Using sequences to test topological properties for topological manifolds

Question

“As a consequence of being second countable,1) topological manifolds can have at most countably many connected components.2) Moreover, every point possesses a countable neighbourhood basis, so that one can use sequences to test topological properties like continuity of mappings or closedness of subsets.”

I think 1) follows from the fact that in a topological manifold every component is open since every topological manifold is locally path connected. Am I correct or do the authors probably have a more simple argument? But I could not understand what the authors mean by 2). In particular, what does countable in countable neighborhood basis have to do with testing topological properties? Only Hausdorff condition must suffice

Answer 1

By "testing topological properties" they mean statements like the following:

• If $A\subseteq X$, then $A$ is closed iff for any sequence $(x_n)$ in $A$ and any $x\in X$, if $x_n\to x$ then $x\in A$.
• If $f:X\to Y$, then $f$ is continuous iff for any sequence $(x_n)$ in $X$ and any $x\in X$, if $x_n\to x$ then $f(x_n)\to f(x)$.

These statements are not true for arbitrary topological spaces (the "only if" parts are true but the "if" parts may not be), but they are true for second-countable spaces. So in this way, the second-countability of manifolds allows you to use sequences to test topological properties in manifolds.

(Actually, though, this assertion is rather misleading, because these properties only require spaces to be first-countable, not second-countable. First-countability is a local property, so any space which is locally homeomorphic to $\mathbb{R}^n$ is first-countable. So, even if manifolds were not required to be second-countable, you could still use sequences in these ways!)

Answer 2

$\mathbb{R}$ with the discrete topology is has path connected components that are clopen. These connected components are each of its singleton subsets. There are uncountably many of them.

It is the second countability what gives you that there are at most countably many connected components. Since it is second countable there must be a countable basis of the topology. At you said the connected components are open, and therefore each one contains an element of the countable basis. Those elements are different for different components, since the components are disjoint. Therefore, the number of components must be at most countable.