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Is there a simple elementary proof of this, which does not assume that $\log$ is a continuous function? This appears to be the core of the issue in completing the proof of $\log$ is continuous.

If $0<a<b$, $-\pi\le c<d\le\pi$, then $\{z\in\Bbb C:a<|z|<b\land c<\arg z<d\}$ is an open subset of $\Bbb C$.

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Mario Carneiro
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Note that $\{z\mid a<|z|<b, c<\arg z<d\} = \{ z \mid a < |z| < b, \operatorname{im} (e^{-i c} z) >0, \operatorname{im} (e^{-i d} z) <0\}$. Since $z \mapsto |z|$, $z \mapsto \operatorname{im} (e^{-i \theta} z)$ are continuous functions, it follows that $W$ is open.

copper.hat
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An open ball $B$ is open, a closed ball $B'$ is closed. Hence a difference $B\smallsetminus B'$ is open. An open annulus is such difference. It suffices you show an open wedge $s< \arg z<r$ is open, since your set is obtained by intersecting an open wedge and an annulus. But a wedge is an intersection of two open half planes determined by the lines $\arg z=s$ and $\arg z=r$, so it suffices to show open planes are... open. If the plane $H$ is determined by a line $\ell$, and $p\in H$, let $\delta$ be the distance from $\ell$ to $p$. Then $B(p,\delta)$ is contained in $H$, and $H$ is open.

Pedro
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  • What do you mean by "ray"? Also, could you put the exact bounds you are thinking of in the answer, so I can use it as a reference in a formal proof? (Actually I think it suffices to consider the case $d=c+\pi$ so that it is actually a half-plane.) – Mario Carneiro Feb 19 '15 at 05:05
  • @MarioCarneiro I mean "ray = wedge". You can show the open half-plane $\Bbb R\times (0,\infty)$ is open simply by taking a ball $B(z,\delta)$ with $\delta$ smaller than they $y$-coordinate of $z$. That's what I mean by measuring distances. – Pedro Feb 19 '15 at 05:07
  • I think you have to include the rotation though. Is there a way to write this as a preimage of $\exp$ and/or $\Re,\Im$ so that it follows from the continuity of these? (A vertical half-plane is open by this method as the $\Re$-preimage of $(0,\infty)$.) – Mario Carneiro Feb 19 '15 at 05:10
  • @MarioCarneiro Well, you can use that $e^z$ being holomorphic is an open mapping, and the region you're describing is the image of an open rectangle in $\Bbb C$ under $e^z$. Then again if you know this I don't understand why you want to prove this. – Pedro Feb 19 '15 at 05:37
  • I don't have most of complex analysis available, since I haven't proved Cauchy's integral theorem which requires a lot of stuff about homotopy groups which I haven't got to yet (working on it, but not yet). The linked question shows that $e^z$ is a local homeomorphism, but that's not good enough without also showing that it maps interiors to interiors, which also needs a lot of algebraic topology. – Mario Carneiro Feb 19 '15 at 05:39
  • Looks like ${z:\theta<\arg z<\theta+\pi}={z:\Im z\cos\theta>\Re z\sin\theta}$, and the latter is clearly an open set. Any idea how you would prove such a thing? – Mario Carneiro Feb 19 '15 at 05:52
  • @MarioCarneiro Forget about my last comment, forget about continuous maps. Can you show a triangle without boundary is open in $\Bbb R^2$? I insists you should simply give a clean geometric argument. Then pick a point in the wedge, and then note you can take a triangle by cutting the wedge and use that triangles are open to get a ball inside the triangle which will be inside the wedge. – Pedro Feb 19 '15 at 05:52
  • If I were showing a triangle is open, I would probably do so by showing it is an intersection of half-planes, which makes it a special case of this approach anyway. Or did you have some other approach in mind? – Mario Carneiro Feb 19 '15 at 05:55
  • @MarioCarneiro I already explained my approach. This discussion is getting too long. If you have more questions, either wait for another answer or make another more specific question. – Pedro Feb 19 '15 at 05:59
  • I don't know that triangles are open. In fact, I'm not even sure I know that $\arg$ describes wedges (where "know" means "have a proof of"). – Mario Carneiro Feb 19 '15 at 06:02