What does it mean for a function to be uniformly continuous on a set?
Is it possible for a function \(f:\RR\to \RR\) to be discontinuous precisely on the rationals \(\QQ\)? If so, produce such a function, if not, why?
- Can the set of discontinuities be precisely the irrationals \(\RR\sm\QQ\)?
Find a sequence of continuous functions that does not converge uniformly, but still has a pointwise limit that is continuous.
Solution
A function \(f\) is uniformly continuous on \(E\) if for every \(\eps > 0\) there is \(\delta > 0\) such that \(|f(x) - f(y)| < \eps\) for all \(x, y \in E\) with \(|x - y| < \delta\).
1. Thomae’s function, \(f(p/q) = 1/q\) for \(p/q\) in lowest terms with \(q \geq 1\) and \(f(x) = 0\) for irrational \(x\), is discontinuous exactly on \(\QQ\).
1.1. \(f\) is continuous at every irrational \(x\).
Given \(\eps > 0\), the set \(\theset{p/q : |p/q - x| < 1,\ 1/q \ge \eps}\) is finite and does not contain \(x\). Choose \(\delta < 1\) smaller than the distance from \(x\) to each of its points. Then \(|f(y)| < \eps = \eps + |f(x)|\) for \(|y - x| < \delta\).
1.2. \(f\) is discontinuous at every rational \(p/q\).
\(f(p/q) = 1/q > 0\), while irrationals \(y \to p/q\) have \(f(y) = 0\).
1.3. Q.E.D.
Steps 1 and 2.
2. No function \(f\colon \RR \to \RR\) is discontinuous exactly on \(\RR \setminus \QQ\).
2.1. The set of discontinuities of \(f\) is an \(F_\sigma\) set.
Let \(\omega_f(x) = \inf_{\delta > 0}\sup\theset{|f(y) - f(z)| : y, z \in B(x,\delta)}\) be the oscillation of \(f\) at \(x\). The set \(\theset{\omega_f < 1/n}\) is open, so \(\theset{\omega_f \ge 1/n}\) is closed, and \(f\) is discontinuous at \(x\) exactly when \(\omega_f(x) \geq 1/n\) for some \(n\).
2.2. \(\RR \setminus \QQ\) is not an \(F_\sigma\) set.
If \(\RR\setminus\QQ = \bigcup_n F_n\) with \(F_n\) closed, then \(\QQ = \bigcap_n U_n\) with \(U_n = \RR \setminus F_n\) open and dense, since \(U_n \supseteq \QQ\). The sets \(U_n\) and \(\RR \setminus \theset q\), \(q \in \QQ\), are countably many open dense sets with empty intersection, contradicting the Baire category theorem.
2.3. Q.E.D.
Steps 1 and 2.
3. \(f_n(x) = \dfrac{nx}{1 + n^2 x^2}\) on \([0,1]\) are continuous and converge pointwise, but not uniformly, to the continuous function \(0\).
\(f_n(0) = 0\), and for \(x > 0\), \(0 \le f_n(x) \le \frac{nx}{n^2 x^2} = \frac{1}{nx} \to 0\). But \(f_n(1/n) = \frac12\) for every \(n\), so \(\sup_{[0,1]}|f_n - 0| \ge \frac12\).