Prove: A set containing limit points of a sequence is a closed set
$begingroup$
The question:
Prove that a set, $S'$, containing the limit points of the sequence $S subset mathbb{R}$, is closed.
What I have so far:
I want to prove this by showing that the complement of $S'$ is open. In other words, $mathbb{R} backslash S' $ is open.
So take any limit point $x$ of $mathbb{R} backslash S' $. Clearly, $x$ is not a limit point of $S$.
I'm at a loss on how to proceed from here.
real-analysis general-topology proof-verification
edited May 12 '14 at 6:05
Jean-Claude Arbaut
14.7k63464
asked May 12 '14 at 0:24
user3025403user3025403
1488
$endgroup$
add a comment |
$begingroup$
The question:
Prove that a set, $S'$, containing the limit points of the sequence $S subset mathbb{R}$, is closed.
What I have so far:
I want to prove this by showing that the complement of $S'$ is open. In other words, $mathbb{R} backslash S' $ is open.
So take any limit point $x$ of $mathbb{R} backslash S' $. Clearly, $x$ is not a limit point of $S$.
I'm at a loss on how to proceed from here.
real-analysis general-topology proof-verification
edited May 12 '14 at 6:05
Jean-Claude Arbaut
14.7k63464
asked May 12 '14 at 0:24
user3025403user3025403
1488
$endgroup$
1
$begingroup$
(1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
$endgroup$
– DanielWainfleet
Aug 7 '16 at 6:54
$begingroup$
The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
$endgroup$
– martin.koeberl
Mar 28 '17 at 13:40
$begingroup$
The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
$endgroup$
– user247327
Jan 2 '18 at 2:44
add a comment |
$begingroup$
The question:
Prove that a set, $S'$, containing the limit points of the sequence $S subset mathbb{R}$, is closed.
What I have so far:
I want to prove this by showing that the complement of $S'$ is open. In other words, $mathbb{R} backslash S' $ is open.
So take any limit point $x$ of $mathbb{R} backslash S' $. Clearly, $x$ is not a limit point of $S$.
I'm at a loss on how to proceed from here.
real-analysis general-topology proof-verification
edited May 12 '14 at 6:05
Jean-Claude Arbaut
14.7k63464
asked May 12 '14 at 0:24
user3025403user3025403
1488
$endgroup$
The question:
Prove that a set, $S'$, containing the limit points of the sequence $S subset mathbb{R}$, is closed.
What I have so far:
I want to prove this by showing that the complement of $S'$ is open. In other words, $mathbb{R} backslash S' $ is open.
So take any limit point $x$ of $mathbb{R} backslash S' $. Clearly, $x$ is not a limit point of $S$.
I'm at a loss on how to proceed from here.
real-analysis general-topology proof-verification
real-analysis general-topology proof-verification
edited May 12 '14 at 6:05
Jean-Claude Arbaut
14.7k63464
asked May 12 '14 at 0:24
user3025403user3025403
1488
edited May 12 '14 at 6:05
Jean-Claude Arbaut
14.7k63464
asked May 12 '14 at 0:24
user3025403user3025403
1488
edited May 12 '14 at 6:05
Jean-Claude Arbaut
14.7k63464
edited May 12 '14 at 6:05
Jean-Claude Arbaut
14.7k63464
edited May 12 '14 at 6:05
Jean-Claude Arbaut
14.7k63464
14.7k63464
asked May 12 '14 at 0:24
user3025403user3025403
1488
asked May 12 '14 at 0:24
user3025403user3025403
1488
asked May 12 '14 at 0:24
user3025403user3025403
1488
1488
1
$begingroup$
(1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
$endgroup$
– DanielWainfleet
Aug 7 '16 at 6:54
$begingroup$
The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
$endgroup$
– martin.koeberl
Mar 28 '17 at 13:40
$begingroup$
The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
$endgroup$
– user247327
Jan 2 '18 at 2:44
add a comment |
1
$begingroup$
(1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
$endgroup$
– DanielWainfleet
Aug 7 '16 at 6:54
$begingroup$
The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
$endgroup$
– martin.koeberl
Mar 28 '17 at 13:40
$begingroup$
The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
$endgroup$
– user247327
Jan 2 '18 at 2:44
1
1
$begingroup$
(1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
$endgroup$
– DanielWainfleet
Aug 7 '16 at 6:54
$begingroup$
(1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
$endgroup$
– DanielWainfleet
Aug 7 '16 at 6:54
$begingroup$
The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
$endgroup$
– martin.koeberl
Mar 28 '17 at 13:40
$begingroup$
The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
$endgroup$
– martin.koeberl
Mar 28 '17 at 13:40
$begingroup$
The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
$endgroup$
– user247327
Jan 2 '18 at 2:44
$begingroup$
The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
$endgroup$
– user247327
Jan 2 '18 at 2:44
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since $x$ is not a limit point of $S$, there is an open neighborhood $U$ of $x$ such that $(Usetminus{x})cap S=emptyset$. Now prove $Ucap S'=emptyset$ and conclude $Usubset(Bbb Rsetminus S')$.
answered May 12 '14 at 0:39
ZirchtZircht
1,623913
$endgroup$
$begingroup$
Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
$endgroup$
– Sayan Bandyapadhyay
May 12 '14 at 1:01
$begingroup$
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
$endgroup$
– Zircht
May 12 '14 at 1:05
add a comment |
$begingroup$
Another nice way to see this would be the following.
Suppose $xin bar{S'}$. then there exists a sequence of numbers in $S'$ which converges to $x$. Let that sequence be $(x_n)$. But again as $x_nin S'$, we have the following,
begin{align*}
x_{11}, &x_{12}, x_{13},cdots rightarrow x_1 && (text{a sequence converging to} x_1) \
x_{21}, &x_{22}, x_{23},cdots rightarrow x_2 && (text{a sequence converging to} x_2) \
x_{31}, &x_{32}, x_{33},cdots rightarrow x_3 && (text{a sequence converging to} x_3) \
vdots \
x_{n1}, &x_{n2}, x_{n3},cdots rightarrow x_n && (text{a sequence converging to} x_n)
end{align*}
Note that all the numbers $x_{ij}in S$. Now use a diagonal argument to get a sequence in $S$ converging to the limit, $x=lim_{nrightarrowinfty}x_n$. This shows that $xin S'$ by definition of $S'$.
answered Mar 28 '17 at 12:53
ParishParish
631315
$endgroup$
add a comment |
$begingroup$
If $X$ is any topological space and $S subset X$, then ${displaystyle {bar {S}}=Xbackslash [ (Xbackslash S)^{circ }]}$. Here, we have
$tag 1 {displaystyle {bar {S}}=mathbb Rbackslash [ (mathbb Rbackslash S)^{circ }]}$
Claim: The OP's $S'$ is equal to $bar {S}$.
It is easy to see that $S' subset bar {S}$.
Using constant sequences, we see that $S subset S'$. Now let $r in bar {S}backslash S$. By (1), for every $n ge 1$ the open interval
$tag 2 (r-frac{1}{n},r+frac{1}{n})$
intersects the set $S$, so we can select $s_n$ in both the interval (2) and $S$. The sequence $(s_n)$ is $S$ converges to $r$ so $r in S'$. Since both $S$ and its $bar {S}backslash S$ are contained in $S'$, $bar {S} subset S'$.
Since $S' = bar {S}$, $S'$ is a closed set.
answered Jan 2 '18 at 3:44
CopyPasteItCopyPasteIt
4,1481628
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $x$ is not a limit point of $S$, there is an open neighborhood $U$ of $x$ such that $(Usetminus{x})cap S=emptyset$. Now prove $Ucap S'=emptyset$ and conclude $Usubset(Bbb Rsetminus S')$.
answered May 12 '14 at 0:39
ZirchtZircht
1,623913
$endgroup$
$begingroup$
Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
$endgroup$
– Sayan Bandyapadhyay
May 12 '14 at 1:01
$begingroup$
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
$endgroup$
– Zircht
May 12 '14 at 1:05
add a comment |
$begingroup$
Since $x$ is not a limit point of $S$, there is an open neighborhood $U$ of $x$ such that $(Usetminus{x})cap S=emptyset$. Now prove $Ucap S'=emptyset$ and conclude $Usubset(Bbb Rsetminus S')$.
answered May 12 '14 at 0:39
ZirchtZircht
1,623913
$endgroup$
$begingroup$
Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
$endgroup$
– Sayan Bandyapadhyay
May 12 '14 at 1:01
$begingroup$
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
$endgroup$
– Zircht
May 12 '14 at 1:05
add a comment |
$begingroup$
Since $x$ is not a limit point of $S$, there is an open neighborhood $U$ of $x$ such that $(Usetminus{x})cap S=emptyset$. Now prove $Ucap S'=emptyset$ and conclude $Usubset(Bbb Rsetminus S')$.
answered May 12 '14 at 0:39
ZirchtZircht
1,623913
$endgroup$
Since $x$ is not a limit point of $S$, there is an open neighborhood $U$ of $x$ such that $(Usetminus{x})cap S=emptyset$. Now prove $Ucap S'=emptyset$ and conclude $Usubset(Bbb Rsetminus S')$.
answered May 12 '14 at 0:39
ZirchtZircht
1,623913
answered May 12 '14 at 0:39
ZirchtZircht
1,623913
answered May 12 '14 at 0:39
ZirchtZircht
1,623913
answered May 12 '14 at 0:39
ZirchtZircht
1,623913
1,623913
$begingroup$
Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
$endgroup$
– Sayan Bandyapadhyay
May 12 '14 at 1:01
$begingroup$
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
$endgroup$
– Zircht
May 12 '14 at 1:05
add a comment |
$begingroup$
Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
$endgroup$
– Sayan Bandyapadhyay
May 12 '14 at 1:01
$begingroup$
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
$endgroup$
– Zircht
May 12 '14 at 1:05
$begingroup$
Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
$endgroup$
– Sayan Bandyapadhyay
May 12 '14 at 1:01
$begingroup$
Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
$endgroup$
– Sayan Bandyapadhyay
May 12 '14 at 1:01
$begingroup$
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
$endgroup$
– Zircht
May 12 '14 at 1:05
$begingroup$
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
$endgroup$
– Zircht
May 12 '14 at 1:05
add a comment |
$begingroup$
Another nice way to see this would be the following.
Suppose $xin bar{S'}$. then there exists a sequence of numbers in $S'$ which converges to $x$. Let that sequence be $(x_n)$. But again as $x_nin S'$, we have the following,
begin{align*}
x_{11}, &x_{12}, x_{13},cdots rightarrow x_1 && (text{a sequence converging to} x_1) \
x_{21}, &x_{22}, x_{23},cdots rightarrow x_2 && (text{a sequence converging to} x_2) \
x_{31}, &x_{32}, x_{33},cdots rightarrow x_3 && (text{a sequence converging to} x_3) \
vdots \
x_{n1}, &x_{n2}, x_{n3},cdots rightarrow x_n && (text{a sequence converging to} x_n)
end{align*}
Note that all the numbers $x_{ij}in S$. Now use a diagonal argument to get a sequence in $S$ converging to the limit, $x=lim_{nrightarrowinfty}x_n$. This shows that $xin S'$ by definition of $S'$.
answered Mar 28 '17 at 12:53
ParishParish
631315
$endgroup$
add a comment |
$begingroup$
Another nice way to see this would be the following.
Suppose $xin bar{S'}$. then there exists a sequence of numbers in $S'$ which converges to $x$. Let that sequence be $(x_n)$. But again as $x_nin S'$, we have the following,
begin{align*}
x_{11}, &x_{12}, x_{13},cdots rightarrow x_1 && (text{a sequence converging to} x_1) \
x_{21}, &x_{22}, x_{23},cdots rightarrow x_2 && (text{a sequence converging to} x_2) \
x_{31}, &x_{32}, x_{33},cdots rightarrow x_3 && (text{a sequence converging to} x_3) \
vdots \
x_{n1}, &x_{n2}, x_{n3},cdots rightarrow x_n && (text{a sequence converging to} x_n)
end{align*}
Note that all the numbers $x_{ij}in S$. Now use a diagonal argument to get a sequence in $S$ converging to the limit, $x=lim_{nrightarrowinfty}x_n$. This shows that $xin S'$ by definition of $S'$.
answered Mar 28 '17 at 12:53
ParishParish
631315
$endgroup$
add a comment |
$begingroup$
Another nice way to see this would be the following.
Suppose $xin bar{S'}$. then there exists a sequence of numbers in $S'$ which converges to $x$. Let that sequence be $(x_n)$. But again as $x_nin S'$, we have the following,
begin{align*}
x_{11}, &x_{12}, x_{13},cdots rightarrow x_1 && (text{a sequence converging to} x_1) \
x_{21}, &x_{22}, x_{23},cdots rightarrow x_2 && (text{a sequence converging to} x_2) \
x_{31}, &x_{32}, x_{33},cdots rightarrow x_3 && (text{a sequence converging to} x_3) \
vdots \
x_{n1}, &x_{n2}, x_{n3},cdots rightarrow x_n && (text{a sequence converging to} x_n)
end{align*}
Note that all the numbers $x_{ij}in S$. Now use a diagonal argument to get a sequence in $S$ converging to the limit, $x=lim_{nrightarrowinfty}x_n$. This shows that $xin S'$ by definition of $S'$.
answered Mar 28 '17 at 12:53
ParishParish
631315
$endgroup$
Another nice way to see this would be the following.
Suppose $xin bar{S'}$. then there exists a sequence of numbers in $S'$ which converges to $x$. Let that sequence be $(x_n)$. But again as $x_nin S'$, we have the following,
begin{align*}
x_{11}, &x_{12}, x_{13},cdots rightarrow x_1 && (text{a sequence converging to} x_1) \
x_{21}, &x_{22}, x_{23},cdots rightarrow x_2 && (text{a sequence converging to} x_2) \
x_{31}, &x_{32}, x_{33},cdots rightarrow x_3 && (text{a sequence converging to} x_3) \
vdots \
x_{n1}, &x_{n2}, x_{n3},cdots rightarrow x_n && (text{a sequence converging to} x_n)
end{align*}
Note that all the numbers $x_{ij}in S$. Now use a diagonal argument to get a sequence in $S$ converging to the limit, $x=lim_{nrightarrowinfty}x_n$. This shows that $xin S'$ by definition of $S'$.
answered Mar 28 '17 at 12:53
ParishParish
631315
answered Mar 28 '17 at 12:53
ParishParish
631315
answered Mar 28 '17 at 12:53
ParishParish
631315
answered Mar 28 '17 at 12:53
ParishParish
631315
631315
add a comment |
add a comment |
$begingroup$
If $X$ is any topological space and $S subset X$, then ${displaystyle {bar {S}}=Xbackslash [ (Xbackslash S)^{circ }]}$. Here, we have
$tag 1 {displaystyle {bar {S}}=mathbb Rbackslash [ (mathbb Rbackslash S)^{circ }]}$
Claim: The OP's $S'$ is equal to $bar {S}$.
It is easy to see that $S' subset bar {S}$.
Using constant sequences, we see that $S subset S'$. Now let $r in bar {S}backslash S$. By (1), for every $n ge 1$ the open interval
$tag 2 (r-frac{1}{n},r+frac{1}{n})$
intersects the set $S$, so we can select $s_n$ in both the interval (2) and $S$. The sequence $(s_n)$ is $S$ converges to $r$ so $r in S'$. Since both $S$ and its $bar {S}backslash S$ are contained in $S'$, $bar {S} subset S'$.
Since $S' = bar {S}$, $S'$ is a closed set.
answered Jan 2 '18 at 3:44
CopyPasteItCopyPasteIt
4,1481628
$endgroup$
add a comment |
$begingroup$
If $X$ is any topological space and $S subset X$, then ${displaystyle {bar {S}}=Xbackslash [ (Xbackslash S)^{circ }]}$. Here, we have
$tag 1 {displaystyle {bar {S}}=mathbb Rbackslash [ (mathbb Rbackslash S)^{circ }]}$
Claim: The OP's $S'$ is equal to $bar {S}$.
It is easy to see that $S' subset bar {S}$.
Using constant sequences, we see that $S subset S'$. Now let $r in bar {S}backslash S$. By (1), for every $n ge 1$ the open interval
$tag 2 (r-frac{1}{n},r+frac{1}{n})$
intersects the set $S$, so we can select $s_n$ in both the interval (2) and $S$. The sequence $(s_n)$ is $S$ converges to $r$ so $r in S'$. Since both $S$ and its $bar {S}backslash S$ are contained in $S'$, $bar {S} subset S'$.
Since $S' = bar {S}$, $S'$ is a closed set.
answered Jan 2 '18 at 3:44
CopyPasteItCopyPasteIt
4,1481628
$endgroup$
add a comment |
$begingroup$
If $X$ is any topological space and $S subset X$, then ${displaystyle {bar {S}}=Xbackslash [ (Xbackslash S)^{circ }]}$. Here, we have
$tag 1 {displaystyle {bar {S}}=mathbb Rbackslash [ (mathbb Rbackslash S)^{circ }]}$
Claim: The OP's $S'$ is equal to $bar {S}$.
It is easy to see that $S' subset bar {S}$.
Using constant sequences, we see that $S subset S'$. Now let $r in bar {S}backslash S$. By (1), for every $n ge 1$ the open interval
$tag 2 (r-frac{1}{n},r+frac{1}{n})$
intersects the set $S$, so we can select $s_n$ in both the interval (2) and $S$. The sequence $(s_n)$ is $S$ converges to $r$ so $r in S'$. Since both $S$ and its $bar {S}backslash S$ are contained in $S'$, $bar {S} subset S'$.
Since $S' = bar {S}$, $S'$ is a closed set.
answered Jan 2 '18 at 3:44
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4,1481628
$endgroup$
If $X$ is any topological space and $S subset X$, then ${displaystyle {bar {S}}=Xbackslash [ (Xbackslash S)^{circ }]}$. Here, we have
$tag 1 {displaystyle {bar {S}}=mathbb Rbackslash [ (mathbb Rbackslash S)^{circ }]}$
Claim: The OP's $S'$ is equal to $bar {S}$.
It is easy to see that $S' subset bar {S}$.
Using constant sequences, we see that $S subset S'$. Now let $r in bar {S}backslash S$. By (1), for every $n ge 1$ the open interval
$tag 2 (r-frac{1}{n},r+frac{1}{n})$
intersects the set $S$, so we can select $s_n$ in both the interval (2) and $S$. The sequence $(s_n)$ is $S$ converges to $r$ so $r in S'$. Since both $S$ and its $bar {S}backslash S$ are contained in $S'$, $bar {S} subset S'$.
Since $S' = bar {S}$, $S'$ is a closed set.
answered Jan 2 '18 at 3:44
CopyPasteItCopyPasteIt
4,1481628
answered Jan 2 '18 at 3:44
CopyPasteItCopyPasteIt
4,1481628
answered Jan 2 '18 at 3:44
CopyPasteItCopyPasteIt
4,1481628
answered Jan 2 '18 at 3:44
CopyPasteItCopyPasteIt
4,1481628
4,1481628
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1
$begingroup$
(1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
$endgroup$
– DanielWainfleet
Aug 7 '16 at 6:54
$begingroup$
The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
$endgroup$
– martin.koeberl
Mar 28 '17 at 13:40
$begingroup$
The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
$endgroup$
– user247327
Jan 2 '18 at 2:44