Prove these statements are equivalent: $a = \limsup a_n$

$a$ is the lowest value such that $\forall \epsilon>0 \exists N \in \mathbb{N}$ such that $a_n <a+\epsilon \forall n\geq N$.

I proved the first implies the second like this:

Let $c=a-\delta, \delta>0$.

$c+\epsilon=a-\delta+\epsilon=a-(\delta-\epsilon)$. Take $\epsilon <\delta$, then $c+\epsilon=a-(\delta-\epsilon)<a$

Let $A_n=\{a_i|i\geq n\}$ and $(b_n)=\sup A_n$ the sequence of sups. Then $\limsup a_n=\lim b_n = \inf b_n$. Since $c+\epsilon <a$ then exists $b_m >c+\epsilon$ so there isn't $c <a$ that satisfies the property.

Now, how I prove the other direction?

  • $\begingroup$ Do you mean $$\lim_{N->\infty}\sup_{n \ge N} a_n$$ ? $\endgroup$ – Astyx Oct 16 '16 at 16:28
  • $\begingroup$ yes, and since the sequence of sup is non increasing the limit will be the inf. $\endgroup$ – Matheus barros castro Oct 16 '16 at 16:36

Let a be the smallest value such that

$$\forall \epsilon \gt0, \exists N \in \Bbb N, \forall n \ge N, a_n \lt a+\epsilon$$

Therefore you get

$$\forall \epsilon \gt0, \exists N_0 \in \Bbb N,\forall N \ge N_0, \sup_{n\ge N} a_n \lt a+\epsilon$$

Now suppose there exists $\epsilon \gt 0$ such that $\forall N_0 \in \Bbb N,\exists N \ge N_0, \sup a_n \lt a-\epsilon$ and call $a' = a - \epsilon\lt a$.

Then, since $\sup a_n$ is decreasing, you get : $\exists N_0 \in \Bbb N, \forall N \gt N_0, \sup_{n\ge N} a_n \lt a'$ which goes against the definition of a. Therefore : $$\forall \epsilon \gt0, \exists N_0 \in \Bbb N,\forall N \ge N_0, a-\epsilon \le \sup_{n\ge N} a_n \lt a+\epsilon$$

ie $$a = \lim_{N-> \infty} \sup_{n \ge N} a_n$$


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