Let $X = \mathbb{R}$ and $X^\mathbb{R}$ denote the set of functions $\mathbb{R} \to \mathbb{R}$. Let $B \subseteq X^\mathbb{R}$ denote the subset of bounded functions $\mathbb{R}\to \mathbb{R}$, i.e. the set of those functions $f$ for which there exists some $M_f$ such that $f(x) < M_f$ for all $x$. Show that $(f,g) \mapsto fg$ does not give a continuous function $X^\mathbb{R} \times X^\mathbb{R}\to X^\mathbb{R}$ when $X^\mathbb{R}$ is equipped with the uniform topology but that the restriction $B \times B \to B$ is continuous.
I'm trying the construct a counterexample using the sequence limit definition of continuity. I understand that for the first part, I am looking for an unbounded function where the preimage of $U$ open in $X^\mathbb{R}$ with the uniform topology, meaning it's not within a distance of $1$, is not open. But I can't think of an explicit example.