Mixing
A sequence converges if and only if it has exactly one limit point - How come is this true?
$begingroup$
I am following Nocedal's Numerical Optimization book; in the Appendix about Analysis and Topology I came accross the highlighted argument:
A sequence converges implies it has one limit point: This is OK, I can show this by contradiction: Assume an arbitrary limit point $x'$ which is different from the converged point $x$ and show that the both cannot coexist.
But the other way around of the equivalence seems hardly true to me: A sequence has exactly one limit implies it converges.
For example, if we think about a sequence which has for every odd $i$, $1/i$ and for every even $i$, $2i+1$, then it has only one limit point, which is $0$ but it is obvious that it does not converge.
Am I missing something here?
real-analysis calculus sequences-and-series
asked Jan 13 at 20:51
Ufuk Can BiciciUfuk Can Bicici
1,22711027
$endgroup$
add a comment |
$begingroup$
I am following Nocedal's Numerical Optimization book; in the Appendix about Analysis and Topology I came accross the highlighted argument:
A sequence converges implies it has one limit point: This is OK, I can show this by contradiction: Assume an arbitrary limit point $x'$ which is different from the converged point $x$ and show that the both cannot coexist.
But the other way around of the equivalence seems hardly true to me: A sequence has exactly one limit implies it converges.
For example, if we think about a sequence which has for every odd $i$, $1/i$ and for every even $i$, $2i+1$, then it has only one limit point, which is $0$ but it is obvious that it does not converge.
Am I missing something here?
real-analysis calculus sequences-and-series
asked Jan 13 at 20:51
Ufuk Can BiciciUfuk Can Bicici
1,22711027
$endgroup$
1
$begingroup$
Is your book considering $infty$ as a possible limit point?
$endgroup$
– mathcounterexamples.net
Jan 13 at 21:04
$begingroup$
There is no information about that in the book.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:08
$begingroup$
If the book had said "a bounded sequence converges if and only if ...," then it would have been fine, wouldn't it?
$endgroup$
– David K
Jan 13 at 21:24
add a comment |
$begingroup$
I am following Nocedal's Numerical Optimization book; in the Appendix about Analysis and Topology I came accross the highlighted argument:
A sequence converges implies it has one limit point: This is OK, I can show this by contradiction: Assume an arbitrary limit point $x'$ which is different from the converged point $x$ and show that the both cannot coexist.
But the other way around of the equivalence seems hardly true to me: A sequence has exactly one limit implies it converges.
For example, if we think about a sequence which has for every odd $i$, $1/i$ and for every even $i$, $2i+1$, then it has only one limit point, which is $0$ but it is obvious that it does not converge.
Am I missing something here?
real-analysis calculus sequences-and-series
asked Jan 13 at 20:51
Ufuk Can BiciciUfuk Can Bicici
1,22711027
$endgroup$
I am following Nocedal's Numerical Optimization book; in the Appendix about Analysis and Topology I came accross the highlighted argument:
A sequence converges implies it has one limit point: This is OK, I can show this by contradiction: Assume an arbitrary limit point $x'$ which is different from the converged point $x$ and show that the both cannot coexist.
But the other way around of the equivalence seems hardly true to me: A sequence has exactly one limit implies it converges.
For example, if we think about a sequence which has for every odd $i$, $1/i$ and for every even $i$, $2i+1$, then it has only one limit point, which is $0$ but it is obvious that it does not converge.
Am I missing something here?
real-analysis calculus sequences-and-series
real-analysis calculus sequences-and-series
asked Jan 13 at 20:51
Ufuk Can BiciciUfuk Can Bicici
1,22711027
asked Jan 13 at 20:51
Ufuk Can BiciciUfuk Can Bicici
1,22711027
asked Jan 13 at 20:51
Ufuk Can BiciciUfuk Can Bicici
1,22711027
asked Jan 13 at 20:51
Ufuk Can BiciciUfuk Can Bicici
1,22711027
asked Jan 13 at 20:51
Ufuk Can BiciciUfuk Can Bicici
1,22711027
1,22711027
1
$begingroup$
Is your book considering $infty$ as a possible limit point?
$endgroup$
– mathcounterexamples.net
Jan 13 at 21:04
$begingroup$
There is no information about that in the book.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:08
$begingroup$
If the book had said "a bounded sequence converges if and only if ...," then it would have been fine, wouldn't it?
$endgroup$
– David K
Jan 13 at 21:24
add a comment |
1
$begingroup$
Is your book considering $infty$ as a possible limit point?
$endgroup$
– mathcounterexamples.net
Jan 13 at 21:04
$begingroup$
There is no information about that in the book.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:08
$begingroup$
If the book had said "a bounded sequence converges if and only if ...," then it would have been fine, wouldn't it?
$endgroup$
– David K
Jan 13 at 21:24
1
1
$begingroup$
Is your book considering $infty$ as a possible limit point?
$endgroup$
– mathcounterexamples.net
Jan 13 at 21:04
$begingroup$
Is your book considering $infty$ as a possible limit point?
$endgroup$
– mathcounterexamples.net
Jan 13 at 21:04
$begingroup$
There is no information about that in the book.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:08
$begingroup$
There is no information about that in the book.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:08
$begingroup$
If the book had said "a bounded sequence converges if and only if ...," then it would have been fine, wouldn't it?
$endgroup$
– David K
Jan 13 at 21:24
$begingroup$
If the book had said "a bounded sequence converges if and only if ...," then it would have been fine, wouldn't it?
$endgroup$
– David K
Jan 13 at 21:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're missing that infinities should be included as possible limit points.
Then your sequence has 2 limit points: $0$ and $+infty$.
answered Jan 13 at 21:04
BerciBerci
60.8k23673
$endgroup$
$begingroup$
But the book defines limit point as a member of $mathbb R^n.$ It seems to me there is a minor technical glitch in the book's exposition.
$endgroup$
– David K
Jan 13 at 21:13
1
$begingroup$
Yes, it seems so. Otherwise, we have to assume that the sequence is bounded to obtain the same.
$endgroup$
– Berci
Jan 13 at 23:54
add a comment |
$begingroup$
A proof goes like this.
proof
We exploit the following very much important property of compact sets:
Any sequence of a compact metric space has a convergent subsequence.
(Most of the corresponding contexts reserve this for the definition of compactness though compactness can be defined in another ways but the result is the same)
Since ${x_k}in Bbb R^n$ eventually falls in ${x | |x-r|le epsilon}$ ( because of the definition of limit point) for any $epsilon>0$ where $r$ is the limit point of $x_n$ and ${x | |x-r|leepsilon}$ is compact , then $x_k$ has a convergent subsequence namely $a_n$. Removing it from $x_n$, we can derive another convergent subsequence namely $b_n$. We can follow this procedure infinitely many times. Now let $x_n$ diverge. Then it must have at least two subsequences convergent to two different values (if not i.e. if all the convergent subsequences tend to same number, then so will do the sequence itself). Then we must have at least two limit points which is obviously a contradiction and the proof finishes up.
answered Jan 13 at 21:15
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
Since when is $mathbb R^n$ compact (other than for $n=0$)?
$endgroup$
– Henning Makholm
Jan 13 at 21:20
$begingroup$
I think there is a mistakte in your proof, $mathbb{R}^n$ is not compact.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:21
$begingroup$
Thank you for the feedback!
$endgroup$
– Mostafa Ayaz
Jan 13 at 21:30
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%2f3072497%2fa-sequence-converges-if-and-only-if-it-has-exactly-one-limit-point-how-come-is%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're missing that infinities should be included as possible limit points.
Then your sequence has 2 limit points: $0$ and $+infty$.
answered Jan 13 at 21:04
BerciBerci
60.8k23673
$endgroup$
$begingroup$
But the book defines limit point as a member of $mathbb R^n.$ It seems to me there is a minor technical glitch in the book's exposition.
$endgroup$
– David K
Jan 13 at 21:13
1
$begingroup$
Yes, it seems so. Otherwise, we have to assume that the sequence is bounded to obtain the same.
$endgroup$
– Berci
Jan 13 at 23:54
add a comment |
$begingroup$
You're missing that infinities should be included as possible limit points.
Then your sequence has 2 limit points: $0$ and $+infty$.
answered Jan 13 at 21:04
BerciBerci
60.8k23673
$endgroup$
$begingroup$
But the book defines limit point as a member of $mathbb R^n.$ It seems to me there is a minor technical glitch in the book's exposition.
$endgroup$
– David K
Jan 13 at 21:13
1
$begingroup$
Yes, it seems so. Otherwise, we have to assume that the sequence is bounded to obtain the same.
$endgroup$
– Berci
Jan 13 at 23:54
add a comment |
$begingroup$
You're missing that infinities should be included as possible limit points.
Then your sequence has 2 limit points: $0$ and $+infty$.
answered Jan 13 at 21:04
BerciBerci
60.8k23673
$endgroup$
You're missing that infinities should be included as possible limit points.
Then your sequence has 2 limit points: $0$ and $+infty$.
answered Jan 13 at 21:04
BerciBerci
60.8k23673
answered Jan 13 at 21:04
BerciBerci
60.8k23673
answered Jan 13 at 21:04
BerciBerci
60.8k23673
answered Jan 13 at 21:04
BerciBerci
60.8k23673
60.8k23673
$begingroup$
But the book defines limit point as a member of $mathbb R^n.$ It seems to me there is a minor technical glitch in the book's exposition.
$endgroup$
– David K
Jan 13 at 21:13
1
$begingroup$
Yes, it seems so. Otherwise, we have to assume that the sequence is bounded to obtain the same.
$endgroup$
– Berci
Jan 13 at 23:54
add a comment |
$begingroup$
But the book defines limit point as a member of $mathbb R^n.$ It seems to me there is a minor technical glitch in the book's exposition.
$endgroup$
– David K
Jan 13 at 21:13
1
$begingroup$
Yes, it seems so. Otherwise, we have to assume that the sequence is bounded to obtain the same.
$endgroup$
– Berci
Jan 13 at 23:54
$begingroup$
But the book defines limit point as a member of $mathbb R^n.$ It seems to me there is a minor technical glitch in the book's exposition.
$endgroup$
– David K
Jan 13 at 21:13
$begingroup$
But the book defines limit point as a member of $mathbb R^n.$ It seems to me there is a minor technical glitch in the book's exposition.
$endgroup$
– David K
Jan 13 at 21:13
1
1
$begingroup$
Yes, it seems so. Otherwise, we have to assume that the sequence is bounded to obtain the same.
$endgroup$
– Berci
Jan 13 at 23:54
$begingroup$
Yes, it seems so. Otherwise, we have to assume that the sequence is bounded to obtain the same.
$endgroup$
– Berci
Jan 13 at 23:54
add a comment |
$begingroup$
A proof goes like this.
proof
We exploit the following very much important property of compact sets:
Any sequence of a compact metric space has a convergent subsequence.
(Most of the corresponding contexts reserve this for the definition of compactness though compactness can be defined in another ways but the result is the same)
Since ${x_k}in Bbb R^n$ eventually falls in ${x | |x-r|le epsilon}$ ( because of the definition of limit point) for any $epsilon>0$ where $r$ is the limit point of $x_n$ and ${x | |x-r|leepsilon}$ is compact , then $x_k$ has a convergent subsequence namely $a_n$. Removing it from $x_n$, we can derive another convergent subsequence namely $b_n$. We can follow this procedure infinitely many times. Now let $x_n$ diverge. Then it must have at least two subsequences convergent to two different values (if not i.e. if all the convergent subsequences tend to same number, then so will do the sequence itself). Then we must have at least two limit points which is obviously a contradiction and the proof finishes up.
answered Jan 13 at 21:15
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
Since when is $mathbb R^n$ compact (other than for $n=0$)?
$endgroup$
– Henning Makholm
Jan 13 at 21:20
$begingroup$
I think there is a mistakte in your proof, $mathbb{R}^n$ is not compact.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:21
$begingroup$
Thank you for the feedback!
$endgroup$
– Mostafa Ayaz
Jan 13 at 21:30
add a comment |
$begingroup$
A proof goes like this.
proof
We exploit the following very much important property of compact sets:
Any sequence of a compact metric space has a convergent subsequence.
(Most of the corresponding contexts reserve this for the definition of compactness though compactness can be defined in another ways but the result is the same)
Since ${x_k}in Bbb R^n$ eventually falls in ${x | |x-r|le epsilon}$ ( because of the definition of limit point) for any $epsilon>0$ where $r$ is the limit point of $x_n$ and ${x | |x-r|leepsilon}$ is compact , then $x_k$ has a convergent subsequence namely $a_n$. Removing it from $x_n$, we can derive another convergent subsequence namely $b_n$. We can follow this procedure infinitely many times. Now let $x_n$ diverge. Then it must have at least two subsequences convergent to two different values (if not i.e. if all the convergent subsequences tend to same number, then so will do the sequence itself). Then we must have at least two limit points which is obviously a contradiction and the proof finishes up.
answered Jan 13 at 21:15
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
Since when is $mathbb R^n$ compact (other than for $n=0$)?
$endgroup$
– Henning Makholm
Jan 13 at 21:20
$begingroup$
I think there is a mistakte in your proof, $mathbb{R}^n$ is not compact.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:21
$begingroup$
Thank you for the feedback!
$endgroup$
– Mostafa Ayaz
Jan 13 at 21:30
add a comment |
$begingroup$
A proof goes like this.
proof
We exploit the following very much important property of compact sets:
Any sequence of a compact metric space has a convergent subsequence.
(Most of the corresponding contexts reserve this for the definition of compactness though compactness can be defined in another ways but the result is the same)
Since ${x_k}in Bbb R^n$ eventually falls in ${x | |x-r|le epsilon}$ ( because of the definition of limit point) for any $epsilon>0$ where $r$ is the limit point of $x_n$ and ${x | |x-r|leepsilon}$ is compact , then $x_k$ has a convergent subsequence namely $a_n$. Removing it from $x_n$, we can derive another convergent subsequence namely $b_n$. We can follow this procedure infinitely many times. Now let $x_n$ diverge. Then it must have at least two subsequences convergent to two different values (if not i.e. if all the convergent subsequences tend to same number, then so will do the sequence itself). Then we must have at least two limit points which is obviously a contradiction and the proof finishes up.
answered Jan 13 at 21:15
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
A proof goes like this.
proof
We exploit the following very much important property of compact sets:
Any sequence of a compact metric space has a convergent subsequence.
(Most of the corresponding contexts reserve this for the definition of compactness though compactness can be defined in another ways but the result is the same)
Since ${x_k}in Bbb R^n$ eventually falls in ${x | |x-r|le epsilon}$ ( because of the definition of limit point) for any $epsilon>0$ where $r$ is the limit point of $x_n$ and ${x | |x-r|leepsilon}$ is compact , then $x_k$ has a convergent subsequence namely $a_n$. Removing it from $x_n$, we can derive another convergent subsequence namely $b_n$. We can follow this procedure infinitely many times. Now let $x_n$ diverge. Then it must have at least two subsequences convergent to two different values (if not i.e. if all the convergent subsequences tend to same number, then so will do the sequence itself). Then we must have at least two limit points which is obviously a contradiction and the proof finishes up.
answered Jan 13 at 21:15
Mostafa AyazMostafa Ayaz
15.6k3939
edited Jan 13 at 21:29
answered Jan 13 at 21:15
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 21:15
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 21:15
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
$begingroup$
Since when is $mathbb R^n$ compact (other than for $n=0$)?
$endgroup$
– Henning Makholm
Jan 13 at 21:20
$begingroup$
I think there is a mistakte in your proof, $mathbb{R}^n$ is not compact.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:21
$begingroup$
Thank you for the feedback!
$endgroup$
– Mostafa Ayaz
Jan 13 at 21:30
add a comment |
$begingroup$
Since when is $mathbb R^n$ compact (other than for $n=0$)?
$endgroup$
– Henning Makholm
Jan 13 at 21:20
$begingroup$
I think there is a mistakte in your proof, $mathbb{R}^n$ is not compact.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:21
$begingroup$
Thank you for the feedback!
$endgroup$
– Mostafa Ayaz
Jan 13 at 21:30
$begingroup$
Since when is $mathbb R^n$ compact (other than for $n=0$)?
$endgroup$
– Henning Makholm
Jan 13 at 21:20
$begingroup$
Since when is $mathbb R^n$ compact (other than for $n=0$)?
$endgroup$
– Henning Makholm
Jan 13 at 21:20
$begingroup$
I think there is a mistakte in your proof, $mathbb{R}^n$ is not compact.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:21
$begingroup$
I think there is a mistakte in your proof, $mathbb{R}^n$ is not compact.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:21
$begingroup$
Thank you for the feedback!
$endgroup$
– Mostafa Ayaz
Jan 13 at 21:30
$begingroup$
Thank you for the feedback!
$endgroup$
– Mostafa Ayaz
Jan 13 at 21:30
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.
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%2f3072497%2fa-sequence-converges-if-and-only-if-it-has-exactly-one-limit-point-how-come-is%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
$begingroup$
Is your book considering $infty$ as a possible limit point?
$endgroup$
– mathcounterexamples.net
Jan 13 at 21:04
$begingroup$
There is no information about that in the book.
$endgroup$
– Ufuk Can Bicici
Jan 13 at 21:08
$begingroup$
If the book had said "a bounded sequence converges if and only if ...," then it would have been fine, wouldn't it?
$endgroup$
– David K
Jan 13 at 21:24