# A problem in real analysis of a topological nature

Let $$f: R \to R$$ be a function such that the closure of its graph contains as a subset the graph of a uniformly continuous function. Does there exist a dense subset $$S$$ of $$R$$ such that the restricted function $$f|S: S \to R$$ is uniformly continuous?

• If $f|_S$ is uniformly continuous then it can be natuarlly extended to a uniformly continuous function on the closure of $S$. Hence $f$ itself has to be uniformly continuous. If you weaken the assumption to "there is a sequence of sets $S_n$ such that their union is dense in $R$ and f restricted to $S_n$ is uniformly continuous on $S_n$" then it's true for measurable functions by Lusin's Theorem. – Martin Kell Dec 26 '18 at 16:27
• @MartinKell: It's true that $f|_S$ will extend to a uniformly continuous function (call it $g$), but $f$ need not agree with $g$ on $S^c$, so we cannot conclude that $f$ is uniformly continuous. – Nate Eldredge Dec 26 '18 at 16:44

Consider the following modification of the Dirichlet "popcorn" function: $$f(x) = \begin{cases} 1/q, & \text{x \in \mathbb{Q}, x=p/q in lowest terms} \\ -1, & x \notin \mathbb{Q},\, x < 0 \\ -2, & x \notin \mathbb{Q}, \, x > 0.\end{cases}$$ Since every real number can be approximated by rationals with arbitrarily large denominator, the closure of the graph of $$f$$ contains the $$x$$-axis, which is the uniformly continuous function $$0$$.
Let $$S \subset \mathbb{R}$$ be dense. If $$f|_S$$ is uniformly continuous, then it extends to a unique uniformly continuous function $$g$$ on all of $$\mathbb{R}$$, and we have $$f=g$$ on $$S$$.
If $$S$$ contains a negative irrational number $$x$$, then $$g(x) = f(x) = -1$$. Let $$y$$ be any positive number in $$S$$. If $$y$$ is rational, we have $$g(y) = f(y) = 0$$. Then by the continuity of $$g$$, there would have to be some $$z \in S$$ with $$f(z) = g(z) \in (-3/4, -1/4)$$ which is impossible. If $$y$$ is irrational, we get a similar contradiction since $$g(y) = f(y) = -2$$. So $$S$$ does not contain any negative irrational. Similarly, $$S$$ does not contain any positive irrational.
So we must have $$S \subset \mathbb{Q}$$. But this is similarly impossible. The rationals in $$S$$ cannot all have the same denominator (in lowest terms), so let $$x_1 = p_1/q_1, x_2 = p_2/q_2 \in S$$, where $$q_1 < q_2$$. Then by the continuity of $$g$$ there must be some $$y \in S$$ with $$f(y) = g(y) \in (\frac{1}{q_1}, \frac{1}{q_1+1})$$, but $$f$$ never takes on any such value.
• Slight simplification of the proof: $S$ cannot contain a single rational number, since $f$ is discontinuous at every rational, hence so is $f|_S$ for any dense $S$. Thus $S\subseteq\mathbb R\setminus\mathbb Q$, but then we have problems around $0$. – Wojowu Dec 26 '18 at 17:08
• @Wojowu: It's not quite that simple, I think. For instance, the function $f=1_\mathbb{Q}$ is discontinuous at every rational (and every irrational), but its restriction to $S=\mathbb{Q}$ is uniformly continuous. So the fact that $f$ is discontinuous at the rationals doesn't immediately rule out the possibility for $S$ to contain rationals. We would have to work a little harder. I guess the key is that $f$ has a "jump" discontinuity at every rational. – Nate Eldredge Dec 26 '18 at 17:16
• You are right, I was a little too quick there. We can either use jump discontinuity as you mention, or we can observe $(q,f(q))$ is an isolated point of the graph. – Wojowu Dec 26 '18 at 17:24