(p)=q$ for all those $n$, and $(p_n,q)=(p_n, sin(frac{1}{p_n})) in S$ and this sequence converges to $(0,q)$ and so $(0,q) in overline{S}$. So we have now shown:
$$Scup ({0}times[-1,1]) subseteq overline{S} subseteq S cup ({0}times[-1,1])$$
and we have equality.
edited Jan 15 at 0:17
thmusic
898
answered Jan 14 at 23:16
Henno BrandsmaHenno Brandsma
110k347116
$endgroup$
Suppose $(x_n, sin(frac{1}{x_n}))$ is a sequence in $S$ that converges (in the plane) to $(p,q)$. This means that $x_n to p$ and $sin(frac{1}{x_n}) to q$.
But $x_n in (0,1]$ so $p in overline{(0,1]}=[0,1]$. If $p >0$ we know by continuity of the sine and the $frac{1}{x}$ function that $q_ntosin( frac1{p})=q$ (as limits are unique) so that $(p,q) in S$. Otherwise $p=0$ and we can only say that $q in [-1,1]$ as all values $sin(frac{1}{x_n}) in [-1,1]$ and $[-1,1]$ is closed.
This shows that $S subseteq overline{S} subseteq Scup ({0}times [-1,1])$.
On the other hand, if $(0,q)$ is in ${0}times [-1,1]$, then pick $p>0$ such that $sin(p) = q$. Then $p_n=frac{1}{p+2npi}$, for $n in mathbb{N}$ is in $(0,1]$ for $nge 1$ and $p_n to 0$ and $sin(frac{1}{p_n}) = sin(p+2npi) = sin(p)=q$ for all those $n$, and $(p_n,q)=(p_n, sin(frac{1}{p_n})) in S$ and this sequence converges to $(0,q)$ and so $(0,q) in overline{S}$. So we have now shown:
$$Scup ({0}times[-1,1]) subseteq overline{S} subseteq S cup ({0}times[-1,1])$$
and we have equality.
edited Jan 15 at 0:17
thmusic
898
answered Jan 14 at 23:16
Henno BrandsmaHenno Brandsma
110k347116
edited Jan 15 at 0:17
thmusic
898
edited Jan 15 at 0:17
thmusic
898
edited Jan 15 at 0:17
thmusic
898
898
answered Jan 14 at 23:16
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 14 at 23:16
Henno BrandsmaHenno Brandsma
110k347116
answered Jan 14 at 23:16
Henno BrandsmaHenno Brandsma
110k347116
110k347116
add a comment |
add a comment |
$begingroup$
Let $x$ be any point in $L = {0}×[-1,1]$.
Show for all open $U$ nhood $x$, $U cap S$ is not empty.
Thus $L$ subset $overline S$ and $L cup S$ subset $overline S$.
To show there are no other points $p$ in $overline S$ find an open ball about $p$ that misses $S$.
edited Jan 14 at 23:46
whiskeyo
1368
answered Jan 14 at 11:22
William ElliotWilliam Elliot
8,1522720
$endgroup$
$begingroup$
Shouldn't your chosen x be a couple ?
$endgroup$
– Pedro Alvarès
Jan 14 at 11:54
$begingroup$
@AnyBany Correction made.
$endgroup$
– William Elliot
Jan 14 at 21:38
add a comment |
$begingroup$
Let $x$ be any point in $L = {0}×[-1,1]$.
Show for all open $U$ nhood $x$, $U cap S$ is not empty.
Thus $L$ subset $overline S$ and $L cup S$ subset $overline S$.
To show there are no other points $p$ in $overline S$ find an open ball about $p$ that misses $S$.
edited Jan 14 at 23:46
whiskeyo
1368
answered Jan 14 at 11:22
William ElliotWilliam Elliot
8,1522720
$endgroup$
$begingroup$
Shouldn't your chosen x be a couple ?
$endgroup$
– Pedro Alvarès
Jan 14 at 11:54
$begingroup$
@AnyBany Correction made.
$endgroup$
– William Elliot
Jan 14 at 21:38
add a comment |
$begingroup$
Let $x$ be any point in $L = {0}×[-1,1]$.
Show for all open $U$ nhood $x$, $U cap S$ is not empty.
Thus $L$ subset $overline S$ and $L cup S$ subset $overline S$.
To show there are no other points $p$ in $overline S$ find an open ball about $p$ that misses $S$.
edited Jan 14 at 23:46
whiskeyo
1368
answered Jan 14 at 11:22
William ElliotWilliam Elliot
8,1522720
$endgroup$
Let $x$ be any point in $L = {0}×[-1,1]$.
Show for all open $U$ nhood $x$, $U cap S$ is not empty.
Thus $L$ subset $overline S$ and $L cup S$ subset $overline S$.
To show there are no other points $p$ in $overline S$ find an open ball about $p$ that misses $S$.
edited Jan 14 at 23:46
whiskeyo
1368
answered Jan 14 at 11:22
William ElliotWilliam Elliot
8,1522720
edited Jan 14 at 23:46
whiskeyo
1368
edited Jan 14 at 23:46
whiskeyo
1368
edited Jan 14 at 23:46
whiskeyo
1368
1368
answered Jan 14 at 11:22
William ElliotWilliam Elliot
8,1522720
answered Jan 14 at 11:22
William ElliotWilliam Elliot
8,1522720
answered Jan 14 at 11:22
William ElliotWilliam Elliot
8,1522720
8,1522720
$begingroup$
Shouldn't your chosen x be a couple ?
$endgroup$
– Pedro Alvarès
Jan 14 at 11:54
$begingroup$
@AnyBany Correction made.
$endgroup$
– William Elliot
Jan 14 at 21:38
add a comment |
$begingroup$
Shouldn't your chosen x be a couple ?
$endgroup$
– Pedro Alvarès
Jan 14 at 11:54
$begingroup$
@AnyBany Correction made.
$endgroup$
– William Elliot
Jan 14 at 21:38
$begingroup$
Shouldn't your chosen x be a couple ?
$endgroup$
– Pedro Alvarès
Jan 14 at 11:54
$begingroup$
Shouldn't your chosen x be a couple ?
$endgroup$
– Pedro Alvarès
Jan 14 at 11:54
$begingroup$
@AnyBany Correction made.
$endgroup$
– William Elliot
Jan 14 at 21:38
$begingroup$
@AnyBany Correction made.
$endgroup$
– William Elliot
Jan 14 at 21:38
add a comment |
3
$begingroup$
If you 'know' it you must have made some attempt. Can you show us what you have done so far?
$endgroup$
– Kavi Rama Murthy
Jan 14 at 9:59
1
$begingroup$
Show that $bar{S} subseteq (left { 0right }times[-1 ,1])cup S $ by reasoning on the image of $sin$, and then show that $ (left { 0right }times[-1 ,1])cup S subseteq bar{S} $ by defining sequences on $S$ whose limits are in $(left { 0right }times[-1 ,1])$.
$endgroup$
– RcnSc
Jan 14 at 10:18
$begingroup$
If $(0,x)in (left { 0 right }times [-1,1])$ , then x=sin(r) for some r, let $(frac{1}{r+2npi },sin(r+2npi ))$ be sequence of S, then it converges to (0,x), so the second inclusion you mentioned is done right ? Or is what I did wrong ?
$endgroup$
– Pedro Alvarès
Jan 14 at 10:47