0

In the Topology by James R Munkres, the definition for basis of a product topology is given as "Collection of all sets of the form $$\prod_{i\epsilon I}^{ }{B_i}$$ where Bi are subset for finitely many i's and Bi equals entire set Xi for remaining indices." Can anyone clarify why this distinction for some i's. Why entire sets than open subsets for remaining i's in defining basis for product topology.

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
Sreeram
  • 9
  • Please see https://math.meta.stackexchange.com/questions/5020/ – Angina Seng Jul 09 '19 at 18:04
  • 2
    Munkres calls the topology with basis the arbitray products of open sets the box topology. – Angina Seng Jul 09 '19 at 18:05
  • The product topology has this crucial property: a map $Y\mapsto\prod_{i\in I}X_i$ is continuous iff each map $Y\to X_i$ got by composing it with the projection to $X_i$ is continuous. – Angina Seng Jul 09 '19 at 18:07
  • The product topology also has the useful property that a sequence of elements $x_n \in \prod_{i\in I} X_i$ converges to $x \in \prod_{i\in I} X_i$ if and only if $x_n \to x$ pointwise, i.e. $\lim_{n\to \infty} x_n(i) = x(i)$ for each $i \in I$. (And if you're familiar with nets, the obvious corresponding condition happens for convergence of nets in general.) In the box topology, on the other hand, this condition is not sufficient to make a convergent sequence. – Daniel Schepler Jul 09 '19 at 18:14
  • 1
    This may help. The product topology is the smallest topology that makes the projection maps continuous. – John Douma Jul 09 '19 at 18:26
  • Is the box topology envisaged as a subspace of product topology? – Sreeram Jul 09 '19 at 19:50
  • No, it's an alternative topology to put on the set $\prod_{i \in I} X_i$. – Henno Brandsma Jul 09 '19 at 21:49

0 Answers0