Partitions of Unity (proof)
So, as we’ve seen we can find a countable atlas ; we use for the coordinate maps since we’ll want the free. We’ve also seen that we have a smooth bump function between the two cubes and in . So let’s define
Now, it’s easily verified that the furthest points from the origin in are the corners, each of which is away. Thus we can tell that , and the support of is contained entirely within . This means that the support of is entirely contained within — by the time we get to the edge it’s already smoothly tailed off to zero, and so even though we define it piecewise, is a smooth function defined on all of .
Now set up the sum
Since is locally finite and is supported within , this sum is guaranteed to be finite at each point, which makes a smooth function on all of . The ball is contained within the cube , so takes the constant value on this ball. Since the preimages form an open cover of , there is always at least one for which . In particular, for which it’s not zero, and thus the whole sum is nonzero at .
Since is never zero, we can divide by it. We define . These are smooth functions on all of , and their sum is everywhere exactly . Thus the form a partition of unity subordinate to . And since refines , the partition is subordinate to this cover as well.