Krdytkyu

I'm currently reading Jeffrey M. Lee Manifolds and Differential Geometry book. I don't understand a part in the proof of Proposition 1.32. (iii).

Proposition 1.32. says:

Let $M$ be a set with a $C^r$ structure given by an atlas
$mathcal{A}$. We have the following:

(i) If for every two distinct points $p, q ∈ M$, we have that either
$p$ and $q$ are respectively in disjoint chart domains $U_alpha$ and
$U_beta$ from the atlas, or they are both in a common chart domain,
then the topology induced by the atlas is Hausdorff.

(ii) If $A$ is countable, or has a countable subatlas, then the
topology induced by the atlas is second countable.

(iii) If the collection of chart domains ${U_alpha}_{alphain A}$
from the atlas $A$ is such that for every fixed $alpha_0in A$ the
set ${alpha in A:U_alphacap U_{alpha_0}neq emptyset}$ is at
most countable, then the topology induced by the atlas is paracompact.
Thus, if this condition holds and if M is connected, then the topology
induced by the atlas is second countable.

Lee starts the proof leaving (i) and (ii) as exercises and then he says:

We prove (iii). Give $M$ the topology induced by the atlas. It is
enough to prove that each connected component has a countable basis.

The bold sentence is what I don't understand. He is probably using this theorem from the appendix:

Proposition B.5. If $X$ is a locally Euclidean Hausdorff space, then
the are following properties equivalent:

(1) X is paracompact.

(2) X is metrizable.

(3) Each connected component of X is second countable.

(4) Each connected component of X is σ-compact.

(5) Each connected component of X is separable.

The problem is that to use B.5 one has to know beforehand that $M$ is hausdorff, which requires verifying (i), but this is not done. So how does one justify the bold sentence?

I know I'm kinda nitpicking here because in practice 99% of the atlases will verify (i) to unsure hausdorffness (otherwise it would be really weird), but I still want to know this precisely.

Part (iii) is in fact false as stated; you have to additionally assume $M$ is Hausdorff (or that the condition of part (i) holds). Here is a counterexample. Let $M=(-infty,0)cup(mathbb{N}times [0,infty))$. For each $ninmathbb{N}$, define $f_n:mathbb{R}to M$ by $f_n(t)=t$ if $t<0$ and $f_n(t)=(n,t)$ if $ngeq 0$. These maps $f_n$ can be taken as an atlas on $M$, which trivially satisfies (iii) since there are only countably many of them. However, I claim that the induced topology on $M$ is not paracompact.

Indeed, let $U_n=f_n(mathbb{R})$; the $U_n$ form an open cover of $M$. Let $mathcal{U}$ be any refinement of this open cover. Note that for each $n$, $mathcal{U}$ contains a neighborhood of $(n,0)$, and these neighborhoods are different for different values of $n$ since they must be contained in $U_n$. But any neighborhood of $(n,0)$ contains the entire interval $(-epsilon,0)$ for sufficiently small $epsilon$. In particular, all of these sets will intersect every neighborhood of $(0,0)in M$ (on the negative side). So there is no neighborhood of $(0,0)$ which intersects only finitely many sets in $mathcal{U}$, so $mathcal{U}$ is not locally finite.