Is Limsup xk < ∞ if and only if the sequence {xk} is bounded above?

[kingwinner]

Claim (?):
limsup x_k < ∞
k->∞

IF AND ONLY IF

the sequence {x_k} is bounded above.


Does anyone know if this is true or not? (note that the claim is "if and only if")
If it is true, why?

Thanks!

 

[kingwinner]

I think I'm OK with the direction:
limsup x_k < ∞ implies the sequence {x_k} is bounded above.
k->∞
(becuase if the sequence is not bounded above, then limsup=∞)




But is the converse direction also true?

The sequence {x_k} is bounded above

implies

limsup x_k < ∞ ?
k->∞

 

[chiro]

Claim (?):
limsup x_k < ∞
k->∞

IF AND ONLY IF

the sequence {x_k} is bounded above.


Does anyone know if this is true or not? (note that the claim is "if and only if")
If it is true, why?

Thanks!

Hey kingwinner.

Think about the idea of the lowest upper bound. Suppose there is a value that is above this upper bound. Then this means the lowest upper bound is that new value.

Now imagine that all values are finite. What does this say about the lowest upper bound? What happens if one value is infinite (positive infinity)?

 

[HallsofIvy]

"lim sup x_n" is, by definition, the supremum of the set of all subequential limits. If it were not finite, then, given any M, there would have to exist subsequences of {x_n having limit larger than M so M could not an upper bound of {x_n}.

 

[kingwinner]

"lim sup x_n" is, by definition, the supremum of the set of all subequential limits. If it were not finite, then, given any M, there would have to exist subsequences of {x_n having limit larger than M so M could not an upper bound of {x_n}.

Does this imply that
the sequence {x_k} is bounded above => limsup x_k < ∞ ?

Are you using contrapositive?

 

[HallsofIvy]

Yes, that is proving "if a then b" by proving the contrapositive, "if b is not true then a is not true".

 

[kingwinner]

Yes, that is proving "if a then b" by proving the contrapositive, "if b is not true then a is not true".

Is it possible to prove "directly" that

sequence {x_k} is bounded above => limsup x_k < ∞ ??

 

[Congruent]

Is it possible to prove "directly" that

sequence {x_k} is bounded above => limsup x_k < ∞ ??

Yes. Using the definition above, let $\{x_{k_n}\}_{n=1}^{\infty}$ be any convergent subsequence. By assumption, there is an $M > 0$ so that $x_k \le M$. In particular, $M \ge x_{k_n}$. Thus $M \ge \lim_{n \to \infty} x_{k_n}$. This shows that M is an upper bound to any subsequential limit. By the definition HoI gave for the limsup as the supremum of all subsequential limits, this shows that limsup x_k < \infty.

 

[kingwinner]

Yes. Using the definition above, let $\{x_{k_n}\}_{n=1}^{\infty}$ be any convergent subsequence. By assumption, there is an $M > 0$ so that $x_k \le M$. In particular, $M \ge x_{k_n}$. Thus $M \ge \lim_{n \to \infty} x_{k_n}$. This shows that M is an upper bound to any subsequential limit. By the definition HoI gave for the limsup as the supremum of all subsequential limits, this shows that limsup x_k < \infty.

How about this? limsup means the limit of the supremum of terms with greater and greater index. If the whole set is bounded above, then sup_{k>0} is finite, right? sup(k>0), sup(k>1), sup(k>2),... forms a decreasing sequence right? The limit is either well defined real number or -infinity, isn't it?


So in some problem, say if we were to show that limsup x_k < infinity, it suffices to prove that sequence {x_k} is bounded above, and vice versa, is that correct?

Thanks.