Problem 1.5

Proof. ( $⟹$) Let $M$ be second-countable. Then $M$ is a topological manifold. By Proposition 1.11, $M$ has countably many connected components. By Theorem 1.1.5, $M$ must be paracompact.

( $⟸$) Let $M$ be paracompact and have countably many connected components. Without loss of generality, we can assume that $M$ is connected (if countably many connected components possess a countable basis, then their union possesses a countable basis as well).

Since $M$ is locally Euclidean, we can cover $M$ by open coordinate domains of $(U_p,{\phi }_p)$ at each $p\in M$. Following the hint, we try to make this cover precompact. Each image ${\phi }_p(U_p)$ possesses a countable basis $B_p$ of precompact balls. The pullback of each basis $B_p$ results in a countable basis $\{{\phi }_p^{-1}(B)\mid B\in B_p\}$ of the corresponding coordinate domain $U_p$. Also notice that the sets ${\phi }_p^{-1}(B)$ are precompact in $U$¹. Therefore, they are also precompact in $M$². In other words, by combining the precompact bases of each coordinate ball $U_p$ of $M$, we obtain a precompact covering of $M$. Since $M$ is paracompact, this precompact covering can be improved to an open, locally finite covering $U$ of $M$; each set equipped with the corresponding coordinate map.
To turn this locally finite covering into a countable basis, we use the following algorithm. Let $V$ be an arbitrary set $U$.

where at each step, $W_n$ is the family of sets in $U$ which intersect the sets in $V_0\cup \dots \cup V_{n-1}$. By the previous exercise, $W_i$ is finite at each step; so is $V_n$. We now argue that the countable family of sets

is a countable basis for $M$. The intersection property is obviously satisfied. We verify that the union of sets in $C$ covers $M$.
Let $y\in M$ be arbitrary. With a slight modification of Proposition 1.11(b), we see that $M$ is path connected. Let $x$ be a point in the initial set $V\in V_0$. Then we can find a path

Since the image $\gamma ([0,1])$ is compact, we can find finitely many sets $U_1,\dots ,U_k\in U$ which cover our path. Equivalently, we can find some numbers $0=a_0<a_1<\dots a_{k-1}<a_k=1$ such that

We now argue that

which would in particular mean that $y\in \gamma ([a_{k-1},a_k])$ is covered by $V_0\cup \dots \cup V_{k-1}$.

• (induction basis) We have $\gamma ([a_0,a_1])=\gamma ([0,a_1])\subseteq U_1=U$ by construction. Thus, $\gamma ([a_0,a_1])$ is covered by $V_0$.
• (induction step) Now suppose that $\gamma ([a_{i-1},a_i])$ is covered by some $V_0\cup \dots \cup V_{i-1}$. In particular, $\gamma (a_i)\in V$ for some $V\in V_0\cup \dots \cup V_i$. Therefore, since $\gamma ([a_i,a_{i+1}])$ intersects the sets in $V_0\cup \dots \cup V_{i-1}$, it must be covered by $V_0\cup \dots \cup V_i$

This closes the induction and completes the proof. □