Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let for each $i\in I$, $(X_i,\mathcal{D}_i)$ be a uniform space. How is the product uniform space defined? Does it produce the product (Tychonoff) topology on $\prod_{i\in I}{X_i}$?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

The product uniformity is by definition the smallest $\mathcal D$ uniformity on $\prod_i X_i$, which makes all projections $\pi_i\colon \prod_i X_i \to X_i$ uniformly continuous. So for each entourage $E_i \in \mathcal D_i$, we want to have $(\pi_i\times\pi_i)^{-1}[E_i] \in \mathcal D$. Taking finite intersections of these sets gives us a basis for $\mathcal D$, that is $\mathcal D$ is the filter generated by $$ \mathcal B := \left\{ \bigcap_{i\in J} (\pi_i\times \pi_i)^{-1}[E_i] \biggm| E_i \in \mathcal D_i, J\subseteq I \text{ finite} \right\} $$ The topology induced by $\mathcal D$ is indeed the product of the topologies induced by the $\mathcal D_i$.

share|improve this answer
    
how do you define $\pi_i \times \pi_i$? –  user59671 Feb 7 '13 at 14:08
1  
@CutieKrait: Let $X=\prod_iX_i$. Then $$\pi_i\times\pi_i:X\times X\to X_i\times X_i:\langle x,y\rangle\mapsto\langle\pi_i(x),\pi_i(x)\rangle\;.$$ –  Brian M. Scott Feb 7 '13 at 21:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.