2
$\begingroup$

In the nlab entry on uniform spaces they speak about an "inherited uniform structure on function spaces". Namely, if $X$ is a set and $(Y,\mathfrak{U})$ is a uniform space, then $Y^X$ can be equipped with the uniform structure generated by: $$ \bigg\{\big\{ (f,g) \,|\, (f(x),g(x))\in U \;\forall x\in X \big\}, U\in \mathfrak{U}\bigg\}\;. $$

Is then this the same as the uniform structure on the $X$-fold Cartesian product? If not, does it hold for $X$ "small" enough, e.g. finite?

Improve this question
$\endgroup$
2
  • $\begingroup$ Really just a comment but I'm not entitled. There is still something wrong with your indexing---$U_i$ is indexed by a mysterious $i$ which is hanging in midair. If you mean, as I suppose, a voisinage $U_x$ which depends on $x$, then the answer to your question is no. To get the product uniformity, you need to assume that $U_x=Y$ for all but finitely many $x$. The analogous construction to the above for topological spaces is called the box topology. $\endgroup$
    – clyde
    Commented Jan 26, 2017 at 16:09
  • $\begingroup$ Oh, the notation may be unclear. Let me correct. $\endgroup$
    – geodude
    Commented Jan 26, 2017 at 16:11

1 Answer 1

Reset to default
3
$\begingroup$

E.g. $C(\mathbb{R} ,[0,1])$ in this function space uniformity is metrisable in this uniformity, using the $\sup$-metric $d(f,g) = \sup \{|f(x) - g(x)|: x \in \mathbb{R}\}$. While the product $\mathbb{R}^{[0,1]}$ is very non-metrisable (not even first countable) in the product topology. These are quite different beasts.. For finite $X$ they amount to the same thing.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .