The intermediate value theorem
The intermediate value theorem (IVT) says: if f is continuous on [a, b] and y lies between f(a) and f(b), then f(c) = y for some c in [a, b]. A continuous function on an interval cannot skip a value. This is the rigorous version of the obvious-looking claim that you cannot cross a road without touching every point in between — but it secretly depends on the completeness of the reals.
Theorem (IVT, root form): f continuous on [a,b], f(a) < 0 < f(b) => exists c with f(c)=0.
Proof sketch (via supremum):
Let S = { x in [a,b] : f(x) < 0 }.
S is nonempty (a is in S) and bounded above by b, so by completeness
c = sup(S) exists in [a,b].
We show f(c) = 0 by ruling out the other cases:
If f(c) < 0: by continuity f stays < 0 on a little interval around c,
so some x slightly > c is still in S -> contradicts c = sup S.
If f(c) > 0: by continuity f stays > 0 just left of c,
so a smaller number is also an upper bound -> contradicts c = sup S.
Both fail, hence f(c) = 0. QEDBoundedness and extreme values
The boundedness theorem says a function continuous on a closed bounded interval [a, b] is bounded there. The extreme value theorem (EVT) goes further: such a function actually attains its maximum and minimum — there exist points p, q in [a, b] with f(p) <= f(x) <= f(q) for all x. The standard proof uses the Bolzano–Weierstrass theorem.
- Boundedness: if f were unbounded above, pick x_n in [a,b] with f(x_n) > n. By Bolzano–Weierstrass a subsequence x_{n_k} -> p in [a,b]. Continuity forces f(x_{n_k}) -> f(p), a finite number — but f(x_{n_k}) > n_k -> infinity, a contradiction. So f is bounded.
- Now let M = sup of f over [a,b], which exists by completeness since f is bounded. Choose x_n with f(x_n) -> M (possible by the definition of supremum).
- Extract x_{n_k} -> q in [a,b] by Bolzano–Weierstrass. Continuity gives f(q) = lim f(x_{n_k}) = M. So the supremum is achieved at q — it is a genuine maximum. The minimum follows by applying the same argument to -f.