1
$\begingroup$

Maybe I've just got a wrong idea of what's going on here, or maybe I just have a definition wrong, but this theorem seems a bit difficult for me to understand, and I think I just need an example. For instance, consider the open set (0,1). This theorem seems like it's saying I can write (0,1)=(0,1/2)U(1/2+epsilon,1) or something? I'm just a bit confused as to how one would concretely form an open set out of disjoint open intervals I guess.

$\endgroup$
2
  • $\begingroup$ No, $(0,1)$ is exactly itself. $\endgroup$
    – Kenny Lau
    Commented Sep 14, 2017 at 15:18
  • $\begingroup$ Related $\endgroup$
    – Kenny Lau
    Commented Sep 14, 2017 at 15:18

1 Answer 1

1
$\begingroup$

A set is countable if it has finitely many elements or if its elements can be put into a one-to-one correspondence with the counting numbers. So the interval $(0,1)$ indeed consists of a countable union of disjoint open sets, namely one such set. The theorem is based upon the fundamental idea that the reals have a countable dense subset, namely the rationals. Then the set of $\epsilon$-balls that are centered on the rational numbers and have rational radii is a countable collection of sets. Any open set in the reals is a union of some of these open sets.

$\endgroup$
2
  • $\begingroup$ +1. It is worth noting that some texts do (sadly) use "countable" to mean "countable and infinite," so the OP's confusion is understandable. $\endgroup$ Commented Sep 14, 2017 at 19:06
  • $\begingroup$ Indeed, you're correct. Even an excellent mathematician may occasionally lapse into ambiguity, but mathematics itself is so beautifully clear. $\endgroup$ Commented Sep 15, 2017 at 1:26

Not the answer you're looking for? Browse other questions tagged .