Prove that any bounded sequence may be split up in countably many sequences having the same limit. The Next CEO of Stack OverflowSuppose $a_n$ is a sequence with $ L_1, L_2, …$ subsequential limits. Suppose $L_n to L$. Prove that $L$ is a subsequential limitLet $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.Let $f$ be a real uniformly continuous function on the bounded set $E$ in $mathbbR^1$. Prove that $f$ is bounded on $E$Prove sequence converges if only some subsequences converge to the same limitShow that if all convergent subsequences of a bounded sequence converge to $l$, the sequence itself must also converge to $l$.Prove that the sequence is bounded and subsequences convergeProve that $f$ has a limit at infinity, provided that $|f(x)-f(x_n)| leq 1/n$Find the subsequential limits for $x_n=left1,1over 10,2over 10,cdots,9over 10,1over 10^2,cdots10^n-1over 10^n,cdotsright$Help with the proof that a sequence is convergent iff it is bounded and has a single subsequential limit.Proof verification. If $x_n$ is a monotone sequence and it has a convergent subsequence $x_n_k$, then $x_n$ is convergent to the same limit.
Why did the Drakh emissary look so blurred in S04:E11 "Lines of Communication"?
Find a path from s to t using as few red nodes as possible
What is a typical Mizrachi Seder like?
That's an odd coin - I wonder why
Does the Idaho Potato Commission associate potato skins with healthy eating?
Is it a bad idea to plug the other end of ESD strap to wall ground?
How did scripture get the name bible?
Prodigo = pro + ago?
Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?
Strange use of "whether ... than ..." in official text
Free fall ellipse or parabola?
Mathematica command that allows it to read my intentions
Ising model simulation
Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?
Avoiding the "not like other girls" trope?
Is a linearly independent set whose span is dense a Schauder basis?
Why did early computer designers eschew integers?
Can a PhD from a non-TU9 German university become a professor in a TU9 university?
Planeswalker Ability and Death Timing
Find the majority element, which appears more than half the time
How should I connect my cat5 cable to connectors having an orange-green line?
How does a dynamic QR code work?
Oldie but Goldie
Which acid/base does a strong base/acid react when added to a buffer solution?
Prove that any bounded sequence may be split up in countably many sequences having the same limit.
The Next CEO of Stack OverflowSuppose $a_n$ is a sequence with $ L_1, L_2, …$ subsequential limits. Suppose $L_n to L$. Prove that $L$ is a subsequential limitLet $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.Let $f$ be a real uniformly continuous function on the bounded set $E$ in $mathbbR^1$. Prove that $f$ is bounded on $E$Prove sequence converges if only some subsequences converge to the same limitShow that if all convergent subsequences of a bounded sequence converge to $l$, the sequence itself must also converge to $l$.Prove that the sequence is bounded and subsequences convergeProve that $f$ has a limit at infinity, provided that $|f(x)-f(x_n)| leq 1/n$Find the subsequential limits for $x_n=left1,1over 10,2over 10,cdots,9over 10,1over 10^2,cdots10^n-1over 10^n,cdotsright$Help with the proof that a sequence is convergent iff it is bounded and has a single subsequential limit.Proof verification. If $x_n$ is a monotone sequence and it has a convergent subsequence $x_n_k$, then $x_n$ is convergent to the same limit.
$begingroup$
Given a bounded sequence $x_n$ prove that it may be split into countably many sequences having the same limit.
The question is stated as above, without any other constraints on the sequence. I'm not sure how to follow with the proof, but here are some thoughts of mine.
First of all $x_n$ is bounded meaning:
$$
exists MinBbb R: |x_n| le M, forall ninBbb N
$$
Recall that a bounded sequence does have subsequential limits. Let's denote the set of subsequential limits as $Bbb L$:
$$
Bbb L = L_1, L_2, L_3, dots
$$
Now from each of the subsequences, we may go one level down and choose subsequences such that their limit is equal to $L_k$, namely:
$$
lim_ptoinfty(x_n)_p^(k) = L_k, forall k inBbb N
$$
Now I was thinking about using an approach similar to proving the countability of rational numbers.
$$
beginarrayc
hline L_1 & L_2 & cdots & L_k \hline
(x_n)_1^(1)& (x_n)_1^(2) &cdots &(x_n)_1^(k) \hline
(x_n)_2^(1) & (x_n)_2^(2) &cdots &(x_n)_2^(k) \hline
(x_n)_3^(1) & (x_n)_3^(2) &cdots &(x_n)_3^(k) \hline
cdots & cdots &cdots &cdots \hline
(x_n)_m^(1) & (x_n)_m^(2) &cdots &(x_n)_m^(k) \hline
endarray
$$
So the first column contains all possible subsequences converging to $L_1$, the second all possible subsequences converging to $L_2$ and etc. We can now assign an index to each element in the table, in the same manner, as it is done while proving countability of rationals.
The problem is I'm not sure whether this approach even works. Especially in the case when $x_n$ is a sequence such that the set of its limit point forms a dense interval.
In case all of the above is a non-sense could someone please suggest a proof of what's in problem statement? Thank you!
real-analysis sequences-and-series limits proof-verification
asked Mar 20 at 11:20
romanroman
2,42021226
$endgroup$
add a comment |
$begingroup$
Given a bounded sequence $x_n$ prove that it may be split into countably many sequences having the same limit.
The question is stated as above, without any other constraints on the sequence. I'm not sure how to follow with the proof, but here are some thoughts of mine.
First of all $x_n$ is bounded meaning:
$$
exists MinBbb R: |x_n| le M, forall ninBbb N
$$
Recall that a bounded sequence does have subsequential limits. Let's denote the set of subsequential limits as $Bbb L$:
$$
Bbb L = L_1, L_2, L_3, dots
$$
Now from each of the subsequences, we may go one level down and choose subsequences such that their limit is equal to $L_k$, namely:
$$
lim_ptoinfty(x_n)_p^(k) = L_k, forall k inBbb N
$$
Now I was thinking about using an approach similar to proving the countability of rational numbers.
$$
beginarrayc
hline L_1 & L_2 & cdots & L_k \hline
(x_n)_1^(1)& (x_n)_1^(2) &cdots &(x_n)_1^(k) \hline
(x_n)_2^(1) & (x_n)_2^(2) &cdots &(x_n)_2^(k) \hline
(x_n)_3^(1) & (x_n)_3^(2) &cdots &(x_n)_3^(k) \hline
cdots & cdots &cdots &cdots \hline
(x_n)_m^(1) & (x_n)_m^(2) &cdots &(x_n)_m^(k) \hline
endarray
$$
So the first column contains all possible subsequences converging to $L_1$, the second all possible subsequences converging to $L_2$ and etc. We can now assign an index to each element in the table, in the same manner, as it is done while proving countability of rationals.
The problem is I'm not sure whether this approach even works. Especially in the case when $x_n$ is a sequence such that the set of its limit point forms a dense interval.
In case all of the above is a non-sense could someone please suggest a proof of what's in problem statement? Thank you!
real-analysis sequences-and-series limits proof-verification
asked Mar 20 at 11:20
romanroman
2,42021226
$endgroup$
$begingroup$
Well the number of such parts has to be countable as the number of terms in whole sequence is countable.
$endgroup$
– Paramanand Singh
Mar 20 at 13:34
add a comment |
$begingroup$
Given a bounded sequence $x_n$ prove that it may be split into countably many sequences having the same limit.
The question is stated as above, without any other constraints on the sequence. I'm not sure how to follow with the proof, but here are some thoughts of mine.
First of all $x_n$ is bounded meaning:
$$
exists MinBbb R: |x_n| le M, forall ninBbb N
$$
Recall that a bounded sequence does have subsequential limits. Let's denote the set of subsequential limits as $Bbb L$:
$$
Bbb L = L_1, L_2, L_3, dots
$$
Now from each of the subsequences, we may go one level down and choose subsequences such that their limit is equal to $L_k$, namely:
$$
lim_ptoinfty(x_n)_p^(k) = L_k, forall k inBbb N
$$
Now I was thinking about using an approach similar to proving the countability of rational numbers.
$$
beginarrayc
hline L_1 & L_2 & cdots & L_k \hline
(x_n)_1^(1)& (x_n)_1^(2) &cdots &(x_n)_1^(k) \hline
(x_n)_2^(1) & (x_n)_2^(2) &cdots &(x_n)_2^(k) \hline
(x_n)_3^(1) & (x_n)_3^(2) &cdots &(x_n)_3^(k) \hline
cdots & cdots &cdots &cdots \hline
(x_n)_m^(1) & (x_n)_m^(2) &cdots &(x_n)_m^(k) \hline
endarray
$$
So the first column contains all possible subsequences converging to $L_1$, the second all possible subsequences converging to $L_2$ and etc. We can now assign an index to each element in the table, in the same manner, as it is done while proving countability of rationals.
The problem is I'm not sure whether this approach even works. Especially in the case when $x_n$ is a sequence such that the set of its limit point forms a dense interval.
In case all of the above is a non-sense could someone please suggest a proof of what's in problem statement? Thank you!
real-analysis sequences-and-series limits proof-verification
asked Mar 20 at 11:20
romanroman
2,42021226
$endgroup$
Given a bounded sequence $x_n$ prove that it may be split into countably many sequences having the same limit.
The question is stated as above, without any other constraints on the sequence. I'm not sure how to follow with the proof, but here are some thoughts of mine.
First of all $x_n$ is bounded meaning:
$$
exists MinBbb R: |x_n| le M, forall ninBbb N
$$
Recall that a bounded sequence does have subsequential limits. Let's denote the set of subsequential limits as $Bbb L$:
$$
Bbb L = L_1, L_2, L_3, dots
$$
Now from each of the subsequences, we may go one level down and choose subsequences such that their limit is equal to $L_k$, namely:
$$
lim_ptoinfty(x_n)_p^(k) = L_k, forall k inBbb N
$$
Now I was thinking about using an approach similar to proving the countability of rational numbers.
$$
beginarrayc
hline L_1 & L_2 & cdots & L_k \hline
(x_n)_1^(1)& (x_n)_1^(2) &cdots &(x_n)_1^(k) \hline
(x_n)_2^(1) & (x_n)_2^(2) &cdots &(x_n)_2^(k) \hline
(x_n)_3^(1) & (x_n)_3^(2) &cdots &(x_n)_3^(k) \hline
cdots & cdots &cdots &cdots \hline
(x_n)_m^(1) & (x_n)_m^(2) &cdots &(x_n)_m^(k) \hline
endarray
$$
So the first column contains all possible subsequences converging to $L_1$, the second all possible subsequences converging to $L_2$ and etc. We can now assign an index to each element in the table, in the same manner, as it is done while proving countability of rationals.
The problem is I'm not sure whether this approach even works. Especially in the case when $x_n$ is a sequence such that the set of its limit point forms a dense interval.
In case all of the above is a non-sense could someone please suggest a proof of what's in problem statement? Thank you!
real-analysis sequences-and-series limits proof-verification
real-analysis sequences-and-series limits proof-verification
asked Mar 20 at 11:20
romanroman
2,42021226
asked Mar 20 at 11:20
romanroman
2,42021226
asked Mar 20 at 11:20
romanroman
2,42021226
asked Mar 20 at 11:20
romanroman
2,42021226
asked Mar 20 at 11:20
romanroman
2,42021226
2,42021226
$begingroup$
Well the number of such parts has to be countable as the number of terms in whole sequence is countable.
$endgroup$
– Paramanand Singh
Mar 20 at 13:34
add a comment |
$begingroup$
Well the number of such parts has to be countable as the number of terms in whole sequence is countable.
$endgroup$
– Paramanand Singh
Mar 20 at 13:34
$begingroup$
Well the number of such parts has to be countable as the number of terms in whole sequence is countable.
$endgroup$
– Paramanand Singh
Mar 20 at 13:34
$begingroup$
Well the number of such parts has to be countable as the number of terms in whole sequence is countable.
$endgroup$
– Paramanand Singh
Mar 20 at 13:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A substantial hint, but not an answer:
You're sort of on the right track. First, pick a subsequence (any one!) that you like, and which converges to some limit $L$. Call the items in this subsequence "good" and the remaining items "bad".
Looking at the good subsequence, can you think of a convergent sub-sequence? Can you think of two disjoint convergent subsequences? If you could, let's call them $a_n$ and $b_n$ (I'm skipping the sequence notation here) and suppose that $P$ and $Q$ are two "bad" items. Then the sequences
$$
P, a_1, a_2, ldots, \
Q, b_1, b_2, ldots,$$
both converge. In fact, if you put "P" in the right place among the $a$s (i.e., in the place it occurs within the $x_n$ sequence), that'd still be find, and the same for $Q$.
Of course, this only deals with two of the "bad" elements, but you have lots of them...but perhaps you can generalize from here.
edited Mar 20 at 13:51
K.Power
3,620926
answered Mar 20 at 11:34
John HughesJohn Hughes
65.1k24293
$endgroup$
add a comment |
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%2f3155304%2fprove-that-any-bounded-sequence-may-be-split-up-in-countably-many-sequences-havi%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A substantial hint, but not an answer:
You're sort of on the right track. First, pick a subsequence (any one!) that you like, and which converges to some limit $L$. Call the items in this subsequence "good" and the remaining items "bad".
Looking at the good subsequence, can you think of a convergent sub-sequence? Can you think of two disjoint convergent subsequences? If you could, let's call them $a_n$ and $b_n$ (I'm skipping the sequence notation here) and suppose that $P$ and $Q$ are two "bad" items. Then the sequences
$$
P, a_1, a_2, ldots, \
Q, b_1, b_2, ldots,$$
both converge. In fact, if you put "P" in the right place among the $a$s (i.e., in the place it occurs within the $x_n$ sequence), that'd still be find, and the same for $Q$.
Of course, this only deals with two of the "bad" elements, but you have lots of them...but perhaps you can generalize from here.
edited Mar 20 at 13:51
K.Power
3,620926
answered Mar 20 at 11:34
John HughesJohn Hughes
65.1k24293
$endgroup$
add a comment |
$begingroup$
A substantial hint, but not an answer:
You're sort of on the right track. First, pick a subsequence (any one!) that you like, and which converges to some limit $L$. Call the items in this subsequence "good" and the remaining items "bad".
Looking at the good subsequence, can you think of a convergent sub-sequence? Can you think of two disjoint convergent subsequences? If you could, let's call them $a_n$ and $b_n$ (I'm skipping the sequence notation here) and suppose that $P$ and $Q$ are two "bad" items. Then the sequences
$$
P, a_1, a_2, ldots, \
Q, b_1, b_2, ldots,$$
both converge. In fact, if you put "P" in the right place among the $a$s (i.e., in the place it occurs within the $x_n$ sequence), that'd still be find, and the same for $Q$.
Of course, this only deals with two of the "bad" elements, but you have lots of them...but perhaps you can generalize from here.
edited Mar 20 at 13:51
K.Power
3,620926
answered Mar 20 at 11:34
John HughesJohn Hughes
65.1k24293
$endgroup$
add a comment |
$begingroup$
A substantial hint, but not an answer:
You're sort of on the right track. First, pick a subsequence (any one!) that you like, and which converges to some limit $L$. Call the items in this subsequence "good" and the remaining items "bad".
Looking at the good subsequence, can you think of a convergent sub-sequence? Can you think of two disjoint convergent subsequences? If you could, let's call them $a_n$ and $b_n$ (I'm skipping the sequence notation here) and suppose that $P$ and $Q$ are two "bad" items. Then the sequences
$$
P, a_1, a_2, ldots, \
Q, b_1, b_2, ldots,$$
both converge. In fact, if you put "P" in the right place among the $a$s (i.e., in the place it occurs within the $x_n$ sequence), that'd still be find, and the same for $Q$.
Of course, this only deals with two of the "bad" elements, but you have lots of them...but perhaps you can generalize from here.
edited Mar 20 at 13:51
K.Power
3,620926
answered Mar 20 at 11:34
John HughesJohn Hughes
65.1k24293
$endgroup$
A substantial hint, but not an answer:
You're sort of on the right track. First, pick a subsequence (any one!) that you like, and which converges to some limit $L$. Call the items in this subsequence "good" and the remaining items "bad".
Looking at the good subsequence, can you think of a convergent sub-sequence? Can you think of two disjoint convergent subsequences? If you could, let's call them $a_n$ and $b_n$ (I'm skipping the sequence notation here) and suppose that $P$ and $Q$ are two "bad" items. Then the sequences
$$
P, a_1, a_2, ldots, \
Q, b_1, b_2, ldots,$$
both converge. In fact, if you put "P" in the right place among the $a$s (i.e., in the place it occurs within the $x_n$ sequence), that'd still be find, and the same for $Q$.
Of course, this only deals with two of the "bad" elements, but you have lots of them...but perhaps you can generalize from here.
edited Mar 20 at 13:51
K.Power
3,620926
answered Mar 20 at 11:34
John HughesJohn Hughes
65.1k24293
edited Mar 20 at 13:51
K.Power
3,620926
edited Mar 20 at 13:51
K.Power
3,620926
edited Mar 20 at 13:51
K.Power
3,620926
3,620926
answered Mar 20 at 11:34
John HughesJohn Hughes
65.1k24293
answered Mar 20 at 11:34
John HughesJohn Hughes
65.1k24293
answered Mar 20 at 11:34
John HughesJohn Hughes
65.1k24293
65.1k24293
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.
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%2f3155304%2fprove-that-any-bounded-sequence-may-be-split-up-in-countably-many-sequences-havi%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
$begingroup$
Well the number of such parts has to be countable as the number of terms in whole sequence is countable.
$endgroup$
– Paramanand Singh
Mar 20 at 13:34