On p.3 of the first volume of Spivak's Comprehensive Introduction to Differential Geometry, he says that it is an "easy exercise" to show that the invariance of domain theorem (if $f:U\subset\mathbb{R}^n\rightarrow\mathbb{R}^n$ is one-to-one and continuous and $U$ is open then $f(U)$ is open) implies that in his definition of a manifold

a metric space $M$ such that every point $x\in M$ has a neighborhood $U$ of $x$ and some integer $n\geq0$ such that $U$ is homeomorphic to $\mathbb{R}^n$,

the neighborhood $U$ in fact must be open.

My question: My proof seems to require a bit of set up, as well as two appeals two the invariance of domain theorem, including once to first prove that the dimension of a manifold is well defined (which Spivak discusses on p.4). In any case, I feel like the argument is longer than what Spivak calls a "complicated little argument" on the previous page. Am I missing something obvious?

Perhaps my real question is whether this kind of comment is to be expected from Spivak, since I am reading this book on my own.

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    $\begingroup$ I wouldn't let perceived contrasts between 'easy exercises' and 'complicated little arguments' bother you, if you can (a) understand the content of the statement and (b) give a proof that makes use of invariance of domain. [You didn't ask, but IMHO both (a) and (b) are nontrivial, and my own definition of "easy" probably wouldn't include them (although my definition of "elementary" would); in my book, if it's easy to write down a bogus proof, then it's not "easy."] Self-study from Spivak (or similarly glib authors) is much improved if you do not read too much into their informal chitchat. $\endgroup$ – anon Feb 18 '11 at 4:29
  • $\begingroup$ It is not the theorem that is easy. What Spivak says is easy is the proof that $U$ must be open in his definition of manifold. You should assume the "invariance of domain" and USE it to prove the simple thing. $\endgroup$ – André Caldas Oct 7 '17 at 1:10

I also find the "it is an easy exercise" remarks very frustrating. I think (but am not sure) that I came up with a proof, probably similar to yours. This assumes the fact stated on page 2 "we can always choose the neighborhood $U$ in our definition to be an open neighborhood." I use this fact in choosing $V$ below.

Let $\varphi : U \to \mathbb{R}^n$ be the homeomorphism given in the definition of the Euclidean neighborhood of $x$, and let $z \in U$.

Since $\, z \in M \quad \exists$ an open set $V$ of $M$ with $z \in V$, and a homeomorphism $\theta:V \to \mathbb{R}^n$.

Consider the set $W \equiv V \cap U$. $W$ is open in the relative topology of $U$, so $\varphi (W)$ is open in $\mathbb{R}^n$. For notational convenience, let $\psi \equiv \varphi \mid_{W}$. The map $\;\theta \circ \psi^{-1} :\varphi (W) \to \mathbb{R}^n$ is continuous, so by Invariance of Domain $\; \theta (W)= \theta \circ \psi^{-1}[\varphi (W)] \;$ is open in $\mathbb{R}^n$, which tells us that $W$ is open in $V$.

Since $V$ is open in $M$, $W$ is open in $M$ and we have $z \in W \subset U$, so $U$ is open.

I wouldn't bet my life that what I did is correct, but I think it is. This is my first time attempting to answer a question, so if I've messed up, please be gentle.

Thanks, Dave

  • $\begingroup$ I think is fine (it is useful to me). @LostInMath please do not forget to upvote this answer and accept it if it's useful to you $\endgroup$ – leo Aug 10 '12 at 3:47
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    $\begingroup$ I have a doubt about whether the value of n is same for both U and V? $\endgroup$ – user84854 Jul 3 '13 at 7:20

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