0
$\begingroup$

Let T be an uncountable set. Let for any finite $S\subset T,S=\{s_1,s_2,...s_n\}$ $\pi_S$ be the map

$\pi_S:R^T \to R^S,\pi_S(w)=\{w(s_1),..w(s_n)\}$. Let $F_S=\{\pi_s^{-1}(B),B\in B^S\}$. Let $F=\cup_{S\subset T,S finite} F_S$. Define $U=\sigma(F)$.

Our professor claimed that this is the smallest $\sigma-field$ that makes all the evaluation maps

$\Phi(t): R^T\to R$

$\Phi(t)(w)=w(t)$ continuous. However,it seems to me that the $\sigma-field$ we would have gotten by taking S to be those subsets of T which have cardinality 1 would be the smallest. What am I understanding wrong?

It also seems to me that the U would make all the evaluation maps $R^T\to R^S$ for finite S measurable.

  • 0
    If $R$ means the reals, and $B^S$ is either just the usual open sets or the Borel $\sigma$-albegra associated then you get the same thing in the end. – 2017-02-04
  • 0
    You mean taking subsets of T with cardinality 1 would suffice? – 2017-02-04
  • 1
    Yes, the sigma field generated by "cardinality 1" sets would include the sigma field generated by finite cardinality sets – 2017-02-04

0 Answers 0