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Jul 20 at 14:00 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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Feb 21 at 12:34 answer added inyo timeline score: 1
Feb 17 at 5:51 comment added Factorial_zero Yes, thank you Sir @Moishe Kohan. Now I understand the possible picture of a simply connected domain so that there are points on the boundary which are not accessible.
Feb 17 at 5:50 comment added Moishe Kohan Right. But you have to take the graph of $\sin(1/x)$ for all nonzero values of $x$ (not just the positive ones). Then points $(0,y)$, $-1<y<1$, are not accessible. Caratheodory's book on conformal maps discusses accessibility in great detail.
Feb 17 at 5:48 comment added Factorial_zero I understand, in that case the points on the imaginary axis (such that imaginary part is bounded by -1 and 1) are not accessible. Am I correct @Moishe Kohan ?
Feb 17 at 5:43 comment added Moishe Kohan Hint: Consider the topologist's sine curve
Feb 17 at 5:40 history edited Factorial_zero CC BY-SA 4.0
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Feb 17 at 4:59 history asked Factorial_zero CC BY-SA 4.0