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 $S = {\{n + \frac{i}{d}:0<i<d \ \forall \ d\in D\}}$
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 $\lim_{x\to c}{f(x)}=0\ \forall c \in \mathbb{R-Z}$
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 $S = {\{(c-1) + \frac{i}{d}:0<i<2d \ \forall \ d\in D\}}$. Then we proceed in the same way as before.
Therefore, we have $\lim_{x\to c}{f(x)}=0 \ \forall \ c \in \mathbb{R}$ 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!!
Best Answer
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 $x\in[0,1]$ such that $|f(x)|\geqslant\varepsilon$. So, if $c\in[0,1]$ and if $\varepsilon>0$, then you take$$\delta=\min\{|x-c|\mid x\in[0,1]\text{ and }|f(x)|\geqslant\varepsilon\}.$$So, $0<|x-c|<\delta\implies|f(x)|<\varepsilon$.
This proves that, as far as the restriction of $f$ to $[0,1]$ is concerned, we have $\lim_{x\to c}f(x)=0$. Since $f$ is periodic with period $1$, the same is true on any interval $[n,n+1]$ (with $n\in\Bbb z$).
Finally, if $n\in\Bbb Z$, since we know that $\lim_{x\to n^-}f(x)=\lim_{x\to n^+}f(x)=0$, we know that $\lim_{x\to c}f(x)=0$.