Mixing
Bolzano-Weierstrass bounded
$begingroup$
Let $a_n$ be a bounded sequence in $R$ that has three subsequences with 1,2 and 3 limit points respectively. Show that there is an $n_0$ in $N$ such that $a_n leq 4$ for all $ngeq n_0$
Now i know that there is an theorem Bolzano-Weierstrass which says that if $a_n$ is bounded above by $M$ then there exists a sub-sequence $a_{n_k}$ That has a $L$ in the interval [-$M$,$M$]. Now should I just state the theorem or do something specific to proof this?
real-analysis sequences-and-series convergence
edited Apr 9 '17 at 11:05
Ali
1,9682520
asked Apr 9 '17 at 10:59
MarkMark
758
$endgroup$
add a comment |
$begingroup$
Let $a_n$ be a bounded sequence in $R$ that has three subsequences with 1,2 and 3 limit points respectively. Show that there is an $n_0$ in $N$ such that $a_n leq 4$ for all $ngeq n_0$
Now i know that there is an theorem Bolzano-Weierstrass which says that if $a_n$ is bounded above by $M$ then there exists a sub-sequence $a_{n_k}$ That has a $L$ in the interval [-$M$,$M$]. Now should I just state the theorem or do something specific to proof this?
real-analysis sequences-and-series convergence
edited Apr 9 '17 at 11:05
Ali
1,9682520
asked Apr 9 '17 at 10:59
MarkMark
758
$endgroup$
$begingroup$
If such $n_0$ does not exist then a subsequence $(a_{n_k})_k$ exists with $a_{n_k}geq4$ for each $k$. This sequence is bounded hence (B-W) will have convergent subsequence with a limit $geq4$.
$endgroup$
– drhab
Apr 9 '17 at 11:08
add a comment |
$begingroup$
Let $a_n$ be a bounded sequence in $R$ that has three subsequences with 1,2 and 3 limit points respectively. Show that there is an $n_0$ in $N$ such that $a_n leq 4$ for all $ngeq n_0$
Now i know that there is an theorem Bolzano-Weierstrass which says that if $a_n$ is bounded above by $M$ then there exists a sub-sequence $a_{n_k}$ That has a $L$ in the interval [-$M$,$M$]. Now should I just state the theorem or do something specific to proof this?
real-analysis sequences-and-series convergence
edited Apr 9 '17 at 11:05
Ali
1,9682520
asked Apr 9 '17 at 10:59
MarkMark
758
$endgroup$
Let $a_n$ be a bounded sequence in $R$ that has three subsequences with 1,2 and 3 limit points respectively. Show that there is an $n_0$ in $N$ such that $a_n leq 4$ for all $ngeq n_0$
Now i know that there is an theorem Bolzano-Weierstrass which says that if $a_n$ is bounded above by $M$ then there exists a sub-sequence $a_{n_k}$ That has a $L$ in the interval [-$M$,$M$]. Now should I just state the theorem or do something specific to proof this?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited Apr 9 '17 at 11:05
Ali
1,9682520
asked Apr 9 '17 at 10:59
MarkMark
758
edited Apr 9 '17 at 11:05
Ali
1,9682520
asked Apr 9 '17 at 10:59
MarkMark
758
edited Apr 9 '17 at 11:05
Ali
1,9682520
edited Apr 9 '17 at 11:05
Ali
1,9682520
edited Apr 9 '17 at 11:05
Ali
1,9682520
1,9682520
asked Apr 9 '17 at 10:59
MarkMark
758
asked Apr 9 '17 at 10:59
MarkMark
758
asked Apr 9 '17 at 10:59
MarkMark
758
758
$begingroup$
If such $n_0$ does not exist then a subsequence $(a_{n_k})_k$ exists with $a_{n_k}geq4$ for each $k$. This sequence is bounded hence (B-W) will have convergent subsequence with a limit $geq4$.
$endgroup$
– drhab
Apr 9 '17 at 11:08
add a comment |
$begingroup$
If such $n_0$ does not exist then a subsequence $(a_{n_k})_k$ exists with $a_{n_k}geq4$ for each $k$. This sequence is bounded hence (B-W) will have convergent subsequence with a limit $geq4$.
$endgroup$
– drhab
Apr 9 '17 at 11:08
$begingroup$
If such $n_0$ does not exist then a subsequence $(a_{n_k})_k$ exists with $a_{n_k}geq4$ for each $k$. This sequence is bounded hence (B-W) will have convergent subsequence with a limit $geq4$.
$endgroup$
– drhab
Apr 9 '17 at 11:08
$begingroup$
If such $n_0$ does not exist then a subsequence $(a_{n_k})_k$ exists with $a_{n_k}geq4$ for each $k$. This sequence is bounded hence (B-W) will have convergent subsequence with a limit $geq4$.
$endgroup$
– drhab
Apr 9 '17 at 11:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose there does not exist such an $n_0$. Then, we claim there is a subsequence $a_{n_k}$ such that $a_{n_k} > 4$ for all $k$. This is because if there are only finitely many $a_n$ having this property, let $N$ be the maximal $n$ for which it happens, and $n geq N$ will imply $a_n leq 4$, a contradiction.
Now, $a_{n_k}$ is bounded, say by $M$, where $M > 4$. Then by Bolzano Weierstrass, there must be a limit point of this subsequence in $[4,M]$. But then, since this is again a subsequence of $a_n$, $a_n$ has a limit point in $[4,M]$, a contradiction since $3$ was said to be the maximal limit point of $a_n$. This contradiction means that $a_{n_k}$ couldn't have existed in the first place.
Note : we did not use the fact that $1,2$ are also limit points of $a_n$.
EDIT : I have a direct proof, but I want you to fill in the details.
1) Let $b_k = sup_{n geq k} a_n$. Then, $b_n$ is a bounded, increasing sequence.
2) $b_n$ must converge, but in addition, converges to $3$, the largest limit point of $a_n$.
3) By definition, taking $epsilon = frac 12$, there exists $K$ such that $|b_k - 3| leq frac 12$ for $k geq K$, since $b_k$ converges to $3$. In particular, by triangle inequality, $|b_k| leq 3 + frac 12 leq 3.5$.
Because $b_K = sup_{k geq K} a_n$, by definition of supremum, for all $m geq K$, $a_m leq b_K$. However, since $b_K leq 3.5$ , we have that for $m geq K$, $a_m leq 3.5$.
With a similar technique, the following extension can be obtained :
Given a sequence $x_n$ with limit points $a_1,...,a_n$, and any $delta_1,...,delta_n > 0$, there exists $n_0$ such that if $n > n_0$ then $x_n in cup_{i=1}^n(a_i-delta_i,a_i+delta_i)$. In words, eventually every $x_i$ is close enough to at least one of the $a_i$.
answered Apr 9 '17 at 11:08
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.5k33376
$endgroup$
$begingroup$
A system flag was raised for the sheer number of comments. It seemed to me that you had concluded, so I purged the whole lot. If you need to review something, @-ping me or flag a moderator for assistance.
$endgroup$
– Jyrki Lahtonen
Apr 9 '17 at 16:10
$begingroup$
@JyrkiLahtonen We have concluded discussion of this question. Thank you for mopping up, there were quite a few comments.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 17:36
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%2f2226064%2fbolzano-weierstrass-bounded%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
$begingroup$
Suppose there does not exist such an $n_0$. Then, we claim there is a subsequence $a_{n_k}$ such that $a_{n_k} > 4$ for all $k$. This is because if there are only finitely many $a_n$ having this property, let $N$ be the maximal $n$ for which it happens, and $n geq N$ will imply $a_n leq 4$, a contradiction.
Now, $a_{n_k}$ is bounded, say by $M$, where $M > 4$. Then by Bolzano Weierstrass, there must be a limit point of this subsequence in $[4,M]$. But then, since this is again a subsequence of $a_n$, $a_n$ has a limit point in $[4,M]$, a contradiction since $3$ was said to be the maximal limit point of $a_n$. This contradiction means that $a_{n_k}$ couldn't have existed in the first place.
Note : we did not use the fact that $1,2$ are also limit points of $a_n$.
EDIT : I have a direct proof, but I want you to fill in the details.
1) Let $b_k = sup_{n geq k} a_n$. Then, $b_n$ is a bounded, increasing sequence.
2) $b_n$ must converge, but in addition, converges to $3$, the largest limit point of $a_n$.
3) By definition, taking $epsilon = frac 12$, there exists $K$ such that $|b_k - 3| leq frac 12$ for $k geq K$, since $b_k$ converges to $3$. In particular, by triangle inequality, $|b_k| leq 3 + frac 12 leq 3.5$.
Because $b_K = sup_{k geq K} a_n$, by definition of supremum, for all $m geq K$, $a_m leq b_K$. However, since $b_K leq 3.5$ , we have that for $m geq K$, $a_m leq 3.5$.
With a similar technique, the following extension can be obtained :
Given a sequence $x_n$ with limit points $a_1,...,a_n$, and any $delta_1,...,delta_n > 0$, there exists $n_0$ such that if $n > n_0$ then $x_n in cup_{i=1}^n(a_i-delta_i,a_i+delta_i)$. In words, eventually every $x_i$ is close enough to at least one of the $a_i$.
answered Apr 9 '17 at 11:08
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.5k33376
$endgroup$
$begingroup$
A system flag was raised for the sheer number of comments. It seemed to me that you had concluded, so I purged the whole lot. If you need to review something, @-ping me or flag a moderator for assistance.
$endgroup$
– Jyrki Lahtonen
Apr 9 '17 at 16:10
$begingroup$
@JyrkiLahtonen We have concluded discussion of this question. Thank you for mopping up, there were quite a few comments.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 17:36
add a comment |
$begingroup$
Suppose there does not exist such an $n_0$. Then, we claim there is a subsequence $a_{n_k}$ such that $a_{n_k} > 4$ for all $k$. This is because if there are only finitely many $a_n$ having this property, let $N$ be the maximal $n$ for which it happens, and $n geq N$ will imply $a_n leq 4$, a contradiction.
Now, $a_{n_k}$ is bounded, say by $M$, where $M > 4$. Then by Bolzano Weierstrass, there must be a limit point of this subsequence in $[4,M]$. But then, since this is again a subsequence of $a_n$, $a_n$ has a limit point in $[4,M]$, a contradiction since $3$ was said to be the maximal limit point of $a_n$. This contradiction means that $a_{n_k}$ couldn't have existed in the first place.
Note : we did not use the fact that $1,2$ are also limit points of $a_n$.
EDIT : I have a direct proof, but I want you to fill in the details.
1) Let $b_k = sup_{n geq k} a_n$. Then, $b_n$ is a bounded, increasing sequence.
2) $b_n$ must converge, but in addition, converges to $3$, the largest limit point of $a_n$.
3) By definition, taking $epsilon = frac 12$, there exists $K$ such that $|b_k - 3| leq frac 12$ for $k geq K$, since $b_k$ converges to $3$. In particular, by triangle inequality, $|b_k| leq 3 + frac 12 leq 3.5$.
Because $b_K = sup_{k geq K} a_n$, by definition of supremum, for all $m geq K$, $a_m leq b_K$. However, since $b_K leq 3.5$ , we have that for $m geq K$, $a_m leq 3.5$.
With a similar technique, the following extension can be obtained :
Given a sequence $x_n$ with limit points $a_1,...,a_n$, and any $delta_1,...,delta_n > 0$, there exists $n_0$ such that if $n > n_0$ then $x_n in cup_{i=1}^n(a_i-delta_i,a_i+delta_i)$. In words, eventually every $x_i$ is close enough to at least one of the $a_i$.
answered Apr 9 '17 at 11:08
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.5k33376
$endgroup$
$begingroup$
A system flag was raised for the sheer number of comments. It seemed to me that you had concluded, so I purged the whole lot. If you need to review something, @-ping me or flag a moderator for assistance.
$endgroup$
– Jyrki Lahtonen
Apr 9 '17 at 16:10
$begingroup$
@JyrkiLahtonen We have concluded discussion of this question. Thank you for mopping up, there were quite a few comments.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 17:36
add a comment |
$begingroup$
Suppose there does not exist such an $n_0$. Then, we claim there is a subsequence $a_{n_k}$ such that $a_{n_k} > 4$ for all $k$. This is because if there are only finitely many $a_n$ having this property, let $N$ be the maximal $n$ for which it happens, and $n geq N$ will imply $a_n leq 4$, a contradiction.
Now, $a_{n_k}$ is bounded, say by $M$, where $M > 4$. Then by Bolzano Weierstrass, there must be a limit point of this subsequence in $[4,M]$. But then, since this is again a subsequence of $a_n$, $a_n$ has a limit point in $[4,M]$, a contradiction since $3$ was said to be the maximal limit point of $a_n$. This contradiction means that $a_{n_k}$ couldn't have existed in the first place.
Note : we did not use the fact that $1,2$ are also limit points of $a_n$.
EDIT : I have a direct proof, but I want you to fill in the details.
1) Let $b_k = sup_{n geq k} a_n$. Then, $b_n$ is a bounded, increasing sequence.
2) $b_n$ must converge, but in addition, converges to $3$, the largest limit point of $a_n$.
3) By definition, taking $epsilon = frac 12$, there exists $K$ such that $|b_k - 3| leq frac 12$ for $k geq K$, since $b_k$ converges to $3$. In particular, by triangle inequality, $|b_k| leq 3 + frac 12 leq 3.5$.
Because $b_K = sup_{k geq K} a_n$, by definition of supremum, for all $m geq K$, $a_m leq b_K$. However, since $b_K leq 3.5$ , we have that for $m geq K$, $a_m leq 3.5$.
With a similar technique, the following extension can be obtained :
Given a sequence $x_n$ with limit points $a_1,...,a_n$, and any $delta_1,...,delta_n > 0$, there exists $n_0$ such that if $n > n_0$ then $x_n in cup_{i=1}^n(a_i-delta_i,a_i+delta_i)$. In words, eventually every $x_i$ is close enough to at least one of the $a_i$.
answered Apr 9 '17 at 11:08
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.5k33376
$endgroup$
Suppose there does not exist such an $n_0$. Then, we claim there is a subsequence $a_{n_k}$ such that $a_{n_k} > 4$ for all $k$. This is because if there are only finitely many $a_n$ having this property, let $N$ be the maximal $n$ for which it happens, and $n geq N$ will imply $a_n leq 4$, a contradiction.
Now, $a_{n_k}$ is bounded, say by $M$, where $M > 4$. Then by Bolzano Weierstrass, there must be a limit point of this subsequence in $[4,M]$. But then, since this is again a subsequence of $a_n$, $a_n$ has a limit point in $[4,M]$, a contradiction since $3$ was said to be the maximal limit point of $a_n$. This contradiction means that $a_{n_k}$ couldn't have existed in the first place.
Note : we did not use the fact that $1,2$ are also limit points of $a_n$.
EDIT : I have a direct proof, but I want you to fill in the details.
1) Let $b_k = sup_{n geq k} a_n$. Then, $b_n$ is a bounded, increasing sequence.
2) $b_n$ must converge, but in addition, converges to $3$, the largest limit point of $a_n$.
3) By definition, taking $epsilon = frac 12$, there exists $K$ such that $|b_k - 3| leq frac 12$ for $k geq K$, since $b_k$ converges to $3$. In particular, by triangle inequality, $|b_k| leq 3 + frac 12 leq 3.5$.
Because $b_K = sup_{k geq K} a_n$, by definition of supremum, for all $m geq K$, $a_m leq b_K$. However, since $b_K leq 3.5$ , we have that for $m geq K$, $a_m leq 3.5$.
With a similar technique, the following extension can be obtained :
Given a sequence $x_n$ with limit points $a_1,...,a_n$, and any $delta_1,...,delta_n > 0$, there exists $n_0$ such that if $n > n_0$ then $x_n in cup_{i=1}^n(a_i-delta_i,a_i+delta_i)$. In words, eventually every $x_i$ is close enough to at least one of the $a_i$.
answered Apr 9 '17 at 11:08
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.5k33376
edited Jan 3 at 17:14
answered Apr 9 '17 at 11:08
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.5k33376
answered Apr 9 '17 at 11:08
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.5k33376
answered Apr 9 '17 at 11:08
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
37.5k33376
37.5k33376
$begingroup$
A system flag was raised for the sheer number of comments. It seemed to me that you had concluded, so I purged the whole lot. If you need to review something, @-ping me or flag a moderator for assistance.
$endgroup$
– Jyrki Lahtonen
Apr 9 '17 at 16:10
$begingroup$
@JyrkiLahtonen We have concluded discussion of this question. Thank you for mopping up, there were quite a few comments.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 17:36
add a comment |
$begingroup$
A system flag was raised for the sheer number of comments. It seemed to me that you had concluded, so I purged the whole lot. If you need to review something, @-ping me or flag a moderator for assistance.
$endgroup$
– Jyrki Lahtonen
Apr 9 '17 at 16:10
$begingroup$
@JyrkiLahtonen We have concluded discussion of this question. Thank you for mopping up, there were quite a few comments.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 17:36
$begingroup$
A system flag was raised for the sheer number of comments. It seemed to me that you had concluded, so I purged the whole lot. If you need to review something, @-ping me or flag a moderator for assistance.
$endgroup$
– Jyrki Lahtonen
Apr 9 '17 at 16:10
$begingroup$
A system flag was raised for the sheer number of comments. It seemed to me that you had concluded, so I purged the whole lot. If you need to review something, @-ping me or flag a moderator for assistance.
$endgroup$
– Jyrki Lahtonen
Apr 9 '17 at 16:10
$begingroup$
@JyrkiLahtonen We have concluded discussion of this question. Thank you for mopping up, there were quite a few comments.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 17:36
$begingroup$
@JyrkiLahtonen We have concluded discussion of this question. Thank you for mopping up, there were quite a few comments.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 17:36
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%2f2226064%2fbolzano-weierstrass-bounded%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
$begingroup$
If such $n_0$ does not exist then a subsequence $(a_{n_k})_k$ exists with $a_{n_k}geq4$ for each $k$. This sequence is bounded hence (B-W) will have convergent subsequence with a limit $geq4$.
$endgroup$
– drhab
Apr 9 '17 at 11:08