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Does anyone have a reference for the fact that the uniformity of pointwise convergence on real functions of $[0,1]$ (that is, the uniformity generated by the sets $\lbrace (f,g) : |f(x) - g(x)| < \epsilon \rbrace$ for every $x\in[0,1]$, $\epsilon>0$) does not have a countable base?

It does not appear that either Bourbaki nor Kelley say this explicitly, although they both use this uniformity repeatedly to illustrate weird behaviour. I guess it is assumed not to have a countable base since it is an uncountable product of euclidean uniformities.

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If it is not in bourbaki, I don't have a reference in mind, but here is a proof :

Let $(B_n)$ for $n \in \mathbb{N}$ be any countable family of entourage.

Then because each $B_n$ is an entourage one has that (for each $n$) there exists a finite set of point $A_n$ of $[0,1]$ and an $\epsilon_n$ such that if for all $x \in A_n$ $|f(x)-g(x)| < \epsilon_n$ then $(f,g)$ is in $B_n$.

The union $A$ of all the set $A_n$ is countable. let $f$ be non zero function which vanish at every point of $A$. then $(0,f)$ is in $B_n$ for all $n$. So $B_n$ cannot be a basis of the uniformity.

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  • $\begingroup$ More generaly a product uniformity has a countable basis if and only if each term of the product has a countable basis and there is at most a countable number of term of the product which are non trivial. (where a uniforme strucure is trivial if it has no entourage other than $X \times X$ ). $\endgroup$ Commented Jun 21, 2013 at 18:48
  • $\begingroup$ Maybe I'm being dense, but why must the $A_n$'s be finite (or countable, even)? $\endgroup$ Commented Jun 21, 2013 at 20:22
  • $\begingroup$ I got it -- I was being dense, thanks! $\endgroup$ Commented Jun 21, 2013 at 20:28

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