# Need Help With Discontinuity Proof

1. Apr 19, 2015

### sonofagun

Let D⊆ℝ be an interval of nonzero length from which at most finitely man points x1,...,xn have been removed and let f: D→ℝ be a function. Then every discontinuity x∈D∪{x1,...,xn} of f is one of four types (removable, jump, infinite or discontinuity by oscillation).

Proof: Let x∈D or let x be one of the finitely many elements that were removed and assume that f is not continuous at x. Let the discontinuity at x be neither a removable discontinuity, nor a jump discontinuity, nor an infinite discontinuity. We will prove that it must be a discontinuity by oscillation. If limz→x-f(x) and limz→x+f(x) both existed, then they would either be equal and the discontinuity would be removable, or not equal and the discontinuity would be a jump discontinuity. Hence at least one of the one-sided limits does not exist at x. If x is the supremum or infimum of D, then f is defined at x and one of the two one-sided limits does not exist by default.

The last sentence is where I need clarification. I don't understand how x being the sup or inf of D implies that f is defined at x. Can someone explain that? The proof goes on:

In this case, the respective other one-sided limit also must not exist, because otherwise there would be a removable discontinuity at x. By symmetry, we can assume without loss of generality that limz→x-f(x) does not exist and f is defined on some interval [x-δ, x).

I'm stumped here as well. Can some one explain in detail why these sentences are true?

2. Apr 19, 2015

### Hawkeye18

What is discontinuity by oscillation?

3. Apr 20, 2015

### sonofagun

Suppose at least one of limz→x-f(x) or limz→x+f(x) does not exist. Then limz→xf(z) does not exist, and therefore ≠ f(x). Hence f is discontinuous at x. Suppose there exists a δ>0 such that f is bounded on {z∈D : distance (z, x) <δ}. If there are sequences zn and wn that converge to x (where zn, wn <x or zn, wn>x ∀n∈ℕ) and limn→∞f(zn) and limn→∞f(wn) both exists, but are not equal, we speak of a discontinuity by oscillation.

By the way, I figured out the first sentence I was having trouble with, but I still don't understand the last two. Anyone?

4. Apr 20, 2015

### Hawkeye18

The proof you presented is a bit confusing and not 100% accurate. The proof should go like that: let point of discontinuity $x$ be inside of the interval where the function is defined. Then if both one-sided limit exist (both finite and infinite limits are allowed) we have either removable discontinuity (both limits exist, finite and equal), or jump (both limits exist, finite and not equal), or infinite (at least one of the limits is infinite).

If one of the one-sided limits does not exist we have discontinuity by oscillation.

Now about the last sentences: if the point of discontinuity $x$ is one of the endpoints, only 1 one-sided omit can be defined, and the one-sided limit coincides with the limit. In this case we cannot have jump discontinuity, because for the jump discontinuity we need 2 one-sided limits, so in this case there are 3 choices: removable discontinuity(limit exists and finite), infinite discontinuity (limit exists and infinite) and discontinuity by oscillation (limit does not exist).

Finally, the statement is a bit confusing in the part where loins $x_1, x_2, \ldots, x_n$ are removed from the domain. I see a typical mistake perpetrated in almost all calculus books here. Namely, if you look at the definition of the continuity at a point, the point should be in the domain of a function, so in real mathematics continuity at a point outside the domain is not discussed at all! So if your points $x_1, x_2, \ldots, x_n$ are not in the domain, continuity at these points should not be discussed.

However, in most modern calculus textbooks the author give correct formal definition of continuity, and then in examples talk about continuity at the points outside the domain.
To give an example, in real mathematics, the function $f:\mathbb R\setminus\{0\}\to \mathbb R$, $f(x)=1/x$ is continuous at all points in the domain: the point $0$ does not belong to the domain, so continuity at this point should not be discussed. But most of the calculus textbooks say (wrongly) that $f$ is discontinuous at $0$.

I know only one calculus text treating this correctly: this is the Strang's calculus (available as free download).

5. Apr 20, 2015

### sonofagun

Thanks for your help. Regarding the points x1...xn:
The definition says at most finitely many points can be removed from the domain. It may be that no points are missing from the domain, in which case the domain is ℝ. But, if there is a set points not in the domain, it can only be finite. Otherwise, we'd have infinitely many points excluded from the domain and hence infinitely many discontinuities. I.e. we couldn't count the number of discontinuities.

By insisting that only finitely many points can be removed, we can count (at least theoretically) all the places where f is not continuous. I assume that's why the authors of these books include this stipulation in the definition, but I'm not totally sure.

6. Apr 21, 2015

### Hawkeye18

I think you are confused here. If a point is not in the domain of a function, the function cannot be continuous or discontinuous at this point: continuity of a function outside of its domain is not discussed at all.

7. Apr 21, 2015

### WWGD

An issue may be
An issue may be that if you remove infinitely-many points, you may be removing a limit point; you do in a bounded domain.

8. Apr 21, 2015

### sonofagun

Maybe I am confused. The definition of continuity says that if f is continuous at x, then f must be defined at x. Hence, I assumed that if f is not defined at x, then f is not continuous at x. For example, the function f(x)=1/x-2 is not defined at 2. Doesn't this imply that f is not continuous at 2?

9. Apr 21, 2015

### sonofagun

I haven't covered limit points yet. I'll keep this in mind though.

10. Apr 22, 2015

### Hawkeye18

The definition of continuity says that we investigate continuity only at points in the domain. Continuity or discontinuity outside of the domain makes no sense, there is no function there: so if you study continuity outside of the domain, you are studying properties of an object that does not exist.

For example, if you have a function defined on an interval $[-1, 1]$, would you say that it is discontinuous at $x=100$? Or is it continuous at this point? Neither answer makes much sense, so in real mathematics, continuity outside of the domain is just not discussed. So in your example the function is not defined at $x=2$, so its continuity there is not discussed.

To learn about continuity, you should look into any analysis textbook. Do not look into calculus texts, continuity is explained horribly there, and often wrongly.

P.S. I think the reason the calculus textbooks explain continuity wrongly, it is because they try to leverage intuition, and the facts about continuous functions are sometimes counterintuitive. For example, a function is always continuous at any isolated point in the domain. In particular, any function defined only on integers is always continuous.

You will probably ask: if $x$ is an isolated point in the domain of $f$, then the limit $\lim_{z\to x} f(z)$ is not defined, so why $f$ is continuous at $x$? The answer is that the definition in terms of limit is applicable only when $x$ is an accumulation point in the domain: in general case you have to use $\varepsilon$-$\delta$ definition, or the definition in terms of sequences.

11. Apr 22, 2015

### sonofagun

Ah, I see. It wouldn't make sense to say f is continuous or discontinuous at a point outside its domain because f isn't defined (doesn't exist) there. Thanks for clearing that up. I'm reading an Analysis book, by the way. It also contains a topological definition which I haven't gotten to yet.