Talk:Uniform space

It is said: "The topology induced by product of uniform spaces is the Tychonoff product of topologies induced by the uniform spaces." How to prove it? --VictorPorton 21:49, 22 May 2012 (CEST)

The product topology is the weakest topology that makes all projections continuous.

The product uniformity is the weakest uniformity that makes all projections uniformly continuous.

The topology generated by the product uniformity makes all projections continuous, since it makes them uniformly continuous. Thus, it is stronger than the product topology. It remains to check that it is weaker...

--Boris Tsirelson 22:33, 22 May 2012 (CEST)

Or maybe fix a point in the product set (not space yet) and consider the correspondence between entourages and neighborhoods; then between filters of entourages and filters of neighborhoods; then between products of the former and products of the latter.

--Boris Tsirelson 22:56, 22 May 2012 (CEST)

I do not understand the notation (particularly use of indexes) used to define product of uniform spaces. Please edit for clarity. --VictorPorton 23:13, 22 May 2012 (CEST)

The proof of the initially stated assertion follows from two facts;

i) If a family $\def\f#1{\mathfrak{#1}}\{\f U_i \}$ of uniform structures on a set $X$ is given, its upper bound (i.e. the weakest structure being finer than each $\f U_i$) has as its topology the upper bound of the topologies of the $\f U_i$.

ii) if $f: X \to X'$ is a map of uniform spaces such that $X$ has the uniform structure induced by $f$, then the topology of $X$ (as a uniform space) is induced by $f$ from the topology of $X'$.

Apply these to $X = \prod X_i$ and the structures induced by the projections $X \to X_i$. --Ulf Rehmann 00:23, 23 May 2012 (CEST)