Continuous at irrational points

  • Thread starter ehrenfest
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[SOLVED] continuous at irrational points

Homework Statement


Every rational x can be written in the form m/n where n>0, and m and n are integers without any common divisors. When x = 0, we take n=1. Consider the function f defined on the reals by

f(x) = 0 if x is irrational and f(x) = 1/n if x = m/n

Prove that f is continuous at every irrational point, and that the right and left-hand limits of f exist at every rational point.


Homework Equations





The Attempt at a Solution


If x is irrational, then when you get closer and closer to it, it will get harder and harder to express it is a rational number and you will need larger and larger n to do it. I need to make that precise somehow.

For the second part, I am guessing that the right and left-hand limits will always be zero. Anything else would be kind of weird. And that is basically for the same reason I gave above, any rational sequence converging to a rational number will get "nastier and nastier" as it gets closer. But I need to make that precise somehow.
 

Answers and Replies

  • #2
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for continuity at an irrational point x_0 you can use epsilon-delta

so let e > 0, you need a d such that |x-x_0| < d => |f(x)-f(x_0)| < e

If x is irrational it's trivial.

Now for x rational, choose M s.t. 1/M < e. What can you say about all the numbers x = m/n which are at most some fixed distance from x_0 and for which n < M.

goodluck!
 
  • #3
HallsofIvy
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Homework Statement


Every rational x can be written in the form m/n where n>0, and m and n are integers without any common divisors. When x = 0, we take n=1. Consider the function f defined on the reals by

f(x) = 0 if x is irrational and f(x) = 1/n if x = m/n

Prove that f is continuous at every irrational point, and that the right and left-hand limits of f exist at every rational point.


Homework Equations





The Attempt at a Solution


If x is irrational, then when you get closer and closer to it, it will get harder and harder to express it is a rational number and you will need larger and larger n to do it. I need to make that precise somehow.

For the second part, I am guessing that the right and left-hand limits will always be zero. Anything else would be kind of weird. And that is basically for the same reason I gave above, any rational sequence converging to a rational number will get "nastier and nastier" as it gets closer. But I need to make that precise somehow.
Yes, that's exactly right. Suppose [itex]m_i/n_i[/itex] is a sequence of rational converging to the real number x. For a fixed N, what is true of the set of all M such that M/N is close to x? What is its size? What does that tell you?
 
  • #4
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Yes, that's exactly right. Suppose [itex]m_i/n_i[/itex] is a sequence of rational converging to the real number x. For a fixed N, what is true of the set of all M such that M/N is close to x? What is its size? What does that tell you?
For a fixed N,
[tex]|\frac{M-Nx}{N}|<\epsilon[/tex]
iff
[tex]|M-Nx| < \epsilon N[/tex]

So the size of the set is less than or equal to [itex]floor(\epsilon N)[/itex] I think. So your point is that the set is finite right? Which means that elements of that set cannot occur an infinite number of times in the sequence. Since this is true for any N, [tex]f(\frac{m_i}{n_i})[/tex] must converge to 0. I see thanks.
 
  • #5
HallsofIvy
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Nicely done! Lovely little problem isn't it?
 

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