English
If s is bounded below and above, then sInf(s) ≤ sSup(s).
Русский
Если множество ограничено снизу и сверху, то infimum не превышает supremum.
LaTeX
$$$\\text{If } BddBelow(s) \\land BddAbove(s), \\; \\operatorname{sInf}(s) \\le \\operatorname{sSup}(s)$$$
Lean4
/-- As `sSup s = 0` when `s` is a set of reals that's either empty or unbounded above,
it suffices to show that all elements of `s` are nonnegative to show that `0 ≤ sSup s`. -/
theorem sSup_nonneg (hs : ∀ x ∈ s, 0 ≤ x) : 0 ≤ sSup s :=
by
obtain rfl | ⟨x, hx⟩ := s.eq_empty_or_nonempty
· exact sSup_empty.ge
· exact sSup_nonneg' ⟨x, hx, hs _ hx⟩