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Using the definition of removable discontinuity from Wikipedia, why can't a monotone function have this type of discontinuity?

In other words, if $x_0\in D(f)$ is a point where the monotone function $f$ is discontinuous, and if $$\lim_{x\to x_{0^-}}f(x)=L^-$$ and $$\lim_{x\to x_{0^+}}f(x)=L^+$$ why cannot be $$L^+=L^-$$

I've been baffled by this for far too long now, thanks for any help!

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If $L^+ = L^-$, then $f$ is continuous (at $x_0$). – DanielM Dec 29 '12 at 12:39
Only if $f(x_0)=L^+=L^-$. – Eckhard Dec 29 '12 at 13:07
up vote 2 down vote accepted

An increasing function can't have a removable discontinuity at points in its domain. Indeed observe that for $\epsilon>0$, $$ \exists \delta>0:0<\left|x-a\right|<\delta\implies \left|f(x)-L\right|<\epsilon\iff L-\epsilon<f(x)<L+\epsilon$$ For $a-\delta<x<a$, $f(x)<f(a)$ and so $$ L-\epsilon<f(x)<f(a)$$ Similarly for $a<x<a+\delta$, $f(x)>f(a)$ and $$f(a)<f(x)< L+\epsilon$$ Therefore, $$\left|f(a)-L\right|<\epsilon$$ for arbitrary $\epsilon>0$ and so $f(a)=L$

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Thanks for the answer. I follow the logic, I am just not sure how does this prove the statement, would appreciate elaboration, thanks! – Dahn Jahn Dec 29 '12 at 18:26
@DahnJahn Sure. I supposed $L^{+}=L^{-}=L$ and proved that $f(a)=L$. This means $f$ is continuous at $a$ – Nameless Dec 29 '12 at 18:29
Thanks, now I understand it completely. – Dahn Jahn Dec 29 '12 at 18:50

Take $f$ and increasing function.

Let $\lim\limits_{x\to x_0^-}f(x) = a$

$f(x_0) = b$

$\lim\limits_{x\to x_0^+}f(x) = c$

Since $f$ is not continuous at $x_0$, you have $a\not=b$ or $b\not=c$. Since it is increasing, you have $a\le b \le c$ so $a<b$ or $b<c$. And you can easily conclude that $a<c$ so $a \not= c$, ie $\lim\limits_{x\to x_0^-}f(x) \not= \lim\limits_{x\to x_0^+}f(x)$

And for a decreasing function, you just use that property for $-f$.

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Right, what I missed is that, of course(!), $x_0$ can't simply be undefined. Stupid mistake from me, thanks for clearing that up! – Dahn Jahn Dec 29 '12 at 12:44
@Dahn Jahn: You probably meant $f(x_0)$ - And you're welcome :) – xavierm02 Dec 29 '12 at 12:46
This does not answer the question. The question is why can't ANY monotone function have a removable discontinuity. You merely showed an example of a monotone function which has a discontinuity that is not removable. – Calvin Lin Dec 29 '12 at 13:39
@CalvinLin Welcome to Math S.E. Calvin! I agree with you wholeheartedly, this does not answer the question. – Nameless Dec 29 '12 at 14:20
I don't get it. You take any monotone function. It's either increasing or decreasing. So we have $a \le b \le c$ or $c\le b\le a$. From the fact it's not continuous, we have at least two of those that aren't equal so $a<c$ or $c>a$ so $c\not=a$ so you can't remove the discontinuity. I don't get why it's wrong... – xavierm02 Dec 29 '12 at 15:26

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