An example of a sequence which does not have any subsequence with a finite limit.
Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.
I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$
But is that true? Also what are other examples of such sequences?
calculus limits examples-counterexamples
asked Dec 10 at 15:45
roman
1,93421221
add a comment |
Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.
I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$
But is that true? Also what are other examples of such sequences?
calculus limits examples-counterexamples
asked Dec 10 at 15:45
roman
1,93421221
10
Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
– астон вілла олоф мэллбэрг
Dec 10 at 15:46
1
@астонвіллаолофмэллбэрг oh my, shame on me
– roman
Dec 10 at 15:48
4
Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
– астон вілла олоф мэллбэрг
Dec 10 at 15:50
@астонвіллаолофмэллбэрг thank you! it's crystal clear now
– roman
Dec 10 at 15:52
1
Great. You can now answer this question yourself and close it.
– астон вілла олоф мэллбэрг
Dec 10 at 15:52
add a comment |
Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.
I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$
But is that true? Also what are other examples of such sequences?
calculus limits examples-counterexamples
asked Dec 10 at 15:45
roman
1,93421221
Find and example of a sequence ${x_n}$ which does not contain any subsequences having a finite limit.
I've been thinking of the following seqeunce:
$$
x_n = sin(n)cdotsin(sqrt{3}n)
$$
But is that true? Also what are other examples of such sequences?
calculus limits examples-counterexamples
calculus limits examples-counterexamples
asked Dec 10 at 15:45
roman
1,93421221
asked Dec 10 at 15:45
roman
1,93421221
asked Dec 10 at 15:45
roman
1,93421221
asked Dec 10 at 15:45
roman
1,93421221
asked Dec 10 at 15:45
roman
1,93421221
1,93421221
10
Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
– астон вілла олоф мэллбэрг
Dec 10 at 15:46
1
@астонвіллаолофмэллбэрг oh my, shame on me
– roman
Dec 10 at 15:48
4
Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
– астон вілла олоф мэллбэрг
Dec 10 at 15:50
@астонвіллаолофмэллбэрг thank you! it's crystal clear now
– roman
Dec 10 at 15:52
1
Great. You can now answer this question yourself and close it.
– астон вілла олоф мэллбэрг
Dec 10 at 15:52
add a comment |
10
Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
– астон вілла олоф мэллбэрг
Dec 10 at 15:46
1
@астонвіллаолофмэллбэрг oh my, shame on me
– roman
Dec 10 at 15:48
4
Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
– астон вілла олоф мэллбэрг
Dec 10 at 15:50
@астонвіллаолофмэллбэрг thank you! it's crystal clear now
– roman
Dec 10 at 15:52
1
Great. You can now answer this question yourself and close it.
– астон вілла олоф мэллбэрг
Dec 10 at 15:52
10
10
Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
– астон вілла олоф мэллбэрг
Dec 10 at 15:46
Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
– астон вілла олоф мэллбэрг
Dec 10 at 15:46
1
1
@астонвіллаолофмэллбэрг oh my, shame on me
– roman
Dec 10 at 15:48
@астонвіллаолофмэллбэрг oh my, shame on me
– roman
Dec 10 at 15:48
4
4
Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
– астон вілла олоф мэллбэрг
Dec 10 at 15:50
Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
– астон вілла олоф мэллбэрг
Dec 10 at 15:50
@астонвіллаолофмэллбэрг thank you! it's crystal clear now
– roman
Dec 10 at 15:52
@астонвіллаолофмэллбэрг thank you! it's crystal clear now
– roman
Dec 10 at 15:52
1
1
Great. You can now answer this question yourself and close it.
– астон вілла олоф мэллбэрг
Dec 10 at 15:52
Great. You can now answer this question yourself and close it.
– астон вілла олоф мэллбэрг
Dec 10 at 15:52
add a comment |
2 Answers
2
active
oldest
votes
Any sequence ${x_n}$ with $|x_n| to infty$ will do.
By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).
EDIT: By the way, if we think of the one point compactification of $mathbb{R}$ where we add $infty$ in, then every sequence has a converging subsequence in this space (which is homeomorphic to circle). So this also shows that the only way to have no subsequence with finite limit is to make that converging subsequence that is quaranteed to exists in the compactified space to have limit $infty$. Well I guess you still have to make the above argument to show that we must have $|x_n|to infty$ (i.e $x to infty$ in the compactified space).
answered Dec 10 at 15:58
ploosu2
4,5771023
add a comment |
The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.
Now if we consider:
$$
x_n = sin(n)sin(sqrt{3}n)
$$
its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.
answered Dec 10 at 15:58
roman
1,93421221
Upvoted. Good job.
– астон вілла олоф мэллбэрг
Dec 10 at 16:00
1
Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
– Henry
Dec 10 at 23:00
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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oldest
votes
Any sequence ${x_n}$ with $|x_n| to infty$ will do.
By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).
EDIT: By the way, if we think of the one point compactification of $mathbb{R}$ where we add $infty$ in, then every sequence has a converging subsequence in this space (which is homeomorphic to circle). So this also shows that the only way to have no subsequence with finite limit is to make that converging subsequence that is quaranteed to exists in the compactified space to have limit $infty$. Well I guess you still have to make the above argument to show that we must have $|x_n|to infty$ (i.e $x to infty$ in the compactified space).
answered Dec 10 at 15:58
ploosu2
4,5771023
add a comment |
Any sequence ${x_n}$ with $|x_n| to infty$ will do.
By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).
EDIT: By the way, if we think of the one point compactification of $mathbb{R}$ where we add $infty$ in, then every sequence has a converging subsequence in this space (which is homeomorphic to circle). So this also shows that the only way to have no subsequence with finite limit is to make that converging subsequence that is quaranteed to exists in the compactified space to have limit $infty$. Well I guess you still have to make the above argument to show that we must have $|x_n|to infty$ (i.e $x to infty$ in the compactified space).
answered Dec 10 at 15:58
ploosu2
4,5771023
add a comment |
Any sequence ${x_n}$ with $|x_n| to infty$ will do.
By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).
EDIT: By the way, if we think of the one point compactification of $mathbb{R}$ where we add $infty$ in, then every sequence has a converging subsequence in this space (which is homeomorphic to circle). So this also shows that the only way to have no subsequence with finite limit is to make that converging subsequence that is quaranteed to exists in the compactified space to have limit $infty$. Well I guess you still have to make the above argument to show that we must have $|x_n|to infty$ (i.e $x to infty$ in the compactified space).
answered Dec 10 at 15:58
ploosu2
4,5771023
Any sequence ${x_n}$ with $|x_n| to infty$ will do.
By Bolzano-Weierstrass Theorem others wont: if $|x_n| not to infty$, there is $M$, such that you can find a subsequence of ${x_n}$ that is bounded by $M$. Then you can find a converging subsequence of that subsequence (and hence of the original sequence).
EDIT: By the way, if we think of the one point compactification of $mathbb{R}$ where we add $infty$ in, then every sequence has a converging subsequence in this space (which is homeomorphic to circle). So this also shows that the only way to have no subsequence with finite limit is to make that converging subsequence that is quaranteed to exists in the compactified space to have limit $infty$. Well I guess you still have to make the above argument to show that we must have $|x_n|to infty$ (i.e $x to infty$ in the compactified space).
answered Dec 10 at 15:58
ploosu2
4,5771023
edited Dec 11 at 9:01
answered Dec 10 at 15:58
ploosu2
4,5771023
answered Dec 10 at 15:58
ploosu2
4,5771023
answered Dec 10 at 15:58
ploosu2
4,5771023
4,5771023
add a comment |
add a comment |
The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.
Now if we consider:
$$
x_n = sin(n)sin(sqrt{3}n)
$$
its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.
answered Dec 10 at 15:58
roman
1,93421221
Upvoted. Good job.
– астон вілла олоф мэллбэрг
Dec 10 at 16:00
1
Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
– Henry
Dec 10 at 23:00
add a comment |
The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.
Now if we consider:
$$
x_n = sin(n)sin(sqrt{3}n)
$$
its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.
answered Dec 10 at 15:58
roman
1,93421221
Upvoted. Good job.
– астон вілла олоф мэллбэрг
Dec 10 at 16:00
1
Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
– Henry
Dec 10 at 23:00
add a comment |
The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.
Now if we consider:
$$
x_n = sin(n)sin(sqrt{3}n)
$$
its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.
answered Dec 10 at 15:58
roman
1,93421221
The example in the OP is wrong. This can be shown by Bolzano-Weierstrass theorem which states that every bounded sequence contains a convergent subsequence.
Now if we consider:
$$
x_n = sin(n)sin(sqrt{3}n)
$$
its clear that the sequence is bounded by $[-1, 1]$. Hence it contains a convergent subsequence.
answered Dec 10 at 15:58
roman
1,93421221
edited Dec 10 at 16:00
answered Dec 10 at 15:58
roman
1,93421221
answered Dec 10 at 15:58
roman
1,93421221
answered Dec 10 at 15:58
roman
1,93421221
1,93421221
Upvoted. Good job.
– астон вілла олоф мэллбэрг
Dec 10 at 16:00
1
Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
– Henry
Dec 10 at 23:00
add a comment |
Upvoted. Good job.
– астон вілла олоф мэллбэрг
Dec 10 at 16:00
1
Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
– Henry
Dec 10 at 23:00
Upvoted. Good job.
– астон вілла олоф мэллбэрг
Dec 10 at 16:00
Upvoted. Good job.
– астон вілла олоф мэллбэрг
Dec 10 at 16:00
1
1
Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
– Henry
Dec 10 at 23:00
Take for example $n=1, 8,77, 146, 721, 5270, 19492 , 33714 ,148357, 263000, 674357$ and you get $ 0.8305543, 0.9506199, 0.9883301, 0.9962933, 0.9996635, 0.9997120, 0.9998930, 0.9999708, 0.9999954, 0.9999987, 0.9999999$
– Henry
Dec 10 at 23:00
add a comment |
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10
Natural numbers $1,2,3,...$? No subsequence is Cauchy, hence convergent?
– астон вілла олоф мэллбэрг
Dec 10 at 15:46
1
@астонвіллаолофмэллбэрг oh my, shame on me
– roman
Dec 10 at 15:48
4
Remember, every bounded sequence of real numbers has a convergent subsequence. Therefore, your sequence being bounded does not have a convergent subsequence. The best bet is to take an unbounded sequence where the distance between any two terms is greater than some fixed number.
– астон вілла олоф мэллбэрг
Dec 10 at 15:50
@астонвіллаолофмэллбэрг thank you! it's crystal clear now
– roman
Dec 10 at 15:52
1
Great. You can now answer this question yourself and close it.
– астон вілла олоф мэллбэрг
Dec 10 at 15:52