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Let $X$ be a (connected) topological space with a $C^\infty$ atlas. It is a known theorem that if $X$ is second-countable and Hausdorff, then it admits partitions of unity. I'm trying to prove the "reverse" theorem:

Let $X$ be a (connected) topological space with a $C^\infty$ atlas. If $X$ admits partitions of unity, then $X$ is second-countable and Hausdorff.

I was able to prove the Hausdorff condition by taking a partition of unity $\{\rho_p,\rho_q\}$ subordinate to $\{M-\{p\},M-\{q\}\}$ and taking neighbourhoods $U,V$ of $p,q$ small enough so that the values of $\rho_p,\rho_q$ in $U$ conflict with the ones in $V$ so that $U\cap V=\emptyset$.

Now I'm stuck with second-countability. Here is my attempt:

For each $p\in M$ take a chart $\varphi_p:U_p\to\mathbb{R}^n$. For a partition of unity $\{\rho_p\}$ subordinate to $\{U_p\}$, let: $$V_p:=\rho_p^{-1}(0,\infty)\subset U_p$$ By definition of partition of unity, $\{V_p\}$ is a locally finite refinement of $\{U_p\}$. Now since $U_p$ is homeomorphic to $\mathbb{R}^n$, $U_p$ is second countable and therefore $V_p$ is second-countable.

I think the natural thing to do is to find countably many points $\{p_n\}_{n\in\mathbb{N}}$ so that $\{V_{p_n}\}_{n\in\mathbb{N}}$ is a cover for $X$, but I can't see how to do that.

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    $\begingroup$ Why do you think $\{M-\{p\},M-\{q\}\}$ is an open cover? $\endgroup$ – Paul Frost Jan 17 at 13:00
  • $\begingroup$ @PaulFrost $M\setminus\{p\}$ is open because $\{p\}$ is closed. Indeed, take $x\in M\setminus\{p\}$ and an open set $V\subset M$ homeomorphic to $\mathbb{R}^n$, containing $x$. If $p\notin V$, we are done. If $p\in V$, then we are in the Euclidean space, so we may take a smaller open $W\subset V$ containing $x$ with $p\notin W$. Right? $\endgroup$ – rmdmc89 Jan 17 at 13:23
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    $\begingroup$ Ah, you are right! You only need to know that $X$ is a $T_1$-space. I thought about the line with two origins which is a non-Hausdorff manifold. $\endgroup$ – Paul Frost Jan 17 at 13:31
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The following is a well-known theorem:

Let $X$ be a $T_1$-space. Then $X$ is paracompact if and only every open cover of $X$ has a subordinated partition if unity.

Here, a paracompact space is a Hausdorff space in which every open cover has a locally finite open refinemmnet.

This shows that your question is answered in the affirmative by the equivalence between paracompactness and second countablity in a locally Euclidean and $T_2$ space.

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This pdf mentioned in Paul Frost's link helped with my idea of proving the cover $\mathcal{C}:=\{V_p\}_p$ is enumerable.

First, we assume each $\overline{V_p}$ is compact (this is possible since $U_p$ is a coordinate neighbourhood).

Now fix $p_0\in X$. Since $X$ is connected, for any $q\in X$ there are points $p_0=q_0,...,q_k=q$ such that $V_{q_i}\cap V_{q_{i+1}}\neq\emptyset$. Let's call $\{V_{q_0},...,V_{q_k}\}$ a "bridge of lenght $k$" between $V_{p_0}$ and $V_q$.

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The minimum length of that bridge will be called $\ell(V_q)$, and this defines a function $\ell:\mathcal{C}\to \mathbb{Z}_{\geq 0}$. We will show that $\ell^{-1}(n)$ is finite $\forall n\geq 0$, which proves $\mathcal{C}$ is enumerable. Of course $\ell^{-1}(0)=\{V_{p_0}\}$ is finite. Assuming $\ell^{-1}(0),...,\ell^{-1}(n)$ are finite, let's prove $\ell^{-1}(n+1)$ is finite. Consider the set: $$K:=\bigcup_{\ell(V_q)\leq n}\overline{V_q}$$

$K$ is compact, since there are finitely many such $V_q$'s by the induction hypothesis. Since $\mathcal{C}$ is locally finite, each $p\in K$ has a neighbourhood $W_p$ which intersectes finitely many $V_q$'s. Taking a finite subcover $W_{p_1},...,W_{p_k}$ of $K$, we conclude that $K$ also intersects finitely many $V_q$'s. Finally, take $V_p\in \ell^{-1}(n+1)$ and a bridge $\{V_{p_0}=V_{q_0},...,V_{q_{n+1}}=V_p\}$. Notice that $\ell(V_{q_n})\leq n$, so $V_{q_n}\subset K$ by construction. Since $V_{q_n}\cap V_p\neq \emptyset$, of course $V_p\cap K\neq \emptyset$, so there are finitely many choices for $V_p$. $_\blacksquare$

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