Real Analysis-Prove Continuous at each irrational and discontinuous at each rational

  • Thread starter Kkathy
  • Start date
  • #1
8
0
Real Analysis--Prove Continuous at each irrational and discontinuous at each rational

The question is, Let {q1, q2...qn} be an enumeration of the rational numbers. Consider the function f(x)=Summation(1/n^2). Prove that f is continuous at each rational and discontinuous at each irrational.

Normally I would try to use Rudin's Th 7.11 (Principles of Mathematics) but I'm not sure if this still works with summation involved...

A very rough proof:
First of all, note that summation(1/n^2) is uniformly continuous by Th 7.10. Let x be an irrational, and x is continuous at all x by Th 7.11 so the sum is also continuous at x.

Is this correct?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
619


The question is, Let {q1, q2...qn} be an enumeration of the rational numbers. Consider the function f(x)=Summation(1/n^2). Prove that f is continuous at each rational and discontinuous at each irrational.

Normally I would try to use Rudin's Th 7.11 (Principles of Mathematics) but I'm not sure if this still works with summation involved...

A very rough proof:
First of all, note that summation(1/n^2) is uniformly continuous by Th 7.10. Let x be an irrational, and x is continuous at all x by Th 7.11 so the sum is also continuous at x.

Is this correct?

I don't have a copy of Rudin, so this is probably not helping my ability to second guess what you are talking about. But that doesn't make much sense. How is Summation(1/n^2) (whatever that means) a function of x? And what does that have to do with your enumeration of the rationals? Do you have a more complete version of the problem statement?
 
  • #3
8
0


I don't think I did a good enough job of describing f(x):
f(x)=∑qn<x(1/n2)

So, the index of summation is qn<x, where the series of q is an enumeration of the rational numbers. What is being summed is one over n-squared. The goal is to prove that f is continuous at each irrational and discontinuous at each irrational.

Also,
Th 7.11 in Rudin is: Let X be a metric space and fn be functions. Suppose that fn converges uniformly to f : X→C. Let {xk} be a sequence in X and x = lim xk. Suppose that an= lim fn(xk) as k → ∞ exists for all n. Then {an} converges and lim fn(xk) as k → ∞ = lim an as n → ∞. In other words limk→∞limn→∞fn(xk) = limn→∞limk→∞ fn(xk).
 
  • #4
Dick
Science Advisor
Homework Helper
26,263
619


I don't think I did a good enough job of describing f(x):
f(x)=∑qn<x(1/n2)

So, the index of summation is qn<x, where the series of q is an enumeration of the rational numbers. What is being summed is one over n-squared. The goal is to prove that f is continuous at each irrational and discontinuous at each irrational.

Also,
Th 7.11 in Rudin is: Let X be a metric space and fn be functions. Suppose that fn converges uniformly to f : X→C. Let {xk} be a sequence in X and x = lim xk. Suppose that an= lim fn(xk) as k → ∞ exists for all n. Then {an} converges and lim fn(xk) as k → ∞ = lim an as n → ∞. In other words limk→∞limn→∞fn(xk) = limn→∞limk→∞ fn(xk).

That's a much better job. You haven't stated 7.10, but if you are claiming you can prove f(x) is uniformly continuous, that's wrong. f isn't even continuous at rationals. You don't really need any theorems like that here. You need to think of how to show you can find a delta for every epsilon or show you can't. Try to use the epsilon-delta definition of continuity. And think hard about what the function f(x) looks like. Here's some hints. The sum of 1/n^2 for n=1,2,3,4,... i.e. for all n is a finite number. And if y>x the f(y)-f(x) is basically the sum of 1/k^2 for each q_k which lies between y and x.
 
  • #5
22,129
3,298


That's a much better job. You haven't stated 7.10, but if you are claiming you can prove f(x) is uniformly continuous, that's wrong. f isn't even continuous at rationals.

To be fair, he wasn't saying that f is uniformly continuous. He was saying that there was a "good" sequence that converges uniformly to f. But I kind of doubt that this is true. So as you said, just finding an appropriate delta for given epsilon seems much more promising.
 
  • #6
8
0


Part of the problem I'm having here is I don't understand what f(x) is. For example, how would I calculate f(1)? The index of summation would be qn < 1 but what is that, is it a finite set of numbers? How about for an irrational number like f(2/3)?
 
  • #7
Dick
Science Advisor
Homework Helper
26,263
619


Part of the problem I'm having here is I don't understand what f(x) is. For example, how would I calculate f(1)? The index of summation would be qn < 1 but what is that, is it a finite set of numbers? How about for an irrational number like f(2/3)?

What f(x) really is depends on which enumeration of the rationals you pick, and I don't think you can really 'calculate' it in any sense. It's always going to be an infinite sum. And 2/3 isn't irrational, is it? But that doesn't mean you can't reason about it anyway. For example, can you show it's always increasing? Here's another definition of continuous to try thinking about. Pick a point a, then if the limit x->a+ f(x) and the limit x->a- f(x) both exist and are equal to f(a), then f(x) is continuous at a.
 
  • #8
8
0


I can see that if I pick two points y and x, then as y and x get closer then f(y) and f(x) will get closer.

I think the reason that irrational numbers are continuous is because the are not dense in the set of rational numbers. Rational numbers are dense so it won't be possible to choose a delta that will always make d(f(y)-f(x)) < epsilon.

Is it enough to just state this, or do I need to pick a particular delta to prove it? How do I figure out which delta to pick?
 
  • #9
Dick
Science Advisor
Homework Helper
26,263
619


I can see that if I pick two points y and x, then as y and x get closer then f(y) and f(x) will get closer.

I think the reason that irrational numbers are continuous is because the are not dense in the set of rational numbers. Rational numbers are dense so it won't be possible to choose a delta that will always make d(f(y)-f(x)) < epsilon.

Is it enough to just state this, or do I need to pick a particular delta to prove it? How do I figure out which delta to pick?

Irrational numbers are also dense, so no, that's not the reason. Your thinking about this is still all fuzzy. Try and prove to me that if y<x, then f(y)<f(x). That will be a useful thing to know. And if you can do that, it will get you on track for the rest of the problem.
 
Last edited:
  • #10
8
0


For any y and x where y>x, then f(y)-f(x) = Ʃx<q<y(1/n2). Because the set of x<qn<y is a finite set of numbers, then as (y-x)→0, f(y)-f(x)→0.
 
  • #11
Dick
Science Advisor
Homework Helper
26,263
619


For any y and x where y>x, then f(y)-f(x) = Ʃx<q<y(1/n2). Because the set of x<qn<y is a finite set of numbers, then as (y-x)→0, f(y)-f(x)→0.

That's a good start. But i) it's not a finite set. You have an enumeration of ALL rationals. Rationals are dense. It's an infinite set. Why is the sum finite and positive? ii) it's x<=qn<y, that 'equal' is very important. Think about why it's true. That's what makes "as (y-x)→0, f(y)-f(x)→0" possibly a false statement.
 
  • #12
8
0


Okay, I think I can see it now.

From the beginning:
Let x and y be rational, and choose y>x. Let qx be the set of rationals < x and qy be the set of rationals < y. Then x=qm for some m (since x is rational, it is part of an enumeration of rationals) and qm∈qy.

Now f(y)=Ʃn∈qy(1/n2)=f(x)+1/m2.

Then since f(y)≥f(x) for every y>x, limy→x+f(y)≥f(x)+1/m2>f(x)

So by definition f isn't continuous at x.

I think that's right. For the other way around:
Let x and y be irrational and y>x. Again, let qx be the set of rationals < x and qy be the set of rationals < y. In this case x≠qm since x is irrational. So there does not have to be a rational number between x and y as y→x; instead, as y→x, qy→qx.

And we get limy→x+f(y)=Ʃn∈qx(1/n2)=f(x). Then f is continuous at x.

Is that it? (And thanks for all your help!)
 
  • #13
Dick
Science Advisor
Homework Helper
26,263
619


Okay, I think I can see it now.

From the beginning:
Let x and y be rational, and choose y>x. Let qx be the set of rationals < x and qy be the set of rationals < y. Then x=qm for some m (since x is rational, it is part of an enumeration of rationals) and qm∈qy.

Now f(y)=Ʃn∈qy(1/n2)=f(x)+1/m2.

Then since f(y)≥f(x) for every y>x, limy→x+f(y)≥f(x)+1/m2>f(x)

So by definition f isn't continuous at x.

I think that's right. For the other way around:
Let x and y be irrational and y>x. Again, let qx be the set of rationals < x and qy be the set of rationals < y. In this case x≠qm since x is irrational. So there does not have to be a rational number between x and y as y→x; instead, as y→x, qy→qx.

And we get limy→x+f(y)=Ʃn∈qx(1/n2)=f(x). Then f is continuous at x.

Is that it? (And thanks for all your help!)

I think you've got a pretty good idea of the idea behind the proof. The presentation could still use a lot of work. Start at the beginning "Let x and y be rational, and choose y>x." Why does y have to be rational? And f(y) isn't equal to f(x)+1/m^2. I find a lot to complain about here. It's not proofy. You are just saying stuff.
 
Last edited:
  • #14
8
0


I see your problem, let me try again.

Let x be rational. Choose y∈R such that y>x. Let qx be the set of rationals < x and qy be the set of rationals < y. Then x=qm for some m (since x is rational, it is part of an enumeration of rationals) and qm∈qy.

Now f(y)=Ʃn∈qy(1/n2)≥f(x)+1/m2.

Then since f(y)≥f(x) for every y>x, limy→x+f(y)≥f(x)+1/m2>f(x)

Therefor f is not continuous at x.

I have a little more trouble showing that f is continuous at irrationals. I'm not sure if I need two separate proofs for if y is rational or irrational or if can I use one proof for either y? It seems obvious that if x and y are irrational, then as y→x, f(y)→f(x) but I'm not sure if it is evident if x is irrational but y is rational. Do I need the delta-epsilon def if y is rational?
 
  • #15
Dick
Science Advisor
Homework Helper
26,263
619


I see your problem, let me try again.

Let x be rational. Choose y∈R such that y>x. Let qx be the set of rationals < x and qy be the set of rationals < y. Then x=qm for some m (since x is rational, it is part of an enumeration of rationals) and qm∈qy.

Now f(y)=Ʃn∈qy(1/n2)≥f(x)+1/m2.

Then since f(y)≥f(x) for every y>x, limy→x+f(y)≥f(x)+1/m2>f(x)

Therefor f is not continuous at x.

I have a little more trouble showing that f is continuous at irrationals. I'm not sure if I need two separate proofs for if y is rational or irrational or if can I use one proof for either y? It seems obvious that if x and y are irrational, then as y→x, f(y)→f(x) but I'm not sure if it is evident if x is irrational but y is rational. Do I need the delta-epsilon def if y is rational?

Ok, that seems ok. Same idea, just better statement. In fact, lim y->x+ f(x) is equal to f(x)+1/m^2. Can you think why that might be true? Just explain in words for now. And lim y->x- f(x)=f(x). Explaining these will tell you how to deal with irrationals. Think about your sets qy and qx.
 
  • #16
8
0


So, qx contains all the q < x. And y > x.
When x is rational, x will be contained in qy because qy is a set of rational numbers less than y and x is a rational number less than y. However, x cannot be the upper bound of y because the numbers are dense, so there will always have to be a number between x and q that is contained in qy. This becomes the 1/m2.

So limy→x+ f(x)=f(x)+1/m2 because 1/m2 is the jump discontinuity at x. However, limy→x- f(x)=f(x) because when you approach x from the left side instead of the right side there is no jump.

But when x is irrational, x will not be contained in qy. As y→x it is possible to contain all of qx in qy by slipping y between x and the first rational number less than x. Then the limits on both sides equal f(x).
 
  • #17
Dick
Science Advisor
Homework Helper
26,263
619


So, qx contains all the q < x. And y > x.
When x is rational, x will be contained in qy because qy is a set of rational numbers less than y and x is a rational number less than y. However, x cannot be the upper bound of y because the numbers are dense, so there will always have to be a number between x and q that is contained in qy. This becomes the 1/m2.

So limy→x+ f(x)=f(x)+1/m2 because 1/m2 is the jump discontinuity at x. However, limy→x- f(x)=f(x) because when you approach x from the left side instead of the right side there is no jump.

But when x is irrational, x will not be contained in qy. As y→x it is possible to contain all of qx in qy by slipping y between x and the first rational number less than x. Then the limits on both sides equal f(x).

Well, that's pretty good. Again you've got the idea ok. But there are factual inaccuracies. There is no "first rational number less than x". Really the idea is that as y gets closer and closer to x from below, qy is a subset of qx and every rational number in qx is eventually contained in qy, right?
 

Related Threads on Real Analysis-Prove Continuous at each irrational and discontinuous at each rational

Replies
4
Views
6K
Replies
7
Views
5K
  • Last Post
Replies
4
Views
6K
Replies
7
Views
2K
Replies
4
Views
3K
Replies
14
Views
8K
Replies
3
Views
1K
Replies
3
Views
3K
  • Last Post
Replies
3
Views
1K
Top