suppose $x_n = \frac{p_n}{q_n} \in Q_n$ where $\frac{p_n}{q_n}$ is in reduced form and $x_n \to a, a \notin \mathbb{Q}$. Prove $q_n \to \infty$ and use the result to deduce the Thomae function is continuous at every irrational point.

$\mathbb{Q}$ is not closed and not open and neither is the set of irrational numbers. And under the assumptions given there's nothing to really prove that the Thomae function is discontinuous at rational points.

$$ \left|\frac{p_n}{q_n} - a \right| = \left| \frac{p_n - a \cdot q_n}{q_n} \right| < \epsilon, \forall \epsilon > 0 $$

I'm not sure what I can use to show $q_n \to \infty$

  • $\begingroup$ If $q_n$ has a finite limit, then so does $p_n$, and both of these limits would be integers. (Assuming that $p_n$ and $q_n$ are integers) $\endgroup$ – Dylan Nov 8 '15 at 3:55

For all $k \in \mathbb{N}$ define $F_k = {1 \over k} \mathbb{Z} = \{\cdots, -{1 \over q}, {0 \over q}, {1 \over q}, \cdots \}$. Note that $d_k=\min_{r \in F_k} |a-r| >0$ for all $n$ (otherwise $a$ must be rational).

Suppose $|q_n|$ is bounded by $N$. This means that $x_n \in F_1 \cup F_2 \cup \cdots \cup F_N$ for all $n$. We have $\delta = \min_{k \in \{1,...,N\}} d_k > 0$, and so $|x_n-a| \ge \delta$ for all $n$, which contradicts $x_n \to a$.

Proof of continuity of $f$ follows from this:

You have $f(a) = 0$. One way to show continuity is to show that for all sequences $y_n \to a$, we have $f(y_n) \to 0$.

Suppose $y_n \to a$. If $y_n$ is irrational, we have $f(y_n) = 0$. If there are only a finite number of rational $y_n$ then we see that $f(y_n) \to 0$.

Suppose there are an infinite number of rational $y_n$ and let $y_{n_k}$ be the subsequence of rationals. and $y_{n_k} = {p_{n_k} \over q_{n_k}}$ with $p_{n_k},q_{n_k}$ coprime, then we must have $q_{n_k} \to \infty$, and so $f(y_{n_k})={1 \over q_{n_k}} \to 0$.

Combining these results we have $f(y_n) \to 0$.

  • $\begingroup$ this makes sense for the most part. My question is though, what is $k$? It seems like $k$ is greatest value such the $ k > |q_n| \implies \frac{1}{k} < \frac{1}{|q_n|}$.I'm also a bit confused because $\delta-epsilon$ style proofs are of the form : if $|x - a| < \delta \implies |f(x) - f(a)| < \epsilon$ which being an implication means that if the antecedent is false but the consequent is true then theres no contradiction. Maybe, I am understanding the proof incorrectly. $\endgroup$ – oliverjones Nov 8 '15 at 18:08
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    $\begingroup$ I have (hopefully) clarified the definition of $F_k$. I didn't use an $\epsilon$-$\delta$ style proof, sequences seem more straightforward here. $\endgroup$ – copper.hat Nov 8 '15 at 18:42

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