# Show that $\{\theta_i\}$ is a differentiable partition of unity, then $\sum i\theta_i$ is a proper map

I have the following problem:

Let M be a (non-compact) manifold and $$\{\theta_i\}$$ a partition of unity subordinate to a countable open cover of relatively compact sets. Then the map $$f=\sum_i i\theta_i$$ is a proper map

So far I only know that using $$f$$ is continous, I can show $$f$$ is proper if I can somehow show that $$f^{-1}([0,N])$$ is bounded, $$\forall N\in\mathbb{N}$$(This makes sense because M is considered as an embedded manifold in some $$\mathbb{R}^p$$).

Any help is appreciated, I'm quite stuck. Thanks everyone!

• Isn't $f^{-1}([0,N])$ contained in the union of the first $N$ relatively compact sets? – Brian Moehring Oct 7 '19 at 19:29
• Not necessarily. Because the value of some $\theta_N$ with big $N$ could positive, but very small – miraunpajaro Oct 7 '19 at 19:31
• Yet outside the first $N$ sets in the cover, $$f = \sum_{i=N+1}^\infty i\theta_i \geq (N+1)\sum_{i=N+1}^\infty \theta_i = N+1$$ – Brian Moehring Oct 7 '19 at 19:42
• Yeah, that seems allright to me. Thank you very much – miraunpajaro Oct 7 '19 at 19:49

Let $$\{V_i\}$$ denote the given countable open cover by relatively compact sets to which the partition of unity $$\{\theta_i\}$$ is subordinate. Then we claim $$f^{-1}([0,N]) \subseteq \bigcup_{i=1}^N V_i$$ for each $$N \in \mathbb{N}.$$
To see this, first note that if $$x\in M,$$ $$f(x) = \sum_{i=1}^\infty i\theta_i(x) \geq \sum_{i=N+1}^\infty i\theta_i(x) \geq \sum_{i=N+1}^\infty (N+1)\theta_i(x) = (N+1)\left(1 - \sum_{i=1}^N \theta_i(x)\right)$$
If $$x \in M \setminus \bigcup_{i=1}^NV_i,$$ then $$\theta_i(x) = 0$$ for each $$i=1,\ldots,N$$ and therefore $$f(x) \geq (N+1)\left(1 - \sum_{i=1}^N\theta_i(x)\right) = N+1.$$
The claim follows. Since further $$f^{-1}([0,N]) \subseteq \bigcup_{i=1}^N \overline{V_i}$$ shows that $$f^{-1}([0,N])$$ is contained in a compact set, it is bounded.