Is Liminf(sn) = -limsup(-sn) for any sequence (s_n)?

[Mr Davis 97]

Homework Statement

Prove that $\lim \inf s_n = - \lim \sup (- s_n)$ for any sequence $(s_n)$, assuming that, for any nonempty set $S$, $\inf S = - \sup (-S).$

Homework Equations

The Attempt at a Solution

Here is my attempt at a solution.

Let $S_N = \{s_n ~|~ n>N \}$. Also, clearly, $-S_N = \{-s_n ~|~ n>N \}$.
By the information in the hypothesis, it is true that $\inf S_N = - \sup (-S_N)$. These are sequences which are equal, so their limits must be equal:
$\lim \inf s_n = - \lim \sup (-s_n)$.

 

[tnich]

Homework Statement

Prove that $\lim \inf s_n = - \lim \sup (- s_n)$ for any sequence $(s_n)$, assuming that, for any nonempty set $S$, $\inf S = - \sup (-S).$

Homework Equations

The Attempt at a Solution

Here is my attempt at a solution.

Let $S_N = \{s_n ~|~ n>N \}$. Also, clearly, $-S_N = \{-s_n ~|~ n>N \}$.
By the information in the hypothesis, it is true that $\inf S_N = - \sup (-S_N)$. These are sequences which are equal, so their limits must be equal:
$\lim \inf s_n = - \lim \sup (-s_n)$.

I think you missed the intent of the assumption that $S$, $\inf S = - \sup (-S).$ The lim inf is the the infimum of what set?

 

[Mr Davis 97]

I think you missed the intent of the assumption that $S$, $\inf S = - \sup (-S).$ The lim inf is the the infimum of what set?

I don't really know what you're asking. Where am I going wrong in my proof?

 

[member 587159]

I think you missed the intent of the assumption that $S$, $\inf S = - \sup (-S).$ The lim inf is the the infimum of what set?

You mean the supremum of what set?

liminf is a limit of decreasing infinums, and for a bounded set converges to a supremum.

 

[tnich]

I don't really know what you're asking. Where am I going wrong in my proof?

What I am asking you for is the definition of $lim inf(s_n)$ for a sequence $s_n$. The definition mentions, or at least implies a set.

The problem with your proof is what you have chosen for your set $S_N$. It doesn't get you any closer to proving anything about a lim inf.

 

[tnich]

You mean the supremum of what set?

liminf is a limit of decreasing infinums, and for a bounded set converges to a supremum.

Yes. Thanks for catching that.

 

[Mr Davis 97]

What I am asking you for is the definition of $lim inf(s_n)$ for a sequence $s_n$. The definition mentions, or at least implies a set.

The problem with your proof is what you have chosen for your set $S_N$. It doesn't get you any closer to proving anything about a lim inf.

I am using the following as the definition of liminf: $\lim \inf s_n = \lim_{N \to \infty} \inf \{s_n ~|~ n>N \}$. Does that help clarify?

 

[member 587159]

I am using the following as the definition of liminf: $\lim \inf s_n = \lim_{N \to \infty} \inf \{s_n ~|~ n>N \}$. Does that help clarify?

Yes. You can actually write this without a limit.

Hint: Assume the sequence is bounded (the unboundef case can be treated separately).

Show that the sequence of infina is increasing. What does this tell you about the limit?

 

[Mr Davis 97]

Yes. You can actually write this without a limit.

Hint: Assume the sequence is bounded (the unboundef case can be treated separately).

Show that the sequence of infina is increasing. What does this tell you about the limit?

That it converges by the monotone convergence theorem?

 

[tnich]

I am using the following as the definition of liminf: $\lim \inf s_n = \lim_{N \to \infty} \inf \{s_n ~|~ n>N \}$. Does that help clarify?

OK. I see where you are going, and I think your proof works OK as is.

I am thinking of the definition of lim inf in terms of subsequential limits.

This is one of these questions where every step seems so obvious, it's hard to know how much to write down.

 

[Ray Vickson]

I am using the following as the definition of liminf: $\lim \inf s_n = \lim_{N \to \infty} \inf \{s_n ~|~ n>N \}$. Does that help clarify?

A criterion that is equivalent to this (and often much easier to use) is: a finite number $U$ is the $\limsup$ of the sequence ${\cal S} = \{s_n\}$ if and only if for any $\epsilon > 0$ we have $s_n < U + \epsilon$ for all $n > N(\epsilon)$ and $U - \epsilon < s_n$ for some $s_n \in {\cal S}.$ (Of course, the latter implies that $U - \epsilon < s_n$ for infinitely many $s_n \in {\cal S}.$)

You can characterize a finite $\liminf$ in a similar way. Then, all you need to do is look at the negatives of the $s_n$.

 

[member 587159]

That it converges by the monotone convergence theorem?

Yes, convergent with limit the supremum of all the terms in the sequence.

Now, you can use that $-\sup(-A) = infA$ and analoguous formulas to easily show the equality.