# Set of discontinuous points.

1. Dec 6, 2006

### MathematicalPhysicist

i have a function f:R->R where f is monotone increasing, i need to show that the set of discontinuous points of f is at most countable.
so i need to find an injective or 1-1 mapping from this set to the naturals, or to the rationals.
i thought perhaps defining the next function g:A->Q, where A is the set of discontinuous points of f, by:
let x0 be in A, so lim(x>x0)f(x)>lim(x<x0)f(x)
g(x0)=x0 if x0 is in Q
but how do i define for points which arent in Q?

2. Dec 6, 2006

### HallsofIvy

Staff Emeritus
First, the correct phrase is "points of discontinuity of f", not "discontinuous points". Points are not "continuous" or "discontinuous", only functions are!

Have you already shown that a discontinuity of an increasing function must be a "jump" discontinuity- that is, that the left and right limits exist but are different? You need that to be able to talk about "lim(x>x0)f(x)>lim(x<x0)f(x)". But I don't see how your idea is going to work. Your g(x) doesn't actually use "lim(x>x0)f(x)>lim(x<x0)f(x)" It just says g(x)= x if f is discontinuous at x- and assumes f is discontinous at x! You mistake is looking at the domain instead of the range.

Try this. Let $lim_{x\rightarrow x_0^-} f(x)= a$, $lim_{x\rightarrow x_0^+} f(x)= b$. Since f has only "jump" discontinuities, those exist and a< b. There exist at least one rational number in the interval (a,b). Let g(x0) be such a rational number. Because f is an increasing function, if x0, x1 are points of discontinuities of f, x1> x0, then the "a" corresponding to x1 is greater than the "b" corresponding to x0: the two intervals do not overlap and so $g(x_0)< g(x_1)$. g is a one-to-one function from A to a subset of Q.

3. Dec 6, 2006

### MathematicalPhysicist

how have you defined the function g, i mean i dont see any explicit definition of the function, and how do you deal with points of discontinuity which are irrational, obviously they arent mapped into Q.

4. Dec 6, 2006

### HallsofIvy

Staff Emeritus
All I can do is repeat what I said- If xi is a point of discontinuity of the increasing function f. Then $lim_{x\rightarrow x_i^-} f(x)= a_i$, $lim_{x\rightarrow x_i^+} f(x)= b_i$ exist and ai< bi.
Choose a rational number yi in the interval (a,b). Such a rational number certainly exists, just choose one. Define g(xi)= yi. Since, if xj> xi, as I said before, since f is increasing, bi< aj so the intervals do not overlap and $y_j\ne y_i$. g is a one-to-one function from the set of points of discontinuity of f into Q. Whether xi is rational or not is irrelevant. The corresponding g(xi) is yi which is, by definition, rational.

5. Dec 7, 2006

### MathematicalPhysicist

shouldn't it be:
b_i>a_j?

6. Dec 7, 2006

### HallsofIvy

Staff Emeritus
No! That's the whole point. If $a_j< b_i< b_j$ then the two intervals overlap so it's possble that $y_i= y_j$ and the function may not be one-to-one.

Because f is increasing, if $x_i< x_j$, then the limit at xi from above must be less than the limit at xj from below: $b_i< a_j$.