linguistic:voice_change
Proof version B

We will only prove the case of f (a) < u < f (b) {\displaystyle f(a)<u<f(b)}, as the f (a) > u > f (b) {\displaystyle f(a)>u>f(b)} case is similar.

Define g (x) = f (x) − u {\displaystyle g(x)=f(x)-u} which is equivalent to f (x) = g (x) + u {\displaystyle f(x)=g(x)+u} and lets us rewrite f (a) < u < f (b) {\displaystyle f(a)<u<f(b)} as g (a) < 0 < g (b) {\displaystyle g(a)<0<g(b)}, and we have to prove, that g (c) = 0 {\displaystyle g(c)=0} for some c ∈ {\displaystyle c\in }, which is more intuitive. We further define the set S = { x ∈: g (x) ≤ 0 } {\displaystyle S=\{x\in:g(x)\leq 0\}}. Because g (a) < 0 {\displaystyle g(a)<0} we know, that a ∈ S {\displaystyle a\in S} so, that S {\displaystyle S} is not empty. Moreover, as S ⊆ {\displaystyle S\subseteq }, we know that S {\displaystyle S} is bounded and non-empty, so by Completeness, the supremum c = sup (S) {\displaystyle c=\sup(S)} exists.

There are 3 cases for the value of g (c) {\displaystyle g(c)}, those being g (c) < 0, g (c) > 0 {\displaystyle g(c)<0,g(c)>0} and g (c) = 0 {\displaystyle g(c)=0}. For contradiction, let us assume, that g (c) < 0 {\displaystyle g(c)<0}. Then, by the definition of continuity, for ϵ = 0 − g (c) {\displaystyle \epsilon =0-g(c)}, there exists a δ > 0 {\displaystyle \delta >0} such that x ∈ (c − δ, c + δ) {\displaystyle x\in (c-\delta,c+\delta)} implies, that | g (x) − g (c) | < − g (c) {\displaystyle |g(x)-g(c)|<-g(c)}, which is equivalent to g (x) < 0 {\displaystyle g(x)<0}. If we just chose x = c + δ N {\displaystyle x=c+{\frac {\delta }{N}}}, where N > δ b − c {\displaystyle N>{\frac {\delta }{b-c}}}, then g (x) < 0 {\displaystyle g(x)<0} and c < x < b {\displaystyle c<x<b}, so x ∈ S {\displaystyle x\in S}. It follows that x {\displaystyle x} is an upper bound for S {\displaystyle S}. However, x > c {\displaystyle x>c}, contradicting the upper bound property of the least upper bound c {\displaystyle c}, so g (c) ≥ 0 {\displaystyle g(c)\geq 0}. Assume then, that g (c) > 0 {\displaystyle g(c)>0}. We similarly chose ϵ = g (c) − 0 {\displaystyle \epsilon =g(c)-0} and know, that there exists a δ > 0 {\displaystyle \delta >0} such that x ∈ (c − δ, c + δ) {\displaystyle x\in (c-\delta,c+\delta)} implies | g (x) − g (c) | < g (c) {\displaystyle |g(x)-g(c)|<g(c)}. We can rewrite this as − g (c) < g (x) − g (c) < g (c) {\displaystyle -g(c)<g(x)-g(c)<g(c)} which implies, that g (x) > 0 {\displaystyle g(x)>0}. If we now chose x = c − δ 2 {\displaystyle x=c-{\frac {\delta }{2}}}, then g (x) > 0 {\displaystyle g(x)>0} and a < x < c {\displaystyle a<x<c}. It follows that x {\displaystyle x} is an upper bound for S {\displaystyle S}. However, x < c {\displaystyle x<c}, which contradict the least property of the least upper bound c {\displaystyle c}, which means, that g (c) > 0 {\displaystyle g(c)>0} is impossible. If we combine both results, we get that g (c) = 0 {\displaystyle g(c)=0} or f (c) = u {\displaystyle f(c)=u} is the only remaining possibility.

Remark: The intermediate value theorem can also be proved using the methods of non-standard analysis, which places "intuitive" arguments involving infinitesimals on a rigorous footing.