Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.



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).

share|improve this question
From what you've written it follows that $f(U)$ is a union of open discs, hence open. Where is the difficulty? –  Qiaochu Yuan Sep 25 '11 at 19:31
@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? –  davin Sep 25 '11 at 19:37
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. –  Olivier Bégassat Sep 25 '11 at 19:45
@OlivierBégassat, that I know. Please see my edit, maybe my question is clearer now. –  davin Sep 25 '11 at 19:51
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. –  Olivier Bégassat Sep 25 '11 at 20:20
show 7 more comments

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
Ah, you've edited. Will try to address that. –  Dylan Moreland Sep 25 '11 at 19:54
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. –  davin Sep 25 '11 at 20:18
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.