F Continuous Rational if and Only if X Irrational

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

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

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).

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Comments

• Define the real-valued function $f$ on $\mathbb{R}$ by setting $f(x)=x$ if $x$ is irrational, and $f(x)=p\sin\frac1q$ if $x=\frac{p}q$ is written in lowest terms. At what points is $f$ continuous?

  I'm pretty sure $f$ is continuous at no point. For any point $x$, we can find $\frac{p}q$ that's really close to $x$, but such that the value $p\sin\frac1q$ is nowhere near $x$. For example, if $x=0$, we have $f(x)=0$, and we can choose $y$ as something like $\frac{34567}{10000000000}$ so that its value is really close to $x$, but $\frac{p}{q}$ is unlikely to be close to $x$. But since the value of $\sin\frac{1}{q}$ fluctuates, I'm unable to turn this into a rigorous proof.

  • @Adriano: No, I don't think $f$ is continuous at rational points except 0, since say $x = \dfrac{p}{q}$ in lowest term, then we can always construct an irrational sequence $\{ x_n \} \to \dfrac{p}{q}$, so $\lim f(x_n) = \dfrac{p}{q} \neq f \left( \dfrac{p}{q} \right) = p \sin \left( \dfrac{1}{p} \right)$

  • @AndréNicolas If the standard example uses $p/q$, the standard example is the function $f(x)=x$ for every $x$ (very much continuous everywhere). Would you be mixing this question with $f(p/q)=p/q$, $f(x)=0$ if $x$ irrational?

  • This function is continuous at $0$. I don't understand your argument purporting to show that it's not. $\qquad$

• Thanks, Henry. I still think the irrational case is not so intuitive for me. Perhaps someone could make it rigorous?

• something rigorous: for every $x$ irrational and $n$ positive integer, let $Q_n^x$ denote the set of all rationals $p/q$ in $(x-1,x+1)$ such that $q\leqslant n$. Then $Q_n^x$ is finite and $Q_n^x$ does not contain $x$ hence $x$ is at positive distance, say $\delta_n$, of $Q_n^x$. For every rational $p/q$ in $(x-\delta_n,x+\delta_n)$, $f(p/q)=f(x)+(p/q-x)+(p/q)(q\sin(1/q)-1)$. Can you finish?

• @Did What do you mean by $x$ is at positive distance of $Q_n^x$? What distance?

• The usual one: $\mathrm{dist}(x,Q)=\inf\{|x-t|\,;\,t\in Q\}$.

• something intuitive: for $x$ irrational, any sequence of rational numbers $p_n/q_n$ converging to $x$, should satisfy $p_n\rightarrow \infty$, and $q_n\rightarrow \infty$.

• Then $p_n\sin(1/q_n)= (p_n/q_n)\sin(1/q_n)/(1/q_n)\rightarrow x$.

• @Henry Why is $|R(x)|\leq 1\frac{(1/q)^2}{2!}$?

• Ah never mind, that bound is loose.

Recents

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Source: https://9to5science.com/f-x-x-if-x-irrational-and-f-x-p-sin-frac1q-if-x-rational

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