0
$\begingroup$

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!

  • 2
    If $L^+ = L^-$, then $f$ is continuous (at $x_0$).2012-12-29
  • 1
    Only if $f(x_0)=L^+=L^-$.2012-12-29

2 Answers 2

3

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

  • 0
    Thanks for the answer. I follow the logic, I am just not sure how does this prove the statement, would appreciate elaboration, thanks!2012-12-29
  • 0
    @DahnJahn Sure. I supposed $L^{+}=L^{-}=L$ and proved that $f(a)=L$. This means $f$ is continuous at $a$2012-12-29
  • 0
    Thanks, now I understand it completely.2012-12-29
0

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 or $b. And you can easily conclude that $a 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 $-f$.

  • 0
    Right, what I missed is that, of course(!), $x_0$ can't simply be undefined. Stupid mistake from me, thanks for clearing that up!2012-12-29
  • 0
    @Dahn Jahn: You probably meant $f(x_0)$ - And you're welcome :)2012-12-29
  • 1
    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.2012-12-29
  • 0
    @CalvinLin Welcome to Math S.E. Calvin! I agree with you wholeheartedly, this does not answer the question.2012-12-29
  • 0
    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 or $c>a$ so $c\not=a$ so you can't remove the discontinuity. I don't get why it's wrong...2012-12-29
  • 0
    @xavierm02 It's mostly due to the way you phrased your argument, and that notation that was chosen/given. Note that he already defined $x_0$ to be a point of removable discontinuity. At a removable discontinuity, you have $a=c$ by definition. What you ended up showing, was that if $x_0$ is a point of discontinuity that is not an essential discontinuity, then it must be a jump discontinuity (and hence not a removable discontinuity). However, this contradicts the notation that was used.2012-12-29
  • 0
    @CalvinLin Have I defined $x_0$ to be a point of *removable* discontinuity? I'm not aware of that! Anyway, this answer helped me understand the general gist of the idea, so thanks for it nontheless.2012-12-29