Unbounded Seq.: 3 is an Upper Bound

[fishturtle1]

Homework Statement

Either give an example or show that no example exists.

An unbounded sequence for which 3 is an upper bound, and no β < 3 is an upper bound.

Homework Equations

The Attempt at a Solution

example: {k}³_k=-∞

and by this notation i mean that k starts at -∞ and ends at 3, and k∈ℤ .

I chose -∞ as the lower bound because I wanted this sequence to be unbounded. Then I chose 3 as the upper bound because that is what the original statement asked for. My questions:
Does this example fulfill the original statement because I am unsure of what the question means by "and no β < 3 is an upper bound." The book uses β as the upper bound in previous pages.

Also does the way I wrote the sequence mean what I want it to mean? Because I am also confused on the notation.

 

[Mark44]

Homework Statement

Either give an example or show that no example exists.

An unbounded sequence for which 3 is an upper bound, and no β < 3 is an upper bound.

How can an unbounded sequence have an upper bound?

Homework Equations

The Attempt at a Solution

example: {k}³_k=-∞

Can you list the first, say, five terms in this sequence?

and by this notation i mean that k starts at -∞ and ends at 3, and k∈ℤ .

I chose -∞ as the lower bound because I wanted this sequence to be unbounded. Then I chose 3 as the upper bound because that is what the original statement asked for.My questions:
Does this example fulfill the original statement because I am unsure of what the question means by "and no β < 3 is an upper bound." The book uses β as the upper bound in previous pages.

Also does the way I wrote the sequence mean what I want it to mean? Because I am also confused on the notation.

 

[Ray Vickson]

Homework Statement

Either give an example or show that no example exists.

An unbounded sequence for which 3 is an upper bound, and no β < 3 is an upper bound.

Homework Equations

The Attempt at a Solution

example: {k}³_k=-∞

and by this notation i mean that k starts at -∞ and ends at 3, and k∈ℤ .

I chose -∞ as the lower bound because I wanted this sequence to be unbounded. Then I chose 3 as the upper bound because that is what the original statement asked for.My questions:
Does this example fulfill the original statement because I am unsure of what the question means by "and no β < 3 is an upper bound." The book uses β as the upper bound in previous pages.

Also does the way I wrote the sequence mean what I want it to mean? Because I am also confused on the notation.

Perhaps an "unbounded sequence" means what everybody else calls an "infinite sequence". An infinite sequence can be bounded or unbounded.

 

[LCKurtz]

Homework Statement

Either give an example or show that no example exists.

An unbounded sequence for which 3 is an upper bound, and no β < 3 is an upper bound.

Homework Equations

The Attempt at a Solution

example: {k}³_k=-∞

and by this notation i mean that k starts at -∞ and ends at 3, and k∈ℤ .

I chose -∞ as the lower bound because I wanted this sequence to be unbounded. Then I chose 3 as the upper bound because that is what the original statement asked for.My questions:
Does this example fulfill the original statement because I am unsure of what the question means by "and no β < 3 is an upper bound." The book uses β as the upper bound in previous pages.

Also does the way I wrote the sequence mean what I want it to mean? Because I am also confused on the notation.

If you mean what I think you do, a better way to write it would be $a_k = 4-k,~k = 1 ..\infty$.

 

[fishturtle1]

This is the definition of an unbounded sequence I've been using from online,

a sequence is bounded if it is bounded above and below <=> if ∃k∈ℝ such that | x_n | ≤ k ∀n∈ℕ.The first five terms in this sequence would be
-∞, -∞+1, -∞+2, -∞+3, -∞+4I was thinking that an unbounded sequence can have an upper bound if it goes to infinity in some direction but converges to a number as well.

 

[fishturtle1]

I checked the answer in the back of the book, and the answer is f_n=3-n which seems similar to a_k=4-k, k=1...∞.

I think I would have gotten this same answer had the original statement been " An unbounded sequence for which 3 is an upper bound ".

I'm still confused by the bold part: "An unbounded sequence for which 3 is an upper bound, and no β < 3 is an upper bound. "
I think that it means no number less than 3 can be an upper bound, but isn't this redundant since we said 3 is an upper bound in the first part of the statement?

 

[Mark44]

This is the definition of an unbounded sequence I've been using from online,

a sequence is bounded if it is bounded above and below <=> if ∃k∈ℝ such that | x_n | ≤ k ∀n∈ℕ.The first five terms in this sequence would be
-∞, -∞+1, -∞+2, -∞+3, -∞+4

No, these aren't numbers.

I was thinking that an unbounded sequence can have an upper bound if it goes to infinity in some direction but converges to a number as well.

 

[LCKurtz]

I checked the answer in the back of the book, and the answer is f_n=3-n which seems similar to a_k=4-k, k=1...∞.

They are the same, assuming $n$ starts at $0$ in $3-n$.

I think I would have gotten this same answer had the original statement been " An unbounded sequence for which 3 is an upper bound ".

I'm still confused by the bold part: "An unbounded sequence for which 3 is an upper bound, and no β < 3 is an upper bound. "
I think that it means no number less than 3 can be an upper bound, but isn't this redundant since we said 3 is an upper bound in the first part of the statement?

No, it isn't redundant. Since $3$ is a term of the sequence, no number $x$ less than $3$ can be an upper bound. How could it be if $x < 3$?

 

[Mark44]

I checked the answer in the back of the book, and the answer is f_n=3-n which seems similar to a_k=4-k, k=1...∞.

This sequence is bounded since all of the terms are less than or equal to 3.

I think I would have gotten this same answer had the original statement been " An unbounded sequence for which 3 is an upper bound ".

I'm still confused by the bold part: "An unbounded sequence for which 3 is an upper bound, and no β < 3 is an upper bound. "
I think that it means no number less than 3 can be an upper bound, but isn't this redundant since we said 3 is an upper bound in the first part of the statement?

I think the problem is poorly worded.

 

[LCKurtz]

This sequence is bounded since all of the terms are less than or equal to 3.
I think the problem is poorly worded.

Aren't you confusing "bounded above" with "bounded"?

 

[Mark44]

Aren't you confusing "bounded above" with "bounded"?

I don't think so. The definition of bounded sequence I am using is that it is a sequence that is bounded above and bounded below. That is, for some M > 0, |a_n| < M for all n in Z⁺ (or similar restriction on n).
See https://proofwiki.org/wiki/Definition:Bounded_Sequence

 

[fishturtle1]

They are the same, assuming $n$ starts at $0$ in $3-n$.No, it isn't redundant. Since $3$ is a term of the sequence, no number $x$ less than $3$ can be an upper bound. How could it be if $x < 3$?

Ok I think I get it. So we're told 3 is an upper bound. That means that the greatest number in this sequence is less than or equal to 3.

we're also told that no number less than 3 is an upper bound.

Therefore 3 must be included in this sequence.

If we were not told that no number less than 3 is an upper bound, then our sequence could have been something like x_n=-10-n, n=0 . . ∞

Am i understanding this correctly?

 

[LCKurtz]

I don't think so. The definition of bounded sequence I am using is that it is a sequence that is bounded above and bounded below. That is, for some M > 0, |a_n| < M for all n in Z⁺ (or similar restriction on n).
See https://proofwiki.org/wiki/Definition:Bounded_Sequence

Then why do you say in post #9 that $a_k = 4-k,~k=1..\infty$ is bounded?

 

[LCKurtz]

Ok I think I get it. So we're told 3 is an upper bound. That means that the greatest number in this sequence is less than or equal to 3.

we're also told that no number less than 3 is an upper bound.

Therefore 3 must be included in this sequence.

If we were not told that no number less than 3 is an upper bound, then our sequence could have been something like x_n=-10-n, n=0 . . ∞

Am i understanding this correctly?

Yes, almost. You could have it true if the sequence just got arbitrarily close to $3$ but less than $3$.

 

[fishturtle1]

Ok, thank you both for your help, helped me a lot on this question .

 

[Mark44]

Then why do you say in post #9 that $a_k = 4-k,~k=1..\infty$ is bounded?

I should have said "bounded above."

 

[LCKurtz]

I should have said "bounded above."

Right. That's what I pointed out in post #10 in the first place.