# Suppose $M$ is Hausdorff and locally Euclidean. Show $M$ is second countable iff it is paracompact and has countably many connected components.

Struggling with paracompact & countably many components $\implies$ second-countable.

We just need to show each component is second-countable. Choose a component $M_i$. $M_i$ is closed so it is paracompact. We can cover $M_i$ with precompact coordinate balls. (Note: A coordinate ball is an open set $U$ with a homeomorphism $\phi:U\rightarrow B$, where $B$ is an open ball in $\mathbb R^n$. A set is precompact if its closure is compact.) By paracompactness, we get a locally finite cover of precompact coordinate balls.

Now I am stuck. A previous exercise proved that, if $\mathcal X$ is a locally finite cover of precompact open sets, then any $X\in\mathcal X$ intersects only finitely many elements of $\mathcal X$. I am trying to use this fact to reduce the cover I obtained to a countable cover, thus proving the component is $\sigma$-compact (and therefore second countable).

HINT: Let $X$ be a connected space, and let $\mathscr{U}$ be a locally finite open cover of $X$ by connected sets. For $x,y\in X$ let $x\sim y$ if and only if there is a finite $\{U_1,\ldots,U_n\}\subseteq\mathscr{U}$ such that $x\in U_1$, $y\in U_n$, and $U_k\cap U_{k+1}\ne\varnothing$ for $k=1,\ldots,n-1$.
• Show that $\sim$ is an equivalence relation on $X$.
• Show that for each $x\in X$, the $\sim$-equivalence class of $x$ is a clopen set.
• I guess if we prove that ${x}$ is clopen, then it is the whole space $X$ and so $X$ can be covered by finitely many of these $U_i$'s, but I don't really see how. – DpS Sep 18 '17 at 4:20
• Moreover, even if we get a countable cover, we still have to show it is a basis. Is that true because these are co-ordinate balls since $X$ is locally Euclidean? – DpS Sep 18 '17 at 4:21