TITLE: Uncountable disjoint union of $\mathbb{R}$
QUESTION [12 upvotes]: I'm doing 1.2 in Lee's Introduction to smooth manifolds: Prove that the disjoint union of uncountably many copies of $\mathbb{R}$ is not second countable.
So first, let $I$ be the set over which we are unioning. Then I believe the disjoint union is just $\mathbb{R}\times I$. Then I believe that sets of the form $\cup_{x\in A}(x,i)$ is open if and only if $A\subset\mathbb{R}$ is open (I know the open sets are defined with the canonical injection though). At first I thought if I let $I=\mathbb{R}$, then $\mathbb{R}\times I=\mathbb{R^2}$, but now I am thinking maybe they just have the same elements, but the topology is different, and this is why $\mathbb{R}\times I$ is not second countable?

REPLY [3 votes]: An attempt to simplify Arthur Fischer's argument:
Each copy of the base space is an open set, and these are pairwise disjoint. Since there uncountably many of them, the countable chain condition
 is not satisfied, therefore the union is not separable or second countable.