I was proving that a manifold has countably many connected components and this question came up. The way I did the original question is to show that components in manifolds are open and then use the fact that second countability implies separable to show that there's only countably many components. The latter argument follows from the fact that, if there's uncountably many components, then there must be some component "left out" but since that component is open, we must get a contradiction.
However, in general, components are not open (in fact, they are always closed). So in the general case, I do not know how to prove that separability implies having a countable number of components or if second countability would imply such (since second countability is a stronger condition). Are those implications true and if not, what are the counter-examples?
Best Answer
Let $\mathbb{R}$ have the usual topology and let $X$ be the set of irrational numbers with the subspace topology inherited from $\mathbb{R}$.
Since $\mathbb{R}$ is second countable, so is $X$.
But the components of $X$ are the singleton subsets of $X$, so $X$ has uncountably many components.