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?
Best Answer
It will be easier if we talk about $(0,1)$ which is homeomorphic to $\mathbb R$, but its diameter is $1$.
First note that the disjoint union is metrizable, by setting the distance between any two points coming from two different copies to be $2$.
For metric spaces second countability implies separability. However a countable subset of the disjoint union cannot meet all the copies, only countably many of them.
Therefore the disjoint union is not separable and thus not second countable.