Math Genius: Calculus – Michael Spivak, Continuous functions Problem 17-(c)

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This is a question from Spivak’s Calculus(2008 edition) on continuous functions.
The statement of the problem is:

Let $f(x)=0$ when $x$ is irrational and $f(frac{p}{q}) = frac{1}{q}$ if $frac{1}{q}$ is in lowest terms. What is the function defined by $g(x) = lim_{y to x}{f(y)}$

My answer:
Claim: $lim_{x to c}{f(x)}=0 forall c in mathbb{R}$

Case 1: When $c in mathbb{R-Z}$, then let $n in mathbb{Z}$ such that $n < c < n + 1$. Let $epsilon > 0$ be given. Let $N in mathbb{N}$ such that $frac{1}{N} < epsilon$.

Consider the following sets: $D={{2, cdots, N}}$ and $S = {{n + frac{i}{d}:0<i<d forall din D}}$

We claim that the set $S$ is precisely the set of all rational numbers in the interval $(n, n+1)$ for which $|f(x)-0|<epsilon$ may fail to hold, for, if $frac{k}{d}$ be any rational number in the interval $(n, n+1)$, then $frac{k}{d}$ can be written as $n + frac{j}{d}$ where $0<j<d$ and hence it belongs to $S$. Any other rational number in the interval $(n, n+1)$ will have a denominator $>N$ and thus the condition $|f(x)-0|<epsilon$ is automatically satisfied. Also note that if $x$ is irrational, $f(x)=0$ and thus $|f(x)-0|< epsilon$

Now, choose $delta$ such that $delta = min({|x-c|:x in S text{and} x neq c})$. Therefore, if $0<|x-c|<delta$, then $x notin S$ and hence $|f(x) – 0|<epsilon$ is satisfied. Hence, by the definition of the limit, we have $lim_{xto c}{f(x)}=0 forall c in mathbb{R-Z}$

Case 2: When $c in mathbb{Z}$
This case is very similar, we just work on the interval $(c-1, c+1)$ and define $S$ such that $S = {{(c-1) + frac{i}{d}:0<i<2d forall din D}}$. Then we proceed in the same way as before.

Therefore, we have $lim_{xto c}{f(x)}=0 forall c in mathbb{R}$ and thus $g(x)=0 forall x in mathbb{R}$

First of all, I would like to know whether my proof is correct.
If yes, then I have a feeling that this proof is needlessly complicated, so I would like to know if there is any other way to solve this problem.

Thanks for any answers!!

Yes, it is correct.

I think that the following proof, which employs the periodicity of $f$, is a bit simpler, but if you don’t agree I will delete it. As you know, for each $varepsilon>0$, there are only finitely many elements $xin[0,1]$ such that $|f(x)|geqslantvarepsilon$. So, if $cin[0,1]$ and if $varepsilon>0$, then you take$$delta=min{|x-c|mid xin[0,1]text{ and }|f(x)|geqslantvarepsilon}.$$So, $0<|x-c|<deltaimplies|f(x)|<varepsilon$.

This proves that, as far as the restriction of $f$ to $[0,1]$ is concerned, we have $lim_{xto c}f(x)=0$. Since $f$ is periodic with period $1$, the same is true on any interval $[n,n+1]$ (with $ninBbb z$).

Finally, if $ninBbb Z$, since we know that $lim_{xto n^-}f(x)=lim_{xto n^+}f(x)=0$, we know that $lim_{xto c}f(x)=0$.

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