While I am reading "An Introduction to Manifolds" by Loring W. Tu, I come to see the above theorem. I followed the proof but got a question on (ii). We are talking about smooth manifolds.
Why do we need to abandon the compact support when $M$ is a general manifold instead a compact manifold?
In the proof, to make the each support subordinate to a certain open set $U_{\alpha}$, the open cover was just re-indexed according to the support. I couldn't get the insight of such necessity. Can you explain the reason for a simple manifold $]1,2[$, an open interval? Thus, $]1,2[$ are open and closed.
Thanks in advance.
Okay, let's take $M=(1,2)$, an open interval. Consider the cover by two sets: $U_1=(1,1.8)$ and $U_2 = (1.2,2)$. A subordinate partition of unity consists of two functions $\rho_1,\rho_2$. If they were compactly supported, then $\rho_1+\rho_2$ would also be compactly supported. But that means the support of $\rho_1+\rho_2$ is a proper subset of $M$ (since $M$ is not compact), contradicting the property $\rho_1+\rho_2\equiv 1$.
It's easy to make a subordinate partition $\{\rho_1, \rho_2\}$, but $\rho_1$ will need to be $1$ at the left end of $M$, and $\rho_2$ will need to be $1$ at the right end. Not compactly supported.
More generally, any partition of unity by compactly supported functions on a non-compact manifold must involve infinitely many functions. In the above example, the compact supports could be the following intervals: $$\left[1+\frac{2}{k+2},1+\frac{2}{k}\right],\quad \left[2-\frac{2}{k+2},2-\frac{2}{k}\right],\quad k =3,\dots,\infty$$