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Take $f$ to be a function over the reals. I want to show that a set of discontinuities of the first kind for $f$ are countable. This is the discontinuity type at point $P \in \mathbb{R}$ where $lim_{x \rightarrow P^{-}} f(x)$ and $\lim_{x \rightarrow P^{+}} f(x)$ both exist but either do not equal each other or do not equal $f(P).$ The suggestion given to me is that if there is a discontinuity but the right and left limits exist, I can slip a rational number between them. Hence, I am considering showing this countability by developing a bijection between the set of discontinuities and the rationals, although I am having difficulty formalizing this in the language of functions. Any assistance will be appreciated.

Tom Brown
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  • Don't look for a bijection between the discontinuities of the 1st kind and the rationals. All you need and want is an injection. But: it seems you need a restriction — $f$ has to be monotone. Otherwise, the rational you choose for a point $P$ might equal the rational you choose for another point of discontinuity of 1st kind. (However, note that for monotone functions, every discontinuity is of the first kind, so the statement simplifies.) – BrianO Jan 19 '16 at 03:57

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This is a minor simplification of the proof of theorem 7.7 from here which contains a slightly more general result.

For each positive integer $k$ let $A_k = \{ a \in \mathbb{R} : f(a+) \text{ and } f(a-) \text{ exist and } |f(a+) - f(a-)| > \frac{1}{k}|\}.$

Since the union of $A_k$'s consists of the set of all jump discontinuities, it is sufficient to show each $A_k$ is countable.

If $A_k$ is not empty for each $t \in A_k$ we can choose a $\delta(t) >0$ such that for any $x \in (t,t+\delta(t))$ we have $|f(t+) - f(x)| < \dfrac{1}{8k},$ note that for any $u,v \in (t,t+\delta(t))$ we have $|f(u) - f(v)| \leq |f(u) - f(t+)| + |f(t+) - f(v)| < \dfrac{1}{4k}.$

If we can show that for distinct $a,b \in A_k$ the intervals $(a,a+\delta(a))$ and $(b,b+\delta(b))$ are disjoint we are done. For then the mapping $a \to (a,a+\delta(a))$ will be 1-1 from $A_k$ to a disjoint collection of open intervals which is countable.

To see this assume $a,b$ with $a < b$ are distinct points in $A_k$. If the intervals $(a,a+\delta(a))$ and $(b,b+\delta(b))$ are not disjoint we would have $ a < b < a + \delta(a) $.

Since there is a jump discontinuity of size $\frac{1}{k}$ at $b$ we can find $c,d$ with $ a < c < b < d < a + \delta(a)$ such that $$|f(c) - f(d)| > \frac{1}{2k}$$ this follows from the fact that $$ \begin{align} |f(c) - f(d)| & \geq |f(b+) - f(b-)| - |(f(b+) -f(d)| - |f(b-) - f(c)| \\ & \geq \frac{1}{k} - |(f(b+) -f(d)| - |f(b-) - f(c)| \end{align} $$

and we can choose $c$ and $d$ to make $|f(b-) - f(c)|$ and $|(f(b+) -f(d)|$ as small as we wish.

However from the definition of $\delta(a)$ we should also have $|f(c) - f(d)| \leq \frac{1}{4k}.$

This contradiction proves $(a,a+\delta(a))$ and $(b,b+\delta(b))$ are disjoint and we are done.

Arin Chaudhuri
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