Definition: A real function $f$ has the intermediate value property on an interval $I$ containing $[a,b]$ if $f(a) < v < f(b)$ or $f(b) < v < f(a)$; that is, if $v$ is between $f(a)$ and $f(b)$, there is between $a$ and $b$ a $c ∈ [a,b]$ such that $f(c) = v$.
Prove that a strictly increasing function $f:[a,b] → R$ which has this property is continuous on $[a,b]$
Attempt:
Let $ε>0$ and $x∈[a,b]$. Consider $(f(x)−ε,f(x)+ε)$. By intermediate value property there is $x′∈f'(f(x)−ε,f(x))$ and $x″∈f'(f(x),f(x)+ε)$. Because $f'$ is increasing $x′<x<x″$. Now let $δ=\min(|x−x′|,|x−x″|)$.
Is this correct? How do I continue?
Best Answer
You can continue as follows.
If $y \in (x-\delta, x+\delta)$, then $|y-x| < \min(|x-x'|, |x-x''|)$. Therefore,
Rearrange the first inequality: $$\begin{aligned} |y-x| &< x-x' \\ \iff -(x-x') < y-x &< x-x' \\ \iff x' < y &< 2x - x' \\ \end{aligned}$$ (in particular, $x' < y$)
Rearrange the second inequality: $$\begin{aligned} |y-x| &< x''-x \\ \iff -(x'' - x) < y-x &< x'' - x \\ \iff 2x - x'' < y &< x'' \\ \end{aligned}$$ (in particular, $y < x''$)
The two inequalities combined therefore imply that $x' < y < x''$. As $f$ is increasing, this means that $f(x') < f(y) < f(x'')$. Recalling that $$f(x') \in (f(x) - \epsilon, f(x))$$ and $$f(x'') \in (f(x), f(x) + \epsilon)$$ we can conclude that $$f(x) - \epsilon < f(x') < f(y) < f(x'') < f(x) + \epsilon$$ and therefore $f(y) \in (f(x) - \epsilon, f(x) + \epsilon)$.
We conclude that $f$ is continuous at $x$.