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?
 A: 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.
