Thomae's function: Difference between revisions

Revision as of 12:39, 17 May 2023

Revision as of 13:46, 9 June 2023

Line 51:

{{Collapse top|title=Proof of continuity at irrational arguments|width=80%}}

Since <math>f</math> is periodic with period <math>1</math> and <math>0 \in \Q,</math> it suffices to check all irrational points in <math>I=(0,\;1).\;</math> Assume now <math>\varepsilon > 0,\; i \in \N</math> and <math>x_0 \in I \setminus \Q.</math> According to the [[Archimedean property]] of the reals, there exists <math>r \in \N</math> with <math>1/r < \varepsilon ,</math> and there exist <math>\; k_i \in \N,</math> such that

for <math>i = 1, \ldots, r</math> we have <math>0 < \frac{k_i}{i} < x_0 < \frac{k_i +1}{i}.</math>