Prove: A set containing limit points of a sequence is a closed set
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
user3025403
1488
add a comment |
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
user3025403
1488
1
(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 $r>0.$)
– DanielWainfleet
Aug 7 '16 at 6:54
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 $S^prime$?
– martin.koeberl
Mar 28 '17 at 13:40
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".
– user247327
Jan 2 '18 at 2:44
add a comment |
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
user3025403
1488
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
user3025403
1488
edited May 12 '14 at 6:05
Jean-Claude Arbaut
14.7k63464
asked May 12 '14 at 0:24
user3025403
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
user3025403
1488
asked May 12 '14 at 0:24
user3025403
1488
asked May 12 '14 at 0:24
user3025403
1488
1488
1
(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 $r>0.$)
– DanielWainfleet
Aug 7 '16 at 6:54
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 $S^prime$?
– martin.koeberl
Mar 28 '17 at 13:40
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".
– user247327
Jan 2 '18 at 2:44
add a comment |
1
(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 $r>0.$)
– DanielWainfleet
Aug 7 '16 at 6:54
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 $S^prime$?
– martin.koeberl
Mar 28 '17 at 13:40
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".
– user247327
Jan 2 '18 at 2:44
1
1
(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 $r>0.$)
– DanielWainfleet
Aug 7 '16 at 6:54
(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 $r>0.$)
– DanielWainfleet
Aug 7 '16 at 6:54
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 $S^prime$?
– martin.koeberl
Mar 28 '17 at 13:40
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 $S^prime$?
– martin.koeberl
Mar 28 '17 at 13:40
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".
– user247327
Jan 2 '18 at 2:44
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".
– user247327
Jan 2 '18 at 2:44
add a comment |
3 Answers
3
active
oldest
votes
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
Zircht
1,623813
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.
– Sayan Bandyapadhyay
May 12 '14 at 1:01
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
– Zircht
May 12 '14 at 1:05
add a comment |
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
Parish
631315
add a comment |
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
CopyPasteIt
4,0301627
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f791041%2fprove-a-set-containing-limit-points-of-a-sequence-is-a-closed-set%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
Zircht
1,623813
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.
– Sayan Bandyapadhyay
May 12 '14 at 1:01
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
– Zircht
May 12 '14 at 1:05
add a comment |
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
Zircht
1,623813
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.
– Sayan Bandyapadhyay
May 12 '14 at 1:01
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
– Zircht
May 12 '14 at 1:05
add a comment |
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
Zircht
1,623813
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
Zircht
1,623813
answered May 12 '14 at 0:39
Zircht
1,623813
answered May 12 '14 at 0:39
Zircht
1,623813
answered May 12 '14 at 0:39
Zircht
1,623813
1,623813
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.
– Sayan Bandyapadhyay
May 12 '14 at 1:01
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
– Zircht
May 12 '14 at 1:05
add a comment |
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.
– Sayan Bandyapadhyay
May 12 '14 at 1:01
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
– Zircht
May 12 '14 at 1:05
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.
– Sayan Bandyapadhyay
May 12 '14 at 1:01
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.
– Sayan Bandyapadhyay
May 12 '14 at 1:01
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
– Zircht
May 12 '14 at 1:05
@SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
– Zircht
May 12 '14 at 1:05
add a comment |
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
Parish
631315
add a comment |
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
Parish
631315
add a comment |
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
Parish
631315
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
Parish
631315
answered Mar 28 '17 at 12:53
Parish
631315
answered Mar 28 '17 at 12:53
Parish
631315
answered Mar 28 '17 at 12:53
Parish
631315
631315
add a comment |
add a comment |
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
CopyPasteIt
4,0301627
add a comment |
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
CopyPasteIt
4,0301627
add a comment |
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
CopyPasteIt
4,0301627
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
CopyPasteIt
4,0301627
answered Jan 2 '18 at 3:44
CopyPasteIt
4,0301627
answered Jan 2 '18 at 3:44
CopyPasteIt
4,0301627
answered Jan 2 '18 at 3:44
CopyPasteIt
4,0301627
4,0301627
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f791041%2fprove-a-set-containing-limit-points-of-a-sequence-is-a-closed-set%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
(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 $r>0.$)
– DanielWainfleet
Aug 7 '16 at 6:54
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 $S^prime$?
– martin.koeberl
Mar 28 '17 at 13:40
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".
– user247327
Jan 2 '18 at 2:44