# The limit of Thomae's function at any a such that 0 < a < 1

I'm currently reading Spivak's Calculus and I'm having trouble understanding the example with the limit of Thomae's function. The function \begin{align} f(x) = \begin{cases} 0 & \text{x irrational, 0 < x < 1}\\ \frac{1}{q} & \text{x = p/q in lowest terms, 0 < x < 1} \end{cases} \end{align} approaches $$0$$ for any $$a \in (0, 1)$$. Looking at this picture, I don't get how that could be true. For instance, if $$a = \frac{1}{2}$$, then it seems to me that for $$\epsilon = \frac{1}{10}$$ there's no $$\delta$$ such that $$0 < \lvert x - a \rvert < \delta \implies \lvert f(x) - 0\rvert < \epsilon$$. What am I missing here?

• There is no a in your definition of $f(x)$.
– Paul
Mar 4 at 15:20
• While $f(1/2) = 1/2$, we can make $f(x)$ as small as we like by taking $x$ sufficiently close to $1/2$ but not equal to $1/2$. Is that what you're asking? Mar 4 at 15:20
• @Paul It is not clear to me what your point is. I think there should not be any $a$ in the definition of $f(x)$. OP is saying that that $\lim_{x\to a} f(x) = 0$ for any $a$ in $(0,1)$.
– MJD
Mar 4 at 15:53

Consider $$\delta = \frac1{20}$$. Other than $$\frac12$$ itself, each rational number in the interval $$(\frac9{20}, \frac{11}{20})$$ has a denominator at least $$11$$, so the value of Thomae's function in this interval is at most $$\frac1{11}$$.

(If you disagree, please say what rational number in the interval has a denominator smaller than $$11$$.)

In fact we can take $$\delta$$ as large as $$\frac1{18}$$ because this is the value that is required to exclude $$\frac49$$ and $$\frac59$$ from the interval.

Note that for any given $$n$$, we can simply list the fractions in (say) $$(0, 1)$$ whose denominators are less than $$n$$. This is a finite set, so there must be an interval around $$\frac12$$ that will exclude them all.

• So if we let $n \in \mathbb{N}$ such that $\frac{1}{n} \leq \epsilon$, there's a finite number of $\frac{p}{q} \in (0, 1)$ such that $q < n$ (let $A$ be that finite set). What I also don't get from Spivak's reasoning is why there's only one $\frac{p}{q}$ closest to $a$, i.e. that there's a fixed $\delta$ such that $\delta = \min{\{\lvert \frac{p}{q} - a \rvert : \frac{p}{q} \in A, \frac{p}{q} \neq a }\}$. What am I missing here? Mar 5 at 17:53
• If $A$ is a finite set, then there is an element of $A$ that is closest to $\frac12$, or possibly two elements that are tied for closest. To find the closest element, order the elements of $A$ by their distance to $\frac12$ and look for the one (possibly two) with the smallest distance. These must exist because a finite set always has a minimum.
– MJD
Mar 5 at 20:15
• So in this case, it's indeed possible for two $\frac{p}{q}$ (not necessarily one element in $A$ as Spivak claims) to be equally close to $a=\frac{1}{2}$? Mar 5 at 21:49
• Sure, but it doesn't matter because any $\delta$ that excludes one will necessarily exclude the other. As long as $\delta$ is less than the distance to the closest $\frac pq$ with a small denominator, all the rational numbers in $(\frac12-\delta, \frac12+\delta)$ will have large denominators.
– MJD
Mar 5 at 22:03