$\liminf x_k \le \limsup x_k$ for every real sequence

For every sequence $(x_k)$ of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences),

$\liminf_{k} x_k \;\le\; \limsup_{k} x_k$

in $\overline{\mathbb{R}}$ (Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$, The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$, its order, and the arithmetic that is left undefined). No hypothesis is placed on $(x_k)$: both sides exist for every sequence (The tail suprema of any real sequence are nonincreasing in $\overline{\mathbb{R}}$, so the limit superior exists for every sequence) and the inequality holds between them in every case, including those in which one or both sides are $\pm\infty$.

• The inequality can be strict, and that is the interesting case. For the alternating sequence the two sides are $-1$ and $1$ ($(-1)^k$ has $\liminf = -1$ and $\limsup = 1$, so it does not converge ↗). Equality is exactly convergence, in the extended sense: that is A real sequence converges to $L \in \mathbb{R}$ iff $\liminf x_k = \limsup x_k = L$, and diverges to $\pm\infty$ iff both equal $\pm\infty$.

• What the proof actually uses is that the two families interleave. Each $i_m$ is below each $s_n$, not merely below $s_m$, and getting that needs a common index $p$ beyond both, which is where totality of the order on $\mathbb{N}$ enters. Without that step one would only know $i_n \le s_n$ for each $n$, which does not by itself compare a supremum of the first family with an infimum of the second.

• No completeness of $\mathbb{R}$ is used here beyond what is already inside Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$. The argument is pure order theory in a totally ordered set with a least and a greatest element.