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In all the proofs I can find of the Open Mapping Theorem (for example here) at the outset it is mentioned that it is enough to prove that for all a in U, f(a) is contained in a disk that is itself contained in f(U).

How is that enough? One needs to prove that for every open set U' (that is a subset of U) the theorem holds, however the U used in that opening statement is a connected open set (and thus a stricter condition).

I find that every way I try and reconcile this I run into something non-trivial (e.g. Considering any open set a countable union of connected open sets. Can you do that in Rn? Does it require a finitely dimensioned space or any other such restriction?), even though the proof's statement suggests it is obvious that proving for U is enough. Stumped.

Thanks.

EDIT

I'm aware that the statement proves that f(U) is open (as a union of open disks), that's not what I'm asking. My question is how is that enough to prove that f is open, i.e. f(V) is open for every V open subset of U (in particular since V may not be connected like U and certainly need not be a disk).

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    It's been my pleasure :D2011-09-25

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But an open subset of $U$ is also open in $\mathbf C$, and hence is a union of elements of the topological base for $\mathbf C$ given by the open balls, which are connected. And $f(\bigcup Y_\alpha) = \bigcup f(Y_\alpha)$ over arbitrary indexing sets.

Note that although you don't need any cardinality argument, it is true that $\mathbf R$ and hence finite products of it are second countable.

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    I suppose this is the answer, that even though V needn't be connected it can be expressed as the union of open and connected sets, each of which we can appl$y$ the original theorem to. That's pretty much what I proposed in the question. Don't know what second countable is, so that bit was beyond me.2011-09-25