$f(x)=x$ if $x$ irrational and $f(x)=p\sin\frac1q$ if $x$ rational

Solution 1:

Let $x \in \mathbb{Q}^+$. One can show that $f(x) < x$. Let $\epsilon = x - f(x)$. For any $\delta > 0$, we can choose an irrational number $y$ such that $x < y < x + \delta$. $f(y) - f(x) = y - f(x) > x - f(x) = \epsilon$. Thus $f$ is not continuous at $x$.

For $x \in \mathbb{Q}^-$, note that $f(-x) = -f(x)$, and negating a function doesn't affect continuity.

For $x = 0$, we want to find $\delta$ such that, if $|y - 0| < \delta$, then $|f(y) - f(0)| < \epsilon$. Let $\delta = \epsilon$. Since $|f(y)| \le |y|$, this is a good choice of $\delta$. Our argument for rationals does not work, because $f(x) \not< x$.

$f$ should be continuous at irrationals, but I can't nail down something rigorous. Informally: a small $\delta$ means that all $y = p/q$ that are less than $\delta$ away from $x$ will have a large $q$. This makes $f(y)$ very close to $y$, and so we only need to pick a $\delta$ slightly smaller than our $\epsilon$. I think one would have to bound $f(y) - y$ somewhere to show that. (Taylor series?)

So the argument would go something like this: choose an arbitrary $\epsilon$, determine a large enough $q$, determine a $\delta$ such that $(x - \delta, x + \delta)$ doesn't contain any smaller $q$s. The "large enough $q$" I have trouble with.

EDIT: Think I have something rigorous.

First, let's find a bound on $|f(y) - y|$. By Taylor series,

$$\sin{\frac{1}{q}} = \frac{1}{q} - \frac{1}{3! \cdot q^3} + \frac{1}{5! \cdot q^5} - \frac{1}{7! \cdot q^7} + \cdots$$

We can truncate this and put a bound on the error:

$$\sin{\frac{1}{q}} = \frac{1}{q} + R(x), \quad |R(x)| \le 1 \frac{(1/q)^2}{2!} = \frac{1}{2q^2}$$

Thus, $|\sin{\frac{1}{q}} - \frac{1}{q}| \le \frac{1}{2q^2} \implies |p\sin{\frac{1}{q}} - \frac{p}{q}| \le \frac{p}{2q^2} \implies |f(\frac{p}{q}) - \frac{p}{q}| \le \frac{p}{2q^2}$

Choose an arbitrary $\epsilon > 0$. We want to find a "large enough" $q$. If we restrict $\delta < 1$, then we know that $\frac{p}{q} < x + 1$. Then $|f(\frac{p}{q}) - \frac{p}{q}| \le \frac{p}{2q^2} < \frac{x + 1}{2q}$. If $q > \frac{x + 1}{2\epsilon}$, then $|f(\frac{p}{q}) - \frac{p}{q}| < \epsilon$. So we take the distance from $x$ to the nearest rational with denominator $\lceil \frac{x + 1}{2\epsilon} \rceil$, and let that be $\delta$ (or $1$, whichever is smaller).