1
$\begingroup$

The definition of a $\sigma$-field $\mathscr{F}$ on a set $X$ (or $\sigma$-ring) requires $\mathscr{F}$ to be a non-empty subset of $\mathscr{P}(X)$. Why is this convention taken? What is the issue with allowing for empty measurable spaces?

Similarly, why is a topology defined to be a non-empty collection of $\mathscr{P}(X)$? Whats the issue with allowing an "empty" topological space?

$\endgroup$
  • 2
    $\begingroup$ For the second: a topology must have empty set and $X$ as open sets (for good reasons), so the fact that it's nonempty follows from that and is not an additional requirement. $\endgroup$ – user147263 Sep 3 '14 at 21:24
  • 3
    $\begingroup$ $\langle \emptyset, \emptyset \rangle$ is not a topology. $\langle \emptyset, \{\emptyset\}\rangle$ is a topological space. $\endgroup$ – William Sep 3 '14 at 21:25
  • $\begingroup$ I know that, that is the first part of the statement of the question, the second part is what are these "good reasons"? $\endgroup$ – AIM_BLB Sep 3 '14 at 21:26
  • 2
    $\begingroup$ If $\langle \emptyset,\emptyset\rangle$ were a topology, there would be no continuous functions from it. On the other hand the empty function $e:\emptyset \to X$ is continuous from $\langle \emptyset,\{\emptyset\}\rangle$ to any topological space $X$. So we have an initial object in the category. $\endgroup$ – Jyrki Lahtonen Sep 3 '14 at 21:34
  • 1
    $\begingroup$ Similarly for the first. Whether or not $X = \varnothing$, the sigma-algebra $\mathcal F$ on it is not empty, since is has (at least) $\varnothing \in \mathcal F$. $\endgroup$ – GEdgar Sep 3 '14 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.