# Partitions of unity $\Leftrightarrow$ Hausdorff + Second-countable (in locally Euclidean space)

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.

• Why do you think $\{M-\{p\},M-\{q\}\}$ is an open cover? – Paul Frost Jan 17 at 13:00
• @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? – rmdmc89 Jan 17 at 13:23
• 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. – Paul Frost Jan 17 at 13:31

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.

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 $$V_q$$.

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 $$V_p$$. $$_\blacksquare$$