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How to prove that a strictly increasing function $f:[a, b]\rightarrow \mathbb{R}$ which has the intermediate value property is continuous on $[a, b]$.

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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.Example of increasing funtion that abbeys the intermediate value property and that is not continuous.

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    But 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
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Let $c$ be a point with $a, and suppose given a small $\epsilon > 0$ (small enough that $c+\epsilon and otherwise arbitrarily small). Use the intermediate value property (and strict increasingness of $f$) to pick $d$ with $c for which $f(d)=f(c)+\epsilon.$ Since $f$ is strictly increasing, for all $x$ in $[c,d]$ we have $f(c) \le f(x) \le f(d).$ And $f(d)-f(c)=\epsilon.$ [Note: In the $\epsilon, \delta$ continuity definition, we have chosen $\epsilon$ first, and then found the point $d$, and are defining $\delta=d-c$.] This given "half" of continuity at $c$, and the left half is similar to show.

At either endpoint $a$ or $b$ we only need the above argument one way.

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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.

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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), Without loss of genarilty we may assume $l:=f(c-\delta), Take a point $y_0 \in [f(c-\epsilon),f(c))$, for this $y_0$ there is $x_0\in [c-\delta,c)$ such that $f(x_0)=y_0$[IVP]. This contradicts (1)