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$I$ is an interval, $I^0$ is the interior of $I$. Let $f:I\to\Bbb R$ be a funcion with intermediate value property on $I$, and $f(x)$ is monotonic on $I^0$. Does it follow that $f(x)$ is continuous on $I$?

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I don't think it's really a duplicate. That question is partly about avoiding using this result as a step in proving that result (I know they're very similar). – Tara B Feb 18 '13 at 16:01
I hastily voted to close; but now wish to remove my vote. The questions are different. In particular, here, $f$ is assumed to be monotone only on the interior of $I$. – David Mitra Feb 18 '13 at 16:11
Formally, the questions are different, but if $I = [a,b]$ it's clear that $f(a) = \lim_{x\to a^-} f(x)$. (The limit exists, possibly as $\pm\infty$, by monotonicity and if the limit does not equal $f(a)$, then $f$ can't satisfy the intermediate value property.) – mrf Feb 18 '13 at 16:26
here $f$ is monotonic, not strictly increasing or strictly decreasing – ziang chen Feb 18 '13 at 17:11
up vote 0 down vote accepted

Hint: Let $x \in I$. Assume $f$ is increasing, since the decreasing case is similar.

We want to show that $\lim_{h\rightarrow 0} f(x+h) = f(x)$. By the intermediate value property, $\forall \epsilon > 0 \exists h_0, h_1 $ such that $f(x+h_0) - f(x) = \epsilon$ and $f(x)-f(x-h_1) = \epsilon$ (note that possibly only one of the $h_i$ exist if $x$ is an end point of I). Can you think what to do next?

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It will be continuous on $I^0$, but not necessarily on $I$.

Let $f\colon[\,0,1\,]\to\mathbb{R}$ be defined as $f(x)=x$ if $0<x\le1$, $f(0)=1$. $f$ is monotonic in $(0,1)$ and satisfies the mean value property on $[\,0,1\,]$, but is discontinuous at $x=0$.

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This is true, but we want $f$ to satisfy the intermediate value property, which your function does not. – Tom Oldfield Feb 18 '13 at 15:59
Thanks! If $f$ sactify intermediate value theorem on every interval $E\subset I$,then it will be continuous on $I$ – ziang chen Feb 18 '13 at 16:00
I does not satisfies the intermediate value property on intervals of the form $[0,a]$ with $0<a<1$, but it satisfies it on $[0,1]$, at least the way I understand it: if $x_0,x_1\in I$ and $f(x_0)<c<f(x_1)$, then there exists $x\in I$ such that $f(x)=c$ (that is, in this definition it is not required hat $x\in[x_0,x_1]$.) – Julián Aguirre Feb 18 '13 at 17:04

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