Help with completing the proof: Any unbounded sequence is either infinitely large or has a convergent...
$begingroup$
Given ${x_n}$ is an unbounded sequence, prove:
$$
lim_{ntoinfty}x_n = infty text{or} lim_{n_k to infty} = A
$$
In other words any unbounded sequence either infinitely large or has a convergent subsequence.
Everything before udpate section is wrong.
Part 1. Proving $|x_n| > M implies lim_{ntoinfty} x_n = infty$
$Box$ Start with the definition of an unbounded sequence:
$$
forall M > 0 in Bbb R exists N in Bbb N: forall n > Nimplies |x_n| > M
$$
Suppose that:
$$
lim_{ntoinfty}x_n ne infty
$$
That means:
$$
forall ninBbb N exists C >0 inBbb R : |x_n| < C
$$
Now taking $M^prime = max{M, C}$ we have that:
$$
begin{cases}
begin{align}
exists N in Bbb N : &forall n>N implies &|x_n| > M^prime \
&forall n in Bbb N: &|x_n| < M^prime
end{align}
end{cases}
$$
So we have arrived to a contradiction which means the assumption is wrong and:
$$
lim_{ntoinfty}x_n = infty
$$
$Box$
Part 2. It's not clear to me how to prove that if unboundedness does not imply infinity limit then it must imply the existence of a convergent subsequence.
Two questions is my mind:
- Is part 1 correct?
- How do I proceed with the second part?
I'm using the following definition.
$$
lim_{ntoinfty}x_n = +infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies x_n > epsilon \
lim_{ntoinfty}x_n = -infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies x_n < -epsilon \
$$
A sequence is considered infinitely large when:
$$
lim_{ntoinfty}x_n = infty text{when} lim_{ntoinfty}x_n = + infty text{or} lim_{ntoinfty}x_n = - infty
$$
Update.
Looks like i've messed up a lot of things. I will try once again.
As shown in comments and answers the statement holds only in case:
$$
lim_{ntoinfty} |x_n| = infty iff forall epsilon > 0 exists N in Bbb N: forall n > N implies |x_n| > epsilon
$$
Consider the following:
$$
lnot P = lnotleft(lim_{ntoinfty}|x_n| = infty right) iff exists epsilon > 0 forall N_1 in Bbb N : exists n > N_1 land |x_n| < epsilon
$$
But on the other hand it is given that $x_n$ is unbounded:
$$
Q = forall M > 0 exists N_2 in Bbb N : |x_{N_2}| > M
$$
Construct a negative expression for boundedness:
$$
lnot P = exists M > 0 forall N_2 in Bbb N : |x_{N_2}| < M
$$
If $S = lnot P implies lnot Q$ then $S = P lor lnot Q$
Let $epsilon = M$, choose $N = max{N_1, N_2}$ then both statements are true and:
$$
exists epsilon > 0 forall N = max{N_1, N_2}: exists n > N land |x_n| < epsilon
$$
Now either $P$ is true, which would mean $lim |x_n| = infty$, or $lnot Q$ is true which would mean the sequence is bounded and hence contains a convergent subsequence.
calculus limits proof-verification
asked Dec 12 '18 at 12:42
romanroman
2,22921224
$endgroup$
|
show 3 more comments
$begingroup$
Given ${x_n}$ is an unbounded sequence, prove:
$$
lim_{ntoinfty}x_n = infty text{or} lim_{n_k to infty} = A
$$
In other words any unbounded sequence either infinitely large or has a convergent subsequence.
Everything before udpate section is wrong.
Part 1. Proving $|x_n| > M implies lim_{ntoinfty} x_n = infty$
$Box$ Start with the definition of an unbounded sequence:
$$
forall M > 0 in Bbb R exists N in Bbb N: forall n > Nimplies |x_n| > M
$$
Suppose that:
$$
lim_{ntoinfty}x_n ne infty
$$
That means:
$$
forall ninBbb N exists C >0 inBbb R : |x_n| < C
$$
Now taking $M^prime = max{M, C}$ we have that:
$$
begin{cases}
begin{align}
exists N in Bbb N : &forall n>N implies &|x_n| > M^prime \
&forall n in Bbb N: &|x_n| < M^prime
end{align}
end{cases}
$$
So we have arrived to a contradiction which means the assumption is wrong and:
$$
lim_{ntoinfty}x_n = infty
$$
$Box$
Part 2. It's not clear to me how to prove that if unboundedness does not imply infinity limit then it must imply the existence of a convergent subsequence.
Two questions is my mind:
- Is part 1 correct?
- How do I proceed with the second part?
I'm using the following definition.
$$
lim_{ntoinfty}x_n = +infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies x_n > epsilon \
lim_{ntoinfty}x_n = -infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies x_n < -epsilon \
$$
A sequence is considered infinitely large when:
$$
lim_{ntoinfty}x_n = infty text{when} lim_{ntoinfty}x_n = + infty text{or} lim_{ntoinfty}x_n = - infty
$$
Update.
Looks like i've messed up a lot of things. I will try once again.
As shown in comments and answers the statement holds only in case:
$$
lim_{ntoinfty} |x_n| = infty iff forall epsilon > 0 exists N in Bbb N: forall n > N implies |x_n| > epsilon
$$
Consider the following:
$$
lnot P = lnotleft(lim_{ntoinfty}|x_n| = infty right) iff exists epsilon > 0 forall N_1 in Bbb N : exists n > N_1 land |x_n| < epsilon
$$
But on the other hand it is given that $x_n$ is unbounded:
$$
Q = forall M > 0 exists N_2 in Bbb N : |x_{N_2}| > M
$$
Construct a negative expression for boundedness:
$$
lnot P = exists M > 0 forall N_2 in Bbb N : |x_{N_2}| < M
$$
If $S = lnot P implies lnot Q$ then $S = P lor lnot Q$
Let $epsilon = M$, choose $N = max{N_1, N_2}$ then both statements are true and:
$$
exists epsilon > 0 forall N = max{N_1, N_2}: exists n > N land |x_n| < epsilon
$$
Now either $P$ is true, which would mean $lim |x_n| = infty$, or $lnot Q$ is true which would mean the sequence is bounded and hence contains a convergent subsequence.
calculus limits proof-verification
asked Dec 12 '18 at 12:42
romanroman
2,22921224
$endgroup$
2
$begingroup$
The statement is incorrect. $x_n = (-1)^ncdot n$ is an unbounded sequence, yet neither of the two properties holds.
$endgroup$
– 5xum
Dec 12 '18 at 12:45
$begingroup$
@5xum since english is not my native language the post may contain translation issues, i've added a description of what i use as a definition of "infinitely large sequence"
$endgroup$
– roman
Dec 12 '18 at 12:54
$begingroup$
No, it's not translation issues. No matter which country you are in, the definition $$ lim_{ntoinfty}x_n = infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies |x_n| > epsilon $$ is wrong. There is no country or language on Earth where the sequence $-1,2,-3,4,-5,6,dots$ converges to $infty$. The absolute values should not be in the definition.
$endgroup$
– 5xum
Dec 12 '18 at 12:56
$begingroup$
@5xum you are right, i've messed things up
$endgroup$
– roman
Dec 12 '18 at 13:05
1
$begingroup$
HINT. Consider that $lim_{nto infty}|x_n|=infty$ means that for every $ r>0,$ the set ${n: |x_n|leq r}$ is finite . So what happens if $neg (lim_{nto infty}|x_n|=infty)?$
$endgroup$
– DanielWainfleet
Dec 12 '18 at 14:04
|
show 3 more comments
$begingroup$
Given ${x_n}$ is an unbounded sequence, prove:
$$
lim_{ntoinfty}x_n = infty text{or} lim_{n_k to infty} = A
$$
In other words any unbounded sequence either infinitely large or has a convergent subsequence.
Everything before udpate section is wrong.
Part 1. Proving $|x_n| > M implies lim_{ntoinfty} x_n = infty$
$Box$ Start with the definition of an unbounded sequence:
$$
forall M > 0 in Bbb R exists N in Bbb N: forall n > Nimplies |x_n| > M
$$
Suppose that:
$$
lim_{ntoinfty}x_n ne infty
$$
That means:
$$
forall ninBbb N exists C >0 inBbb R : |x_n| < C
$$
Now taking $M^prime = max{M, C}$ we have that:
$$
begin{cases}
begin{align}
exists N in Bbb N : &forall n>N implies &|x_n| > M^prime \
&forall n in Bbb N: &|x_n| < M^prime
end{align}
end{cases}
$$
So we have arrived to a contradiction which means the assumption is wrong and:
$$
lim_{ntoinfty}x_n = infty
$$
$Box$
Part 2. It's not clear to me how to prove that if unboundedness does not imply infinity limit then it must imply the existence of a convergent subsequence.
Two questions is my mind:
- Is part 1 correct?
- How do I proceed with the second part?
I'm using the following definition.
$$
lim_{ntoinfty}x_n = +infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies x_n > epsilon \
lim_{ntoinfty}x_n = -infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies x_n < -epsilon \
$$
A sequence is considered infinitely large when:
$$
lim_{ntoinfty}x_n = infty text{when} lim_{ntoinfty}x_n = + infty text{or} lim_{ntoinfty}x_n = - infty
$$
Update.
Looks like i've messed up a lot of things. I will try once again.
As shown in comments and answers the statement holds only in case:
$$
lim_{ntoinfty} |x_n| = infty iff forall epsilon > 0 exists N in Bbb N: forall n > N implies |x_n| > epsilon
$$
Consider the following:
$$
lnot P = lnotleft(lim_{ntoinfty}|x_n| = infty right) iff exists epsilon > 0 forall N_1 in Bbb N : exists n > N_1 land |x_n| < epsilon
$$
But on the other hand it is given that $x_n$ is unbounded:
$$
Q = forall M > 0 exists N_2 in Bbb N : |x_{N_2}| > M
$$
Construct a negative expression for boundedness:
$$
lnot P = exists M > 0 forall N_2 in Bbb N : |x_{N_2}| < M
$$
If $S = lnot P implies lnot Q$ then $S = P lor lnot Q$
Let $epsilon = M$, choose $N = max{N_1, N_2}$ then both statements are true and:
$$
exists epsilon > 0 forall N = max{N_1, N_2}: exists n > N land |x_n| < epsilon
$$
Now either $P$ is true, which would mean $lim |x_n| = infty$, or $lnot Q$ is true which would mean the sequence is bounded and hence contains a convergent subsequence.
calculus limits proof-verification
asked Dec 12 '18 at 12:42
romanroman
2,22921224
$endgroup$
Given ${x_n}$ is an unbounded sequence, prove:
$$
lim_{ntoinfty}x_n = infty text{or} lim_{n_k to infty} = A
$$
In other words any unbounded sequence either infinitely large or has a convergent subsequence.
Everything before udpate section is wrong.
Part 1. Proving $|x_n| > M implies lim_{ntoinfty} x_n = infty$
$Box$ Start with the definition of an unbounded sequence:
$$
forall M > 0 in Bbb R exists N in Bbb N: forall n > Nimplies |x_n| > M
$$
Suppose that:
$$
lim_{ntoinfty}x_n ne infty
$$
That means:
$$
forall ninBbb N exists C >0 inBbb R : |x_n| < C
$$
Now taking $M^prime = max{M, C}$ we have that:
$$
begin{cases}
begin{align}
exists N in Bbb N : &forall n>N implies &|x_n| > M^prime \
&forall n in Bbb N: &|x_n| < M^prime
end{align}
end{cases}
$$
So we have arrived to a contradiction which means the assumption is wrong and:
$$
lim_{ntoinfty}x_n = infty
$$
$Box$
Part 2. It's not clear to me how to prove that if unboundedness does not imply infinity limit then it must imply the existence of a convergent subsequence.
Two questions is my mind:
- Is part 1 correct?
- How do I proceed with the second part?
I'm using the following definition.
$$
lim_{ntoinfty}x_n = +infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies x_n > epsilon \
lim_{ntoinfty}x_n = -infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies x_n < -epsilon \
$$
A sequence is considered infinitely large when:
$$
lim_{ntoinfty}x_n = infty text{when} lim_{ntoinfty}x_n = + infty text{or} lim_{ntoinfty}x_n = - infty
$$
Update.
Looks like i've messed up a lot of things. I will try once again.
As shown in comments and answers the statement holds only in case:
$$
lim_{ntoinfty} |x_n| = infty iff forall epsilon > 0 exists N in Bbb N: forall n > N implies |x_n| > epsilon
$$
Consider the following:
$$
lnot P = lnotleft(lim_{ntoinfty}|x_n| = infty right) iff exists epsilon > 0 forall N_1 in Bbb N : exists n > N_1 land |x_n| < epsilon
$$
But on the other hand it is given that $x_n$ is unbounded:
$$
Q = forall M > 0 exists N_2 in Bbb N : |x_{N_2}| > M
$$
Construct a negative expression for boundedness:
$$
lnot P = exists M > 0 forall N_2 in Bbb N : |x_{N_2}| < M
$$
If $S = lnot P implies lnot Q$ then $S = P lor lnot Q$
Let $epsilon = M$, choose $N = max{N_1, N_2}$ then both statements are true and:
$$
exists epsilon > 0 forall N = max{N_1, N_2}: exists n > N land |x_n| < epsilon
$$
Now either $P$ is true, which would mean $lim |x_n| = infty$, or $lnot Q$ is true which would mean the sequence is bounded and hence contains a convergent subsequence.
calculus limits proof-verification
calculus limits proof-verification
asked Dec 12 '18 at 12:42
romanroman
2,22921224
asked Dec 12 '18 at 12:42
romanroman
2,22921224
edited Dec 12 '18 at 15:21
roman
asked Dec 12 '18 at 12:42
romanroman
2,22921224
asked Dec 12 '18 at 12:42
romanroman
2,22921224
asked Dec 12 '18 at 12:42
romanroman
2,22921224
2,22921224
2
$begingroup$
The statement is incorrect. $x_n = (-1)^ncdot n$ is an unbounded sequence, yet neither of the two properties holds.
$endgroup$
– 5xum
Dec 12 '18 at 12:45
$begingroup$
@5xum since english is not my native language the post may contain translation issues, i've added a description of what i use as a definition of "infinitely large sequence"
$endgroup$
– roman
Dec 12 '18 at 12:54
$begingroup$
No, it's not translation issues. No matter which country you are in, the definition $$ lim_{ntoinfty}x_n = infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies |x_n| > epsilon $$ is wrong. There is no country or language on Earth where the sequence $-1,2,-3,4,-5,6,dots$ converges to $infty$. The absolute values should not be in the definition.
$endgroup$
– 5xum
Dec 12 '18 at 12:56
$begingroup$
@5xum you are right, i've messed things up
$endgroup$
– roman
Dec 12 '18 at 13:05
1
$begingroup$
HINT. Consider that $lim_{nto infty}|x_n|=infty$ means that for every $ r>0,$ the set ${n: |x_n|leq r}$ is finite . So what happens if $neg (lim_{nto infty}|x_n|=infty)?$
$endgroup$
– DanielWainfleet
Dec 12 '18 at 14:04
|
show 3 more comments
2
$begingroup$
The statement is incorrect. $x_n = (-1)^ncdot n$ is an unbounded sequence, yet neither of the two properties holds.
$endgroup$
– 5xum
Dec 12 '18 at 12:45
$begingroup$
@5xum since english is not my native language the post may contain translation issues, i've added a description of what i use as a definition of "infinitely large sequence"
$endgroup$
– roman
Dec 12 '18 at 12:54
$begingroup$
No, it's not translation issues. No matter which country you are in, the definition $$ lim_{ntoinfty}x_n = infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies |x_n| > epsilon $$ is wrong. There is no country or language on Earth where the sequence $-1,2,-3,4,-5,6,dots$ converges to $infty$. The absolute values should not be in the definition.
$endgroup$
– 5xum
Dec 12 '18 at 12:56
$begingroup$
@5xum you are right, i've messed things up
$endgroup$
– roman
Dec 12 '18 at 13:05
1
$begingroup$
HINT. Consider that $lim_{nto infty}|x_n|=infty$ means that for every $ r>0,$ the set ${n: |x_n|leq r}$ is finite . So what happens if $neg (lim_{nto infty}|x_n|=infty)?$
$endgroup$
– DanielWainfleet
Dec 12 '18 at 14:04
2
2
$begingroup$
The statement is incorrect. $x_n = (-1)^ncdot n$ is an unbounded sequence, yet neither of the two properties holds.
$endgroup$
– 5xum
Dec 12 '18 at 12:45
$begingroup$
The statement is incorrect. $x_n = (-1)^ncdot n$ is an unbounded sequence, yet neither of the two properties holds.
$endgroup$
– 5xum
Dec 12 '18 at 12:45
$begingroup$
@5xum since english is not my native language the post may contain translation issues, i've added a description of what i use as a definition of "infinitely large sequence"
$endgroup$
– roman
Dec 12 '18 at 12:54
$begingroup$
@5xum since english is not my native language the post may contain translation issues, i've added a description of what i use as a definition of "infinitely large sequence"
$endgroup$
– roman
Dec 12 '18 at 12:54
$begingroup$
No, it's not translation issues. No matter which country you are in, the definition $$ lim_{ntoinfty}x_n = infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies |x_n| > epsilon $$ is wrong. There is no country or language on Earth where the sequence $-1,2,-3,4,-5,6,dots$ converges to $infty$. The absolute values should not be in the definition.
$endgroup$
– 5xum
Dec 12 '18 at 12:56
$begingroup$
No, it's not translation issues. No matter which country you are in, the definition $$ lim_{ntoinfty}x_n = infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies |x_n| > epsilon $$ is wrong. There is no country or language on Earth where the sequence $-1,2,-3,4,-5,6,dots$ converges to $infty$. The absolute values should not be in the definition.
$endgroup$
– 5xum
Dec 12 '18 at 12:56
$begingroup$
@5xum you are right, i've messed things up
$endgroup$
– roman
Dec 12 '18 at 13:05
$begingroup$
@5xum you are right, i've messed things up
$endgroup$
– roman
Dec 12 '18 at 13:05
1
1
$begingroup$
HINT. Consider that $lim_{nto infty}|x_n|=infty$ means that for every $ r>0,$ the set ${n: |x_n|leq r}$ is finite . So what happens if $neg (lim_{nto infty}|x_n|=infty)?$
$endgroup$
– DanielWainfleet
Dec 12 '18 at 14:04
$begingroup$
HINT. Consider that $lim_{nto infty}|x_n|=infty$ means that for every $ r>0,$ the set ${n: |x_n|leq r}$ is finite . So what happens if $neg (lim_{nto infty}|x_n|=infty)?$
$endgroup$
– DanielWainfleet
Dec 12 '18 at 14:04
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The statement is incorrect. $x_n = (-1)^ncdot n$ is an unbounded sequence, yet neither of the two properties holds.
Also, your proof is wrong here:
Suppose that: $$ lim_{ntoinfty}x_n ne infty $$ That means: $$forall ninBbb N exists C >0 inBbb R : |x_n| < C $$
This is false. EVERY sequence satisfies the condition $$forall ninBbb N exists C >0 inBbb R : |x_n| < C$$ (since you can always set $C=|x_n|+1$) however not every sequence satisfies the condition $$lim_{ntoinfty}x_n ne infty$$
In fact, the condition $$lim_{ntoinfty}x_n = infty$$
is written as:
$$forall C exists Nforall n:n>Nimplies x_n>C$$
which means that the negation of that is written as:
$$exists Cforall Nexists n>N: n>Nland x_n<C$$ which is different from your statement in that there is no absolute value, and the orders of the quantifiers are different.
answered Dec 12 '18 at 12:49
5xum5xum
90.6k394161
$endgroup$
$begingroup$
Thank you for pointing that out, i've reworked the OP
$endgroup$
– roman
Dec 12 '18 at 13:33
$begingroup$
@roman But the statement you are "proving" is still false!
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
I do not what to say, this problem is taken from a book i'm solving. I've tried to google-translate the problem statement: "Prove every unbounded sequence is either infinitely large, or has a finite partial limit". Since you say it is wrong in the first place, i guess the only way is to abandon that problem
$endgroup$
– roman
Dec 12 '18 at 13:45
$begingroup$
@roman The statement is true if you restrict it to positive sequences, or if you define "being infinitely large" as $lim_{ntoinfty} |x_n| =infty$, so you might be misunderstanding some part of the problem.
$endgroup$
– 5xum
Dec 12 '18 at 13:53
add a comment |
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The statement is incorrect. $x_n = (-1)^ncdot n$ is an unbounded sequence, yet neither of the two properties holds.
Also, your proof is wrong here:
Suppose that: $$ lim_{ntoinfty}x_n ne infty $$ That means: $$forall ninBbb N exists C >0 inBbb R : |x_n| < C $$
This is false. EVERY sequence satisfies the condition $$forall ninBbb N exists C >0 inBbb R : |x_n| < C$$ (since you can always set $C=|x_n|+1$) however not every sequence satisfies the condition $$lim_{ntoinfty}x_n ne infty$$
In fact, the condition $$lim_{ntoinfty}x_n = infty$$
is written as:
$$forall C exists Nforall n:n>Nimplies x_n>C$$
which means that the negation of that is written as:
$$exists Cforall Nexists n>N: n>Nland x_n<C$$ which is different from your statement in that there is no absolute value, and the orders of the quantifiers are different.
answered Dec 12 '18 at 12:49
5xum5xum
90.6k394161
$endgroup$
$begingroup$
Thank you for pointing that out, i've reworked the OP
$endgroup$
– roman
Dec 12 '18 at 13:33
$begingroup$
@roman But the statement you are "proving" is still false!
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
I do not what to say, this problem is taken from a book i'm solving. I've tried to google-translate the problem statement: "Prove every unbounded sequence is either infinitely large, or has a finite partial limit". Since you say it is wrong in the first place, i guess the only way is to abandon that problem
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– roman
Dec 12 '18 at 13:45
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@roman The statement is true if you restrict it to positive sequences, or if you define "being infinitely large" as $lim_{ntoinfty} |x_n| =infty$, so you might be misunderstanding some part of the problem.
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– 5xum
Dec 12 '18 at 13:53
add a comment |
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The statement is incorrect. $x_n = (-1)^ncdot n$ is an unbounded sequence, yet neither of the two properties holds.
Also, your proof is wrong here:
Suppose that: $$ lim_{ntoinfty}x_n ne infty $$ That means: $$forall ninBbb N exists C >0 inBbb R : |x_n| < C $$
This is false. EVERY sequence satisfies the condition $$forall ninBbb N exists C >0 inBbb R : |x_n| < C$$ (since you can always set $C=|x_n|+1$) however not every sequence satisfies the condition $$lim_{ntoinfty}x_n ne infty$$
In fact, the condition $$lim_{ntoinfty}x_n = infty$$
is written as:
$$forall C exists Nforall n:n>Nimplies x_n>C$$
which means that the negation of that is written as:
$$exists Cforall Nexists n>N: n>Nland x_n<C$$ which is different from your statement in that there is no absolute value, and the orders of the quantifiers are different.
answered Dec 12 '18 at 12:49
5xum5xum
90.6k394161
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Thank you for pointing that out, i've reworked the OP
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– roman
Dec 12 '18 at 13:33
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@roman But the statement you are "proving" is still false!
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
I do not what to say, this problem is taken from a book i'm solving. I've tried to google-translate the problem statement: "Prove every unbounded sequence is either infinitely large, or has a finite partial limit". Since you say it is wrong in the first place, i guess the only way is to abandon that problem
$endgroup$
– roman
Dec 12 '18 at 13:45
$begingroup$
@roman The statement is true if you restrict it to positive sequences, or if you define "being infinitely large" as $lim_{ntoinfty} |x_n| =infty$, so you might be misunderstanding some part of the problem.
$endgroup$
– 5xum
Dec 12 '18 at 13:53
add a comment |
$begingroup$
The statement is incorrect. $x_n = (-1)^ncdot n$ is an unbounded sequence, yet neither of the two properties holds.
Also, your proof is wrong here:
Suppose that: $$ lim_{ntoinfty}x_n ne infty $$ That means: $$forall ninBbb N exists C >0 inBbb R : |x_n| < C $$
This is false. EVERY sequence satisfies the condition $$forall ninBbb N exists C >0 inBbb R : |x_n| < C$$ (since you can always set $C=|x_n|+1$) however not every sequence satisfies the condition $$lim_{ntoinfty}x_n ne infty$$
In fact, the condition $$lim_{ntoinfty}x_n = infty$$
is written as:
$$forall C exists Nforall n:n>Nimplies x_n>C$$
which means that the negation of that is written as:
$$exists Cforall Nexists n>N: n>Nland x_n<C$$ which is different from your statement in that there is no absolute value, and the orders of the quantifiers are different.
answered Dec 12 '18 at 12:49
5xum5xum
90.6k394161
$endgroup$
The statement is incorrect. $x_n = (-1)^ncdot n$ is an unbounded sequence, yet neither of the two properties holds.
Also, your proof is wrong here:
Suppose that: $$ lim_{ntoinfty}x_n ne infty $$ That means: $$forall ninBbb N exists C >0 inBbb R : |x_n| < C $$
This is false. EVERY sequence satisfies the condition $$forall ninBbb N exists C >0 inBbb R : |x_n| < C$$ (since you can always set $C=|x_n|+1$) however not every sequence satisfies the condition $$lim_{ntoinfty}x_n ne infty$$
In fact, the condition $$lim_{ntoinfty}x_n = infty$$
is written as:
$$forall C exists Nforall n:n>Nimplies x_n>C$$
which means that the negation of that is written as:
$$exists Cforall Nexists n>N: n>Nland x_n<C$$ which is different from your statement in that there is no absolute value, and the orders of the quantifiers are different.
answered Dec 12 '18 at 12:49
5xum5xum
90.6k394161
answered Dec 12 '18 at 12:49
5xum5xum
90.6k394161
answered Dec 12 '18 at 12:49
5xum5xum
90.6k394161
answered Dec 12 '18 at 12:49
5xum5xum
90.6k394161
90.6k394161
$begingroup$
Thank you for pointing that out, i've reworked the OP
$endgroup$
– roman
Dec 12 '18 at 13:33
$begingroup$
@roman But the statement you are "proving" is still false!
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
I do not what to say, this problem is taken from a book i'm solving. I've tried to google-translate the problem statement: "Prove every unbounded sequence is either infinitely large, or has a finite partial limit". Since you say it is wrong in the first place, i guess the only way is to abandon that problem
$endgroup$
– roman
Dec 12 '18 at 13:45
$begingroup$
@roman The statement is true if you restrict it to positive sequences, or if you define "being infinitely large" as $lim_{ntoinfty} |x_n| =infty$, so you might be misunderstanding some part of the problem.
$endgroup$
– 5xum
Dec 12 '18 at 13:53
add a comment |
$begingroup$
Thank you for pointing that out, i've reworked the OP
$endgroup$
– roman
Dec 12 '18 at 13:33
$begingroup$
@roman But the statement you are "proving" is still false!
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
I do not what to say, this problem is taken from a book i'm solving. I've tried to google-translate the problem statement: "Prove every unbounded sequence is either infinitely large, or has a finite partial limit". Since you say it is wrong in the first place, i guess the only way is to abandon that problem
$endgroup$
– roman
Dec 12 '18 at 13:45
$begingroup$
@roman The statement is true if you restrict it to positive sequences, or if you define "being infinitely large" as $lim_{ntoinfty} |x_n| =infty$, so you might be misunderstanding some part of the problem.
$endgroup$
– 5xum
Dec 12 '18 at 13:53
$begingroup$
Thank you for pointing that out, i've reworked the OP
$endgroup$
– roman
Dec 12 '18 at 13:33
$begingroup$
Thank you for pointing that out, i've reworked the OP
$endgroup$
– roman
Dec 12 '18 at 13:33
$begingroup$
@roman But the statement you are "proving" is still false!
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
@roman But the statement you are "proving" is still false!
$endgroup$
– 5xum
Dec 12 '18 at 13:33
$begingroup$
I do not what to say, this problem is taken from a book i'm solving. I've tried to google-translate the problem statement: "Prove every unbounded sequence is either infinitely large, or has a finite partial limit". Since you say it is wrong in the first place, i guess the only way is to abandon that problem
$endgroup$
– roman
Dec 12 '18 at 13:45
$begingroup$
I do not what to say, this problem is taken from a book i'm solving. I've tried to google-translate the problem statement: "Prove every unbounded sequence is either infinitely large, or has a finite partial limit". Since you say it is wrong in the first place, i guess the only way is to abandon that problem
$endgroup$
– roman
Dec 12 '18 at 13:45
$begingroup$
@roman The statement is true if you restrict it to positive sequences, or if you define "being infinitely large" as $lim_{ntoinfty} |x_n| =infty$, so you might be misunderstanding some part of the problem.
$endgroup$
– 5xum
Dec 12 '18 at 13:53
$begingroup$
@roman The statement is true if you restrict it to positive sequences, or if you define "being infinitely large" as $lim_{ntoinfty} |x_n| =infty$, so you might be misunderstanding some part of the problem.
$endgroup$
– 5xum
Dec 12 '18 at 13:53
add a comment |
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$begingroup$
The statement is incorrect. $x_n = (-1)^ncdot n$ is an unbounded sequence, yet neither of the two properties holds.
$endgroup$
– 5xum
Dec 12 '18 at 12:45
$begingroup$
@5xum since english is not my native language the post may contain translation issues, i've added a description of what i use as a definition of "infinitely large sequence"
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– roman
Dec 12 '18 at 12:54
$begingroup$
No, it's not translation issues. No matter which country you are in, the definition $$ lim_{ntoinfty}x_n = infty stackrel{text{def}}{iff} forall epsilon > 0 exists N in Bbb N :forall n > N implies |x_n| > epsilon $$ is wrong. There is no country or language on Earth where the sequence $-1,2,-3,4,-5,6,dots$ converges to $infty$. The absolute values should not be in the definition.
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– 5xum
Dec 12 '18 at 12:56
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@5xum you are right, i've messed things up
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– roman
Dec 12 '18 at 13:05
1
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HINT. Consider that $lim_{nto infty}|x_n|=infty$ means that for every $ r>0,$ the set ${n: |x_n|leq r}$ is finite . So what happens if $neg (lim_{nto infty}|x_n|=infty)?$
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– DanielWainfleet
Dec 12 '18 at 14:04