Prove that F is discontinuous at every rational number

  • #1
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Homework Statement



Let [tex]x_{1}, x_{2}, ...[/tex] be a sequence of rational numbers in which each rational number in (0,1) occurs exactly once. Define the function,

[tex]H(x) = 0 [/tex] if [tex]x \leq 0[/tex], and 1 if [tex]x > 0.[/tex]

Next, define the function

[tex]F(x)= \sum^{\infty}_{k=1} 2^{-k} H(x - x_{k}). [/tex]

Prove that F is strictly increasing on (0,1), discontinuous at every rational number in (0,1) (Bonus: prove that F is continuous at every irrational number in (0,1)).



Ok, so my colleague and I understand, heuristically, why it is strictly increasing. This is because for any y1 and y2 in (0,1), such that y2>y1, the series of F(y2) will have more terms that are non zero than the series of F(y1). Thus F(y2)>F(y1). Now we need to prove that it is discontinuous at every rational number, which, up to this point, continue's to baffle us.

I understand that since it is strictly increasing, then for any point c in (0,1), the left and right hand limits must exist (i.e. F(c-) and F(c+) exist and are finite). But I can't quite make them unequal to each other at each rational point. Any insight?

thanks,

M
 
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Answers and Replies

  • #2
Office_Shredder
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At a rational number, when you look at the limit as x approaches y of the sum of H(x-xk)s, only one of these H's is going to have a different value depending on which side you approach it from. Which one is it?
 
  • #3
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I don't think I understand what you mean.

So at some rational number y, we have the sum of H(y-xk). Now if we are approaching y from the left, then, depending on where the sequence terms xk is, H will be either 1 or 0. Now as we are approaching from the right, the opposite will happen, meaning some of the H=1 will turn to H=0 as we approach y from the right. But what does this have to do with y being rational?
 
  • #4
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Hey micho,

Try just looking at a single term of the series, [itex]H(x-x_k)[/itex], where k is now fixed. The question is this, as x approaches from the left side of [itex]x_k[/itex] what does H equal? And as we approach from the right side, what does H equal?

Now, consider that your summation is over all of the rationals, as you approach any given rational [itex]x_k[/itex], how does this change the series if we approach from the left side versus the right side?
 
  • #5
122
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Hey micho,

Try just looking at a single term of the series, [itex]H(x-x_k)[/itex], where k is now fixed. The question is this, as x approaches from the left side of [itex]x_k[/itex] what does H equal? And as we approach from the right side, what does H equal?

Now, consider that your summation is over all of the rationals, as you approach any given rational [itex]x_k[/itex], how does this change the series if we approach from the left side versus the right side?


Hi Coto,

Thanks for the reply. I understand the first paragraph clearly. So for a given xk, as x approach from the left, H = 0. Whereas as x approach from the right H=1.

Now for the second paragraph, for a given rational q, this rational will also appear in the sequence x1, x2, x3,.... But I can't see how this changes anything. So the only H that will be different depending on where we approach it is H evaluated at the rational q.

So the limit of H(x-xk) as x---->q- will equal 0, whereas the limit of H(x-xk) as x---->q+ is 1. Is this what you meant?
 
  • #6
307
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Now for the second paragraph, for a given rational q, this rational will also appear in the sequence x1, x2, x3,.... But I can't see how this changes anything. So the only H that will be different depending on where we approach it is H evaluated at the rational q.

With some qualifications, you have it. For any rational [itex]q \in [0,1][/itex], there is an [itex]x_k = q[/itex]. This means that if you evaluate the limit of the series at any rational number, the series will always differ with the k'th term depending if you approach the point from the right or the left. The series differs with only this single term, but what does that mean for the continuity of the series at that point?
 
  • #7
122
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This means that F(y3-)<F(y3+), since F(y3+) will be greater by 1/2^3. This implies a discontinuity at y3 (where y is a rational).

Now I'm trying to prove that F is continuous at each rational. I understand it heuristically, I just need to write it out in symbols.
 
  • #8
307
3
Well good luck. If you understand it heuristically, write out your thoughts into logical sentences and then translate the sentences into logical constructions.
 

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