Continuity Counterexample 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.
 A: Let $f$ be any unbounded function that is never $0$.  Let $g$ be its inverse with respect to multiplication, i.e., $g(x)=1/f(x)$.
Now $fg$ is constantly $1$, but multiplication is not continuous at $(f,g)$.
To prove this, use the $\epsilon$-$\delta$-definition of continuity rather than the sequential definition.

From Asaf's comments I conclude that I was not very clear.
I will fill in some details.  I assume that by "uniform topology" you mean the topology generated by sets of the form $$U_{\varepsilon}(h)=\{e\in\mathbb R^{\mathbb R}:\sup_{x\in\mathbb R}|e(x)-h(x)|<\varepsilon\}$$ where $h:\mathbb R\to\mathbb R$ and $\varepsilon>0.$  
Now let $f$ and $g$ be as above, i.e., $f$ unbounded and never $0$, $g$ its pointwise inverse.
Let $\varepsilon=1$.
$fg$ is constantly $1$.  I will show that there is no $\delta>0$ such that for all $f'\in U_\delta(f)$ and all $g'\in U_\delta(g)$ we have $f'g'\in U_\varepsilon(fg)$.
Namely, let $\delta>0$.  Let $x\in\mathbb R$ be such that $|f(x)|>10/\delta$.
Now $|g(x)|<\delta/10$.  Choose $h\in U_\delta(g)$ such that $h(x)>\delta/2$.
We have $|f(x)h(x)|>10/2=5$.  In particular, $fh\not\in U_\varepsilon(fg)$.
This shows that pointwise multiplication is not continuous on $\mathbb R^{\mathbb R}$ with the uniform topology.
