Let f, g, h, and k be defined on [0,1] as follows: f(x)= 0 if x ∉ℚ 1 if x ∈ℚ

Let f, g, h, and k be defined on [0,1] as follows: ...

Key Concepts

Continuity

Continuity is a fundamental concept in calculus and real analysis that assesses whether small changes in the input of a function lead to small changes in the output. A function is continuous at a point if the limit of the function as the input approaches that point equals the function’s value at that point. This definition is critical when analyzing functions that have different definitions on dense subsets of the domain.

Density of Rational and Irrational Numbers

The fact that the rational numbers and the irrational numbers are both dense in the real numbers means that in any interval, no matter how small, there are both rational and irrational numbers. This density is crucial in studying functions that are defined differently on rational and irrational numbers, as it allows the construction of sequences from each subset converging to any given point, which is often used to test for continuity.

Sequential Criterion for Continuity

The sequential criterion for continuity states that a function is continuous at a point if and only if for every sequence converging to that point, the sequence of function values converges to the function's value at that point. This approach is especially useful for functions that have different behaviors on different subsets of their domain, as sequences can be chosen from different parts (such as rational and irrational numbers) to reveal discontinuities.

Piecewise-Defined Functions on Dense Sets

Many functions in real analysis are defined piecewise on dense subsets of their domain, such as defining different expressions on rational and irrational numbers. Analyzing the continuity of such functions typically involves evaluating the behavior of the function along sequences drawn from each subset. This approach can expose differences in the limit values when approaching a point, which in turn shows discontinuity.

Transcript

00:01 We are given a special function when you are asked to prove some properties about the continuity and discontinuities of this function.

00:13 The function is f of x equals 1 over q if x is a rational number, with x equal p over q being in lowest terms, and the function equals zero if x is not rational.

00:34 In part a, we are asked to show that the function is discontinuous at all rational points.

00:53 So, show that the function is discontinuous.

01:03 Let's, first of all, suppose that the limit, before we do this actually, let's make c be a rational number, and we'll suppose that limit as x approaches c, on f of x exists then we have that the limit as x approaches c on f of x is equal to l for some l now in order to prove this is discontinuous we need to have that l cannot be equal to the value of f of c so in order to do this i'm actually going to prove an even stronger statement not only that the function is discontinuous at all rational numbers, whether the function has no limit at each rational number.

03:10 So we're going to, first, suppose that the limit l is not equal to 0.

03:27 And we're going to take epsilon to be equal to the absolute value of l over 2, which we know is going to be strictly greater than 0.

03:55 Let delta be any positive number.

04:02 Now we want to pick a point x not inside the delta neighborhood of c such that the gap f of x not minus l is going to be greater than or equal to epsilon.

04:26 So let's think about the different possibilities we have for x not.

04:31 If we choose an x not, which is a rational number, it's not not entirely clear how we will be able to write our gap simply in terms of l.

04:44 But if we choose an x not which is irrational, then f of x not will be zero, and our gap will simply be the value of l.

04:54 And we can use this to show that our gap will be strictly larger than epsilon.

05:01 But in order to do this, we need to show that we can indeed pick an irrational number from this neighborhood.

05:08 So this is actually a property of the real numbers called the density of irrational numbers.

05:18 And we can prove this with this quick lemma using another property of the real numbers, which is the density of the rational numbers and the real numbers.

05:31 So we'll prove two quick lemmas and then return to the main proof.

05:37 So in lemma 1, q is dense.

05:46 In r.

05:53 And so what this means is suppose that we have two real numbers x and y with y greater than x.

06:11 And take epsilon to be equal to the difference between y minus x, which we know is going to be greater than zero.

06:31 And now we can find this is more intuitive, but you can prove this using properties of real numbers, find a natural number n such that 1 over n is going to be less than epsilon, then it follows that n y minus n x is going to be greater than 1.

07:06 And so if you think about the real number line, if you have two points, the distance between which is going to be strictly greater than 1, there must be some integer between these points.

07:19 So find another integer, let's want an integer not a natural, number m, such that nx is going to be less than m, which is less than n y.

07:39 Then we've shown that x is less than m over n, which is less than y.

07:46 So we found a rational number n, which lies between two arbitrary real numbers.

07:55 This proves that q is dense in the real numbers.

08:00 But we don't just want to prove that q is dense, we want to prove that your rational numbers is dense.

08:07 So for the second lemma, we have the irrational numbers, which will denote by r minus q, is dense in r.

08:29 To prove this, again, suppose that we have some y greater than x in the real numbers.

08:47 And by lemma 1, we're going to find some rational number, we'll call it r, such that x is going to be less than r is less than y and this implies that y minus r is greater than zero now using this information it's not entirely clear how we're going to obtain an irrational number but so let's just continue and i think i'll show you where we need to change so we have y minus r is greater than zero so again we can find some natural number n such that y minus r is going to be greater than 1 over n, which then implies that r is going to be less than y minus 1 over n.

10:26 We know that y minus 1 over n is going to be less than y.

10:30 So we have found the number that lies between x and y, which is different from r.

10:35 However, it's not clear how this number is going to be irrational.

10:41 But consider instead of y minus r being greater than we also know that y minus r over 2 is greater than 0 so that we can find a natural number n such that y minus r over 2 is going to be greater than 1 over n this implies that r is going to be less than y minus 2 over n or instead of writing it this way we can write it as r plus 2 over n will be less than y so now we have that x is going to be less than r this is less than r plus and here's where we're going to introduce in your rational number so we know that root two is positive so this is going to be r is less than r plus root 2 over n and we know that square root function is increasing so this is going to be less than r plus root 4 over n which is going to be r plus 2 over n and this is going to be less than y, as our previous inequality showed.

12:24 And so now we have obtained what is ostensibly any irrational number between x and y.

12:33 So r plus root 2 over n is not a rational number.

12:49 In order to prove that this is so, if you were to suppose it's rational number, you would show that root 2 is rational, which is a contradiction.

12:56 So now we have also shown that irrational numbers are dense in r.

13:05 We're going to use this dilemma to prove that f is discontinuous at each rational number.

13:13 In fact, the limit doesn't exist, each rational number, at least a non -zero limit doesn't exist.

13:22 So we're going to pick x not from the delta neighborhood of x.

13:36 So this is perceived to me.

13:38 And then make sure that we don't pick c itself, i'll just do, well, we can just do this, because c is irrational.

13:55 So i can pick an x not from this delta neighborhood of c, such that x not is irrational.

14:08 Then we have that the size of the gap, f of x, not minus l, it's going to be absolute value of l, which is greater than absolute value of l over 2, which we have is equal to epsilon.

14:22 So you've shown that the size of the gap is strictly greater than epsilon.

14:27 So this is a contradiction.

14:36 So another possibility is that l is going to be equal to 0.

14:46 But we have that f of c is going to be equal to, well, if c is m over n, it's going to be some 1 over n.

15:01 So ffc is not equal to 0.

15:06 So even if the limit were to exist, if the limit were 0, we would have that value of the function in the limit would be different at c.

15:16 So it follows that f is discontinuous at c in the rationales.

15:36 For part b, it has to show that the function is continuous at every irrational point.

15:50 I'm going to use a method different from the one given in the book.

15:54 I think the method i'll use is more direct and a little more intuitive.

16:06 So first, let's consider what the graph of this function might look like.

16:25 So if the f of x is going to be zero, x is irrational.

16:31 Since the irrational numbers are dense in the real numbers, we're going to have a graph something like this, a bunch of dots in the line y equals zero.

16:49 But then, all the rational points, points, we have f is 1 over q.

16:59 So for example, consider 0.

17:03 Well, in lowest terms, this is going to be 0 over 1.

17:09 So we'll have point at 0 1.

17:17 What about, say, 1? well, in lowest terms, this is 1 over 1.

17:25 So again, you have a point at 1 1.

17:33 And what about 2? well, in lowest terms, 2 over 1.

17:41 So we're actually going to expand.

17:42 Scale here a little bit, just to show a little more detail.

17:48 So we have one here and then two here.

17:55 These are both the same.

17:57 So already we're starting to see a pattern in the rational numbers.