How to prove that a strictly increasing function $f:[a, b]\rightarrow \mathbb{R}$ which has the intermediate value property is continuous on $[a, b]$.
Intermediate value property problem and continuous function
4 Answers
You cannot prove that because it is not true, there are functions that are not continuous that have this property. Just think of a function that is strictly increasing but with discontinuities and that is defined in the hole interval that you will notice that it can still obey the property.
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0But since $g(x)=2(x-1)$ already covers $[0,2]$ and is strictly increasing there, the limit as $x \to 1^-$ of the first function must be $0$ in order to maintain that the piecewise function is strictly increasing and continuous. And (I think) that makes the function continuous at 1. – 2012-11-09
Let $c$ be a point with $a
At either endpoint $a$ or $b$ we only need the above argument one way.
Hint: Since $f$ is increasing, for every $c\in[a,b]$, $f(c^{\pm}):=\lim_{x\to c^\pm}f(x)$ exists, and it suffices to show that $f(c)=f(c^\pm)$, where for $c$=$a$(resp. $b$), only consider $c^+$(resp. $c^-$). If it is not the case, you may use the intermediate value property to derive a contradiction.
We have infact, more general result:
$f: \mathbb R \to \mathbb R$ strictly increasing and has IVP is continuous in $\mathbb R$.
Pf: If possible $f$ be not continuous in $\mathbb R$. So, there exists $c \in \mathbb R$ where $f$ is not continuous. So, there exists $\epsilon >0 ,\forall \delta>0$ with $|x-c|<\delta$ implies $|f(x)-f(c)|\ge \epsilon.$------(1).
Now as $f$ is strictly increasing $f(c-\delta)