Limit of an increasing and bounded sequence [closed]Show that every monotonic increasing and bounded sequence...
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Limit of an increasing and bounded sequence [closed]
Show that every monotonic increasing and bounded sequence is Cauchy.what is the difference between bounded and convergent?Proof about limit of a monotone increasing sequence bounded aboveIf a sequence is not bounded above, there exists an increasing subsquence, $b_n$ such that $1/b_n$ convergest to zeroIncreasing sequence and subsequenceQuestion on proof concerning : A bounded, monotone sequence convergesEquality of limit and supremum of a sequenceUsing monotone convergence to prove that the limit of a measure of increasing sets is the measure of the unionProve the limit superior of a bounded sequence convergesProof Verification: A monotonically increasing sequence that is bounded above always has a LUB
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An increasing sequence in $mathbb{R}$ which is bounded above always converges to its supremum. Does this supremum always equal this upper bound?
real-analysis
asked Mar 12 at 14:29
TwistTwist
586
$endgroup$
closed as off-topic by RRL, Alex Provost, Eric Wofsey, mrtaurho, Cesareo Mar 13 at 8:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Alex Provost, mrtaurho, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
An increasing sequence in $mathbb{R}$ which is bounded above always converges to its supremum. Does this supremum always equal this upper bound?
real-analysis
asked Mar 12 at 14:29
TwistTwist
586
$endgroup$
closed as off-topic by RRL, Alex Provost, Eric Wofsey, mrtaurho, Cesareo Mar 13 at 8:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Alex Provost, mrtaurho, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
What do you mean by "this upper bound"? The definition of supremum is the least upper bound of the set.
$endgroup$
– Mark
Mar 12 at 14:30
$begingroup$
@Mark some upper bound that one can find.
$endgroup$
– Twist
Mar 12 at 14:34
$begingroup$
No, note some, specifically the smallest (which is the supremum, by definition). The sequence $0.9,,,0.99,,,0.999,,,ldots$ has many upper bounds such as $3$, $pi$ and $10$, but it converges to its least upper bound, which is $1$.
$endgroup$
– StackTD
Mar 12 at 14:41
add a comment |
$begingroup$
An increasing sequence in $mathbb{R}$ which is bounded above always converges to its supremum. Does this supremum always equal this upper bound?
real-analysis
asked Mar 12 at 14:29
TwistTwist
586
$endgroup$
An increasing sequence in $mathbb{R}$ which is bounded above always converges to its supremum. Does this supremum always equal this upper bound?
real-analysis
real-analysis
asked Mar 12 at 14:29
TwistTwist
586
asked Mar 12 at 14:29
TwistTwist
586
asked Mar 12 at 14:29
TwistTwist
586
asked Mar 12 at 14:29
TwistTwist
586
asked Mar 12 at 14:29
TwistTwist
586
586
closed as off-topic by RRL, Alex Provost, Eric Wofsey, mrtaurho, Cesareo Mar 13 at 8:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Alex Provost, mrtaurho, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Alex Provost, Eric Wofsey, mrtaurho, Cesareo Mar 13 at 8:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Alex Provost, mrtaurho, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
What do you mean by "this upper bound"? The definition of supremum is the least upper bound of the set.
$endgroup$
– Mark
Mar 12 at 14:30
$begingroup$
@Mark some upper bound that one can find.
$endgroup$
– Twist
Mar 12 at 14:34
$begingroup$
No, note some, specifically the smallest (which is the supremum, by definition). The sequence $0.9,,,0.99,,,0.999,,,ldots$ has many upper bounds such as $3$, $pi$ and $10$, but it converges to its least upper bound, which is $1$.
$endgroup$
– StackTD
Mar 12 at 14:41
add a comment |
3
$begingroup$
What do you mean by "this upper bound"? The definition of supremum is the least upper bound of the set.
$endgroup$
– Mark
Mar 12 at 14:30
$begingroup$
@Mark some upper bound that one can find.
$endgroup$
– Twist
Mar 12 at 14:34
$begingroup$
No, note some, specifically the smallest (which is the supremum, by definition). The sequence $0.9,,,0.99,,,0.999,,,ldots$ has many upper bounds such as $3$, $pi$ and $10$, but it converges to its least upper bound, which is $1$.
$endgroup$
– StackTD
Mar 12 at 14:41
3
3
$begingroup$
What do you mean by "this upper bound"? The definition of supremum is the least upper bound of the set.
$endgroup$
– Mark
Mar 12 at 14:30
$begingroup$
What do you mean by "this upper bound"? The definition of supremum is the least upper bound of the set.
$endgroup$
– Mark
Mar 12 at 14:30
$begingroup$
@Mark some upper bound that one can find.
$endgroup$
– Twist
Mar 12 at 14:34
$begingroup$
@Mark some upper bound that one can find.
$endgroup$
– Twist
Mar 12 at 14:34
$begingroup$
No, note some, specifically the smallest (which is the supremum, by definition). The sequence $0.9,,,0.99,,,0.999,,,ldots$ has many upper bounds such as $3$, $pi$ and $10$, but it converges to its least upper bound, which is $1$.
$endgroup$
– StackTD
Mar 12 at 14:41
$begingroup$
No, note some, specifically the smallest (which is the supremum, by definition). The sequence $0.9,,,0.99,,,0.999,,,ldots$ has many upper bounds such as $3$, $pi$ and $10$, but it converges to its least upper bound, which is $1$.
$endgroup$
– StackTD
Mar 12 at 14:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The supremum is the lowest possible upper bound by definition, but nothing says that an upper bound has to be strict. For instance, the sequence $a_n$ given by
$$
a_n = 1-frac1n
$$
is increasing and bounded above, and it clearly converges to its supremum $1$. However, one could, entirely correctly, say that $a_n$ is bounded above by $10$. And clearly $a_n$ doesn't converge to $10$.
This is kind of a construed example, because it's really easy to see what the supremum is. But take something like
$$
b_1 = 1, b_n = sqrt{b_{n-1} + 5}
$$
It's quite easy to show by induction that it's bounded, because assuming $b_n<10$, we immediately see that $b_{n+1}<10$ as well. However, finding the actual supremum and limit takes a bit of observational skill and calculation.
edited Mar 12 at 14:58
J. W. Tanner
3,4601320
answered Mar 12 at 14:31
ArthurArthur
118k7118202
$endgroup$
add a comment |
$begingroup$
No,
$$a_n<1implies a_n<2.$$
answered Mar 12 at 14:45
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The supremum is the lowest possible upper bound by definition, but nothing says that an upper bound has to be strict. For instance, the sequence $a_n$ given by
$$
a_n = 1-frac1n
$$
is increasing and bounded above, and it clearly converges to its supremum $1$. However, one could, entirely correctly, say that $a_n$ is bounded above by $10$. And clearly $a_n$ doesn't converge to $10$.
This is kind of a construed example, because it's really easy to see what the supremum is. But take something like
$$
b_1 = 1, b_n = sqrt{b_{n-1} + 5}
$$
It's quite easy to show by induction that it's bounded, because assuming $b_n<10$, we immediately see that $b_{n+1}<10$ as well. However, finding the actual supremum and limit takes a bit of observational skill and calculation.
edited Mar 12 at 14:58
J. W. Tanner
3,4601320
answered Mar 12 at 14:31
ArthurArthur
118k7118202
$endgroup$
add a comment |
$begingroup$
The supremum is the lowest possible upper bound by definition, but nothing says that an upper bound has to be strict. For instance, the sequence $a_n$ given by
$$
a_n = 1-frac1n
$$
is increasing and bounded above, and it clearly converges to its supremum $1$. However, one could, entirely correctly, say that $a_n$ is bounded above by $10$. And clearly $a_n$ doesn't converge to $10$.
This is kind of a construed example, because it's really easy to see what the supremum is. But take something like
$$
b_1 = 1, b_n = sqrt{b_{n-1} + 5}
$$
It's quite easy to show by induction that it's bounded, because assuming $b_n<10$, we immediately see that $b_{n+1}<10$ as well. However, finding the actual supremum and limit takes a bit of observational skill and calculation.
edited Mar 12 at 14:58
J. W. Tanner
3,4601320
answered Mar 12 at 14:31
ArthurArthur
118k7118202
$endgroup$
add a comment |
$begingroup$
The supremum is the lowest possible upper bound by definition, but nothing says that an upper bound has to be strict. For instance, the sequence $a_n$ given by
$$
a_n = 1-frac1n
$$
is increasing and bounded above, and it clearly converges to its supremum $1$. However, one could, entirely correctly, say that $a_n$ is bounded above by $10$. And clearly $a_n$ doesn't converge to $10$.
This is kind of a construed example, because it's really easy to see what the supremum is. But take something like
$$
b_1 = 1, b_n = sqrt{b_{n-1} + 5}
$$
It's quite easy to show by induction that it's bounded, because assuming $b_n<10$, we immediately see that $b_{n+1}<10$ as well. However, finding the actual supremum and limit takes a bit of observational skill and calculation.
edited Mar 12 at 14:58
J. W. Tanner
3,4601320
answered Mar 12 at 14:31
ArthurArthur
118k7118202
$endgroup$
The supremum is the lowest possible upper bound by definition, but nothing says that an upper bound has to be strict. For instance, the sequence $a_n$ given by
$$
a_n = 1-frac1n
$$
is increasing and bounded above, and it clearly converges to its supremum $1$. However, one could, entirely correctly, say that $a_n$ is bounded above by $10$. And clearly $a_n$ doesn't converge to $10$.
This is kind of a construed example, because it's really easy to see what the supremum is. But take something like
$$
b_1 = 1, b_n = sqrt{b_{n-1} + 5}
$$
It's quite easy to show by induction that it's bounded, because assuming $b_n<10$, we immediately see that $b_{n+1}<10$ as well. However, finding the actual supremum and limit takes a bit of observational skill and calculation.
edited Mar 12 at 14:58
J. W. Tanner
3,4601320
answered Mar 12 at 14:31
ArthurArthur
118k7118202
edited Mar 12 at 14:58
J. W. Tanner
3,4601320
edited Mar 12 at 14:58
J. W. Tanner
3,4601320
edited Mar 12 at 14:58
J. W. Tanner
3,4601320
3,4601320
answered Mar 12 at 14:31
ArthurArthur
118k7118202
answered Mar 12 at 14:31
ArthurArthur
118k7118202
answered Mar 12 at 14:31
ArthurArthur
118k7118202
118k7118202
add a comment |
add a comment |
$begingroup$
No,
$$a_n<1implies a_n<2.$$
answered Mar 12 at 14:45
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
No,
$$a_n<1implies a_n<2.$$
answered Mar 12 at 14:45
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
No,
$$a_n<1implies a_n<2.$$
answered Mar 12 at 14:45
Yves DaoustYves Daoust
131k676229
$endgroup$
No,
$$a_n<1implies a_n<2.$$
answered Mar 12 at 14:45
Yves DaoustYves Daoust
131k676229
edited Mar 12 at 14:56
answered Mar 12 at 14:45
Yves DaoustYves Daoust
131k676229
answered Mar 12 at 14:45
Yves DaoustYves Daoust
131k676229
answered Mar 12 at 14:45
Yves DaoustYves Daoust
131k676229
131k676229
add a comment |
add a comment |
3
$begingroup$
What do you mean by "this upper bound"? The definition of supremum is the least upper bound of the set.
$endgroup$
– Mark
Mar 12 at 14:30
$begingroup$
@Mark some upper bound that one can find.
$endgroup$
– Twist
Mar 12 at 14:34
$begingroup$
No, note some, specifically the smallest (which is the supremum, by definition). The sequence $0.9,,,0.99,,,0.999,,,ldots$ has many upper bounds such as $3$, $pi$ and $10$, but it converges to its least upper bound, which is $1$.
$endgroup$
– StackTD
Mar 12 at 14:41