I'm reading part of Lee's Introduction to manifolds. I have come to the following proposition.

$\textbf{Proposition 14.6 (Local Structure of Integral Manifolds).}$ Let $D$ be an involutive $k$-dimensional distribution on a smooth manifold $M$, and let $(U,\varphi)$ be a flat chart for $D$. If $N$ is any integral manifold of $D$, then $N\cap U$ is a countable disjoint union of open subsets of $k$-dimensional slices of $U$, each of which is open in $N$ and embedded in $M$.

Proof. Because the inclusion map $\eta:N\hookrightarrow M$ is continuous, $N\cap U=\eta^{-1}(U)$ is open in $N$, and thus consists of a countable disjoint union of connected components, each of which is open in $N$.

The proof then continues, and I will read the rest shortly. I was just wondering: why can't $\iota^{-1}(U)$ be an uncountable union of disjoint connected open components?


I don't have the book handy, but I am going to guess that the author takes manifolds to be separable and/or second countable as part of the definition. It's easy to check that in either case, a manifold can't contain uncountably many pairwise disjoint open sets.

  • $\begingroup$ Let me try to deduce this. Suppose we have a second countable space, i.e. the topology has a countable basis. If there were uncountably many pairwise disjoint open subsets of our space, each of them would nonetheless have to be the union of the sets of the basis contained in it, by definition of basis, but that would need uncountably many distinct elements of the basis - at least one for each of the disjoint open subsets - which is a contradiction. If the space is separable, we have a dense countable subspace, so those open subsets must contain each an element of the dense subspace. (cont…) $\endgroup$ – MickG Aug 9 '15 at 19:37
  • $\begingroup$ These elements must be pairwise distinct as the open subsets are pairwise disjoint, so they must be uncountable, but they must be elements of a countable subspace, which is again a contradiction. Could you include this in your answer for future readers' sake? $\endgroup$ – MickG Aug 9 '15 at 19:38
  • $\begingroup$ BTW for the book, a topological manifold is a Hausdorff second countable locally Euclidean topological space. So one of your two guesses was correct :). $\endgroup$ – MickG Aug 9 '15 at 19:40
  • 2
    $\begingroup$ @MickG: My idea was to give a hint rather than details - if you would like to write them out for posterity, please feel free to post your own answer! $\endgroup$ – Nate Eldredge Aug 9 '15 at 22:35

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