The Unapologetic Mathematician

Mathematics for the interested outsider

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

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