4

Possible Duplicate:
Showing properties of discontinuous points of a strictly increasing function
How to show that a set of discontinuous points of an increasing function is at most countable

I'm struggling to find an elegant proof of the following problem

Let $f : (a, b) \to \mathbb R$ be non-decreasing, $a, b \in \mathbb R$, then $f$ only has countably many discontinuities.

My intuition was to show by contradiction that the set of discontinuities $N \subseteq (a,b)$ is discrete, i.e. all discontinuities are isolated. From there on it's easy to prove that there is an injective function $N \to \mathbb Q$.

But does my first step make sense? Say we had non-isolated discontinuities like $\epsilon > 0, x_0 \in N$ such that $B_\epsilon(x_0) \subseteq N$ - how could one derive a contradiction?

I've already shown that $$\lim_{x \nearrow x_0} f(x) \text{ and } \lim_{x \searrow x_0} f(x)$$ exist for all points $x_0 \in (a, b)$ and that $f$ is continuous at $x_0$ iff both limits equal. I just somehow fail to do the final step properly. Any thoughts, please?

Community
  • 1
Aaron W.
  • 43
  • 3
  • 2
    Discontinuities need not be isolated. If the left and right limits are not equal, then they define a non-empty open interval. In that interval is a rational number. So...? – Qiaochu Yuan Dec 28 '11 at 20:23
  • @Alex: the Cantor set doesn't have the discrete topology as a subset of the reals. – Qiaochu Yuan Dec 28 '11 at 20:24
  • It's not true that $N$ needs to be discrete. It can even be dense: Chose an injection $h:\mathbb Q\to \mathbb N$, and let $f(x)=x+\sum_{y\in \mathbb Q \cap (a,x)} 2^{-h(y)}$ which is strictly increasing but discontinuous at every rational number. – hmakholm left over Monica Dec 28 '11 at 20:27
  • I realized that and deleted my comment. Sorry to leave yours hanging. – Alex Becker Dec 28 '11 at 20:28
  • http://math.stackexchange.com/questions/84870/how-to-show-that-a-set-of-discontinuous-points-of-an-increasing-function-is-at-m – David Mitra Dec 28 '11 at 20:41
  • 1
    In addition to 84870 which David Mitra links to, there are: http://math.stackexchange.com/questions/56831/is-there-an-everywhere-discontinuous-increasing-function and http://math.stackexchange.com/questions/14458/showing-properties-of-discontinuous-points-of-a-strictly-increasing-function – Jonas Meyer Dec 28 '11 at 22:17
  • Since the question was already explained, here is an intuitive explanation of this phenomena: Any monotonic function only has jump discontinuities. If the set of jump discontinuities would be uncountable, you can find uncountably many jump discontinuities in some compact interval $[c,d]$. A simple countability argument shows that there is an n so that infinitely many of "jumps" have to be at least $\frac{1}{n}$, which contradicts the fact that $f$ only changes $|f(d)-f(c)|$ over that compact interval... – N. S. Dec 28 '11 at 23:19

2 Answers2

2

No, the set of discontinuities need not be discrete. For example, it's quite possible to have a discontinuity at $0$ and also at $1/n$ for each positive integer $n$.

Hint: for each discontinuity, there are rational numbers that are not in $f((a,b))$.

Robert Israel
  • 448,999
  • Okay well, thanks for helping me out of this discrete-set dead-end. The solution is much easier without. – Aaron W. Dec 28 '11 at 21:10
2

Certainly $\lim_{x \nearrow x_0} f(x) \leq \lim_{x \searrow x_0} f(x)$. Also, $\sum\limits_{x_0\in (a,b)}\lim_{x \searrow x_0} f(x) - \lim_{x \nearrow x_0} f(x)$ is finite. It is not hard to show that since this is finite, all but countably many terms are $0$.

EDIT: When I shoot from the hip I forget that if $f(x)$ is unbounded you need to normalize it as $f(x)/\max\{|f(y)| : y\in (\frac{x+a}{2},\frac{x+b}{2})\}$ which is defined because $f(x)$ is monotonic and has all the discontinuities of $f(x)$.

Alex Becker
  • 60,569