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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".2011-09-30

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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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    Yes, all subsets of $\mathbb R$ are metrizable. But also $\mathbb R^{\mathbb R}$ with the uniform topology is metrizable, hence you can use $\varepsilon$-$\delta$-arguments here as well. The natural "metric" is not a metric since it has infinite values, but you can just cut of at $1$ or at $10000$ and get a metric that induces the topology (since we only need to know $\varepsilon$-balls for small $\varepsilon$. I believe that is this case proofs using sequences are just more difficult than $\varepsilon$-$\delta$-proofs.2011-10-03