Let $A$ be a bounded sequence in $\mathbb{R}$ and $c > \lim\sup A$. Show that only finitely many terms of $A$ are greater than or equal to $c$.

Question

I need to show that at most finitely many terms of this sequence are greater than or equal to $c$.

I don't know if it is the wording of the problem but I don't know what this is asking me to do. Help on this would be amazing! And thank you in advance.

Answer 1

Here's what the problem means:

You have that $A=\{x_1,x_2,\ldots\}\subset\mathbb{R}$ is a bounded sequence. That it is bounded means that $|x_i|\leq K$ for all $i=1,2,\ldots$, and for some constant $K>0$.

You then take the $\limsup$ of this sequence. You can define this in different ways, but let me know if you feel shaky about this.

So let $c>\limsup A$ be some constant. You need to prove that only a finite number of elements from $A$ can be larger than $c$.

Answer 2

I think you, perhaps, need a more intuitive understanding of $\limsup_{n\to\infty} s_n=s$.

[I would much prefer to write it this way than the way you did, since then you can relate to what you remember about sequence limits.]

The phrase $\lim_{n\to\infty} s_n=s$ can be thought of in the following way when $s$ is a finite number. Instead of the $\epsilon$, $N$ version imagine it this way:

1. $\lim_{n\to\infty} s_n=s$ is equivalent to saying that if you choose any numbers $u$ and $v$ with $u<s< v$ then the inequality $$u< s_n < v$$ fails for only the first few terms of the sequence [i.e., with only finitely many exceptions].

In the version you would have learned we always used $u=s-\epsilon$ and $v=s+\epsilon$ but if you are addicted to using $\epsilon$ all the time you get messy details sometimes and miss out on the intuition as well.

Then the limsup and liminf versions are just half of this, like right and left hand continuity is half of continuity, and like almost every mathematical idea involving real numbers gets split into a "one-sided" version.

2. $\limsup_{n\to\infty} s_n\leq s$ is equivalent to saying that if you choose any number $v$ with $ s< v$ then the inequality $$ s_n < v$$ fails for only the first few terms of the sequence [i.e., with only finitely many exceptions].

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3. $\liminf_{n\to\infty} s_n\geq s$ is equivalent to saying that if you choose any number $u$ with $u<s $ then the inequality $$u< s_n $$ fails for only the first few terms of the sequence [i.e., with only finitely many exceptions].