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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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    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.2011-02-18
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    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.2017-10-07

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