Proving a function is discontinuous on Q and continuous on R\Q

  • #1
michael.wes
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I have cast my post in LaTeX here:

http://www.michaelwesolowski.com/asdjhf.pdf [Broken]

Any and all help is appreciated!
 
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  • #2
Dick
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Don't use the hint directly if you find it's confusing to use it. If f(x)>=1/M then that means x=n/m where m<=M, right? For x in (0,1) that's a finite set of n/m for a given M. So min(|alpha-n/m|) is greater than 0 if alpha is irrational since the set is finite. Use that to construct an easy epsilon delta argument.
 
  • #3
michael.wes
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I'm still spinning my wheels for the first part.. but I will ask my next substantive question now.

We need to show that [tex]f(x_n)\rightarrow f(\alpha) = 0[/tex] for all sequences [tex]x_n, \ where \ x_n \rightarrow\alpha \ as \ n\rightarrow\infty[/tex] in order to prove continuity on R\Q. I didn't ask this in the original post: why is it sufficient to consider all rational sequences only? Irrational sequences obviously don't need treatment, as their image is zero and thus they converge to zero immediately.

But aren't there sequences that have not all rational and not all irrational for all natural numbers n? I can't think of any that would converge to an arbitrary irrational number, but consider [tex]x_n=1/(\sqrt{2})^n[/tex]. With a little work we can show this converges to zero (rational), but alternates between rational and irrational numbers.

Another example: we know that the sum of a rational and an irrational is irrational. Consider the sequence [tex]x_n=\frac{1+((-1)^n+(-1)^{2n})\sqrt{2}}{n}[/tex]. This sequence also converges to zero, but alternates between rational and irrational numbers.

My point is: is there an error of omission? I realize there are pure irrational and pure rational subsequences to each of these examples, but they still make me think about whether we only need consider rational sequences overall.
 
  • #4
Dick
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If you are going to do it by sequences, you need to consider ALL sequences. As I said before, there are are only a finite number of points of the form n/m in (0,1) where m<=M for some fixed M. For any other point x, f(x)<1/M be it rational or irrational. If you take delta to be the minimum of |alpha-n/m| then any point within a distance of delta from alpha must have a value of f less than 1/M. Does that help?
 

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