The Unapologetic Mathematician

Atlases Refining Covers, part 2

Again with the late posts…

Now, armed with our two new technical assumptions, we can prove the existence of the refining covers we asserted yesterday.

Since $M$ is (now) known to be locally compact, Hausdorff, and second countable, there must exist a countable basis $\{Z_k\}$ for the topology of $M$ with each closure $\bar{Z}_k$ compact. Basically we can start with a neighborhood of each point that has compact closure and whittle it down to a countable basis, using the Hausdorff property to make sure we keep compact closure.

We will construct a sequence of compact sets inductively. Let $A_1=\bar{Z}_1$, which is compact by assumption. Given $A_i$ already defined, let $j$ be the first index for which $A_i\subseteq Z_1\cup\dots\cup Z_j$, and define $A_{i+1}=\bar{Z}_1\cup\dots\cup\bar{Z}_j\cup\bar{Z}_{j+1}$. Then $\{A_k\}$ is a sequence of compact sets with $A_k\subseteq\mathrm{int}(A_{k+1})$, and whose union is all of $M$. Define $A_0$ to be the empty set.

Now, we can write

$\displaystyle M=\bigcup\limits_{i\geq0}\left(A_{i+1}\setminus\mathrm{int}(A_i)\right)$

so for every point $p$ we can find a chart $(V_p,\phi_p)$ sending $p$ to $0\in\mathbb{R}^n$ and with $\phi_p(V_p)=B_{3n}(0)$, $V_p\subseteq U_\alpha$ for some $\alpha$, and $V_p\subseteq\mathrm{int}(A_{i+2})\setminus A_{i-1}=W_i$ for some $i$.

Indeed, we can surely find some chart around $p$, and intersecting it with some open $U_\alpha$ — which should contain $p$ — and with the open $W_i$ — likewise — still gives us a chart. We can subtract off whatever offset we need to make sure that this chart sends $p$ to $0$. Then we can take a ball of some radius around $0$ and let $V_p$ be its preimage. Scaling up the coordinate map lets us expand this ball until its radius is ${3n}$. Messy, no?

So now the collection of all the preimages $\phi_p^{-1}(B_1(0))$ as $p$ runs over $A_{i+1}\setminus\mathrm{int}(A_i)$ is an open cover of this compact set, and thus it contains a finite subcover, which we write as $P_i$. Taking the union of all of the $P_i$ gives a countable cover $V_k$ of $M$ refining $U_\alpha$. Each $V_k$ is the domain of a chart with $\phi_k(V_k)=B_{3n}(0)$, and the collection of preimages $\phi_k^{-1}(B_1(0))$ covers $M$, as asserted.

The only thing we haven’t shown here is that $V_k$ is locally finite. But since each point $p\in M$ must lie in one of the $A_{i+1}\setminus\mathrm{int}(A_i)$, so $W_i$ is an open neighborhood of $p$ that intersects at most finitely many $A_i$, and each $A_i$ can intersect at most finitely many $V_k$, so $W_i$ touches at most finitely many of them itself.

Got all that? We’re not out of the woods yet…

March 11, 2011 - Posted by | Differential Topology, Topology

1. Not sure why, but for me the LaTeX V_k is rendering as “f(z) = z(z-1)(z-2) = z^3 -z” in both this post and the last.

Comment by Robert | March 11, 2011 | Reply

2. clear cache? I’m not seeing it…

Comment by John Armstrong | March 11, 2011 | Reply

3. Can you help me a bit?
Why A_k subset int(A_k+1) holds?
I can’t figure it out…

Comment by NikosP | April 8, 2012 | Reply

4. Does it help if I remind you that the sets $Z_k$ in a basis are open?

Comment by John Armstrong | April 8, 2012 | Reply

5. Hello John. I do not see how you obtain a countable basis for which each subset $Z_k$ has compact closure. Could you please explain the process that you use to “whittle it down”?

Besides, don’t you simply need $\sigma$-compactness in order to build the sequence $A_n$? This seems to be much simpler to obtain.

Comment by Robin | March 1, 2013 | Reply