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).

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    $\begingroup$ From what you've written it follows that $f(U)$ is a union of open discs, hence open. Where is the difficulty? $\endgroup$ Sep 25, 2011 at 19:31
  • $\begingroup$ @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? $\endgroup$
    – davin
    Sep 25, 2011 at 19:37
  • $\begingroup$ 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. $\endgroup$ Sep 25, 2011 at 19:45
  • $\begingroup$ @OlivierBégassat, that I know. Please see my edit, maybe my question is clearer now. $\endgroup$
    – davin
    Sep 25, 2011 at 19:51
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    $\begingroup$ 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. $\endgroup$ Sep 25, 2011 at 20:20

1 Answer 1


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.

  • $\begingroup$ Ah, you've edited. Will try to address that. $\endgroup$ Sep 25, 2011 at 19:54
  • $\begingroup$ 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. $\endgroup$
    – davin
    Sep 25, 2011 at 20:18

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