Suppose that $(x_n)_{n=1}^\infty$ is a real sequence such that $\limsup_n x_n$ exists. I wish to show that
$\limsup_n x_n\le\beta \iff \forall\varepsilon>0\ \ \exists N\ \ \forall n\ge N, x_n < \beta + \varepsilon$.
I do understand the quantifiers, and have two equivalent definitions of $\limsup_n x_n$:
- $\lim_{n\to\infty}\left(\sup_{m\ge n} x_m\right)$;
- the greatest number that any subsequence can converge to.
However, this proof looks quite finicky, and would like a suggestion for an effective approach.
Suppose the left hand side holds, and let $(x_{n_k})_k$ be a subsequence that converges to $p = \limsup_n x_n$, so that we have $p \le \beta$.
Suppose the right hand side does not hold, then there exists some $\varepsilon_0 > 0$ such that for all $N$ there exists some $n > N$ with $x_n \ge \beta + \varepsilon_0$ (using the negation rules for quantifiers).
This means that we can find a subsubsequence $(x_{n_{k_l}})_l$ such that all terms are $\ge \beta + \varepsilon_0$, but the limit of this subsubsequence, which is still $p$, is also $\ge \beta + \varepsilon_0$, which contradicts $p \le \beta$. So the right hand side does hold.
Suppose the right hand side holds, and again let $(x_{n_k})_k$ be a subsequence converging to $p = \limsup _n x_n$. Suppose $p > \beta$. Then take $\varepsilon_0 = \frac{(p - \beta)}{2} > 0$. Using the right hand side we find a $N_0$ such that for all $k \ge N_0$ all terms $x_{n_k}$ are $< \beta + \varepsilon_0$, so $p$, still the limit, obeys $p \le \beta + \varepsilon_0$. But by how $\varepsilon_0$ was defined, $p > \beta + \varepsilon_0$, so we have a contradiction. Hence $p \le \beta$.