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.

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As a general tip, if $X=\mathbb R$, just use $\mathbb R$. No need for additional notations; the word "belongs" is ambiguous: from the context it is clear you mean subset, but it could mean "a member of". –  Asaf Karagila Sep 30 '11 at 4:40

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.

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Stefan, are you sure you are reading this correctly? I'd think that the function of pointwise multiplication $(f,g)\mapsto fg$ is not continuous; furthermore $\mathbb R^\mathbb R$ is not metrizable so $\epsilon-\delta$ is meaningless there. –  Asaf Karagila Sep 30 '11 at 8:47
Asaf, the topology on $\mathbb R^{\mathbb R}$ that we are talking about is the uniform topology, so I assume that means generated by $\varepsilon$-balls in the sup-"metric". –  Stefan Geschke Sep 30 '11 at 13:03
Moreover, I am claiming that pointwise multiplication is not continuous, just like you say. –  Stefan Geschke Sep 30 '11 at 13:22
Hi, thanks, that's really helpful. –  jamie coulter Sep 30 '11 at 13:38
Stefan, actually the answer was quite clear. It was the question I found unclear, making me uncertain that you answered the actual question. However, if the OP says that you've been helpful - who am I to judge? –  Asaf Karagila Sep 30 '11 at 15:08