If $t:=\limsup a_n$, then for all $\epsilon>0$, exists $N\in\mathbb{N}$ such that $a_n<t+\epsilon$ for all $n\geq N$

Question

This question I will translate as

If $t:=\limsup a_n$, then for all $\epsilon>0$, exists $N\in\mathbb{N}$ such that $a_n<t+\epsilon$ for all $n\geq N$

Proof by contradiction

If $t\geq a_n$ and exists some $\epsilon>0$, for all $N\in\mathbb{N}$ such that $a_n\geq t+\epsilon$ for some $n\geq N\,$ will lead to contradiction

$$t\geq a_n\geq t+\epsilon\\0\geq a_n-t\geq\epsilon>0$$

contradiction built as $0>0$

Is this valid proof?

Thank you for your comment

Answer 1

I really can't follow what you're doing in your proof, which is probably a bad indicator. What we can do is the following:

Set $b_n=\sup_{m\geq n}a_m$, and recall that

$$t=\limsup_{n\to\infty}a_n=\lim_{n\to\infty}\left(\sup_{m\geq n}a_m\right)=\lim_{n\to\infty}b_n.$$

Notice that $b_n$ defines a decreasing sequence, and so $b_n\geq t$ for all $n\in\mathbb{N}$. By the definition of the limit, we can, for each $\varepsilon>0$, find some $N\in\mathbb{N}$ such that

$$b_n-t=\lvert b_n-t\rvert<\varepsilon$$

whenever $n\geq N$, which we can rewrite as

$$b_n<t+\varepsilon.$$

Finally, by the definition of the supremum,

$$a_n\leq\sup_{m\geq n}a_m=b_n<t+\varepsilon,$$

and we are done.