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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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    From what you've written it follows that $f(U)$ is a union of open discs, hence open. Where is the difficulty?2011-09-25
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    @QiaochuYuan, the theorem requires that to be the case for all open subsets of the original U. Proving it for a single open set would be fine (because every subset could be considered like the original U with f maintaining its smoothness etc.), but U is a *connected* open set. So my question is: if the theorem is true for all connected open sets, what makes it true for all general open sets?2011-09-25
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    By definition of an open set $U\subset\mathbb{C}$, for every $z\in U$ there exists a positive $r_z$ such that the open disk of radius $r_z$ and center $z$, let's call this $D(z,r_z)$, is contained in $U$: $D(z,r_z)\subset U$. Thus $U=\cup_{z\in U}D(z,r_z)$ is the union of open disks.2011-09-25
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    @OlivierBégassat, that I know. Please see my edit, maybe my question is clearer now.2011-09-25
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    Because $f(\cup_{i\in I}X_i)=\cup_{i\in I}f(X_i)$ is always true, and you have, returning to your example, $f(U)=f(\cup_{z\in U}D(z,r_z))=\cup_{z\in U}f(D(z,r_z))$ is open as the union of open sets.2011-09-25
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    @OlivierBégassat, I must be explaining myself badly, because it looks like you're still explaining why f(U) is open. I understand that point and don't disagree. But that isn't enough to prove that f is an open mapping. There are, for example, many V open subset U that are not disks, and I don't understand how to prove in that case that f(V) is open.2011-09-25
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    Okay, I understand. So what you know is that $f(U)$ is open whenever $f$ is holomorphic on the open set $U$. If $V\subset U$ is an open subset of $U$, then by definition of the induced topology, it is of the form $V=U\cap V'$ where $V'\subset\mathbb{C}$ is an open subset of $\mathbb{C}$. Thus it actually is an open subset of $\mathbb{C}$. Restrict $f$ to the open subset $V$ and it remains holomorphic (call the restricted function $g:V\to\mathbb{C}$). Apply the previous theorem to $g$ and you get $\big(f(V)=\big)g(V)$ open in $\mathbb{C}$.2011-09-25
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    An open set in $\mathbb{C}$ is a union of connected open sets (indeed, of open balls!).2011-09-25
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    @OlivierBégassat, I would agree, except for the fact that we proved the theorem with U being connected and we allow V not to be, in which case you can't "apply the previous theorem" because not all the conditions hold.2011-09-25
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    Then say $V$ is the union of its connected components, and all of them are open (because disks are connected). Let's call them $(V_i)_{i\in I}$. Take any of these connected components, say $V_i$. What I wrote above implies that $f(V_i)$ is open. Finally, $f(V)=f(\cup_{i\in I}V_i)=\cup_{i\in I}f(V_i)$ is open as the union of open sets.2011-09-25
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    @OlivierBégassat, thank you very much. I appreciate your patience.2011-09-25
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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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    Ah, you've edited. Will try to address that.2011-09-25
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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 apply 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