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Interested by this question in math.SE, which shares a link to planetmath about definition of a simple boundary point. This link gives reference to the book Functions of one complex variable II of Conway. In this book, there is an exercise which I find interested in:

If $w$ is a simple boundary point of $\Omega$, then there is a $\delta > 0$ such that $D(w;\delta)\cap \Omega$ is connected.

I think the author means that we can find $\delta$ small arbitrary such that the connectedness happens (not so sure about this). If we can solve this problem, then we can easily determine which point is not a simple boundary point. As an example, let $\Omega = D(0;1) \backslash \{x:0\leq x < 1\}$, and $0<\beta \leq 1$, then $\beta$ cannot be a simple boundary point of $\Omega$ because if we choose $\epsilon>0$ small enough, we can't find any $\delta \in (0,\epsilon)$ such that $D(\beta;\delta)\cap \Omega$ is connected.

Could anyone give me a hint? It is hardly for me to see which direction I should go, to use the condition that $D(w;\delta)\cap \Omega$ is connected.


Base on Seub's answer, I put here some details.

For Lemma 1: the idea is picking two sequences in two connected components of $\Omega$, combining them into a sequence whose even subsequence is one sequence, and odd subsequence is the other sequence.

This lemma together with Lemma 3 (to prove, use the fact that there is a positive distance from a closed set to a point outside it) solves the problem.

About Lemma 2, for the $(\Leftarrow)$ side, let a sequence in $\Omega$ converges to $w$, then eventually that sequence will be inside the disc $D(w,\delta)$. Because $w$ is a simple boundary point of $\Omega \cap D(w;\delta)$, the "latter" part of that sequence can be connected by a curve converges to $w$. In addition, $\Omega$ is connected leads to the "former" part can be connected by a curve. Combining two curves with a reparemeterizing, we get a desired curve.

For the $(\Rightarrow)$ side, let a sequence in $\Omega \cap D(w;\delta)$ converges to $w$, then it can be connected by a curve in $\Omega$. Eventually, this curve will be inside the disc $D(w;\delta)$. So the "latter" part of the curve (which is in $\Omega \cap D(w;\delta)$) will connect the "latter" part of the sequence. We may choose (at first) $\delta$ small enough for $\Omega \cap D(w;\delta)$ connected. Then the "former" part of the sequence can be connected by a curve in $\Omega \cap D(w;\delta)$. We conclude.


Edit: Seub gives a counterexample in his answer!

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  • $\begingroup$ Just a comment: It could be that the author assumes $\Omega$ to be a region (open and connected) although he does not state it explicitly. It is always like this in the whole Chapter 5 in Conway's book as far as I can tell. If this is the case, Saub's example cannot be used since the set $D(0,2)\setminus\cup_{n\geq1}L_{n}$ is not open. $\endgroup$ – Twi Oct 30 '16 at 20:15
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I apologize for my first answer which was incorrect.


First a couple of remarks:

  • I agree with you that the author probably meant

"(...) then there exists $\delta~$ arbitrarily small such that (...)"

otherwise it is kind of silly. For example, if $\Omega$ is connected and bounded, just take $\delta$ such that $\Omega \subset D(\omega, \delta)$.

  • I think the definition of simple boundary point should be

Let $\Omega$ be an open set in $\mathbb{C}$ and $\omega$ be a point in the boundary of $\Omega$. Then we call $\omega$ a simple boundary point if whenever $(\omega_n)$ is a sequence of points of $\Omega$ converging to $\omega$ there is a continuous path $\gamma:[0,1]\rightarrow \mathbb{C}$ such that $\gamma(t) \in \Omega$ for $0 \leqslant t < 1$, $\gamma(1) = \omega$ and there is a sequence $(t_n)$ in $[0,1)$ such that $t_n \rightarrow 1$ and $\gamma(t_n) = \omega_n$ for all $n~$ sufficiently large.

This would make the definition of a simple boundary point a local one, meaning that $\omega$ is a simple boundary point of $\Omega$ iff $\forall \delta>0$, $\omega$ is a simple boundary point of $\Omega \cap D(\omega, \delta)$.


While I believe these remarks are important, they actually don't matter for what I am about to say: after thinking about it, I believe the result is false!

Here's my counter-example: let $L_n$ be the vertical line segment $\frac{1}{n+1} + i \left[-\frac{1}{n},\frac{1}{n}\right]$ and $\Omega = D(0,2) \setminus \bigcup_{n=1}^{+\infty} L_n$. I really encourage you to draw a picture (of $\Omega$ and some "small" disk $D(0,\delta)$).

I claim that $0$ is a simple boundary point of $\Omega$. It might be a bit tedious to prove it rigorously, but it's certainly believable.

On the other hand, as soon as $\delta \leqslant 1$, $D(0, \delta) \cap \Omega$ is never connected. Indeed, let $n$ be the integer such that $\frac{1}{n+1} < \delta \leqslant \frac{1}{n}$. Then $L_n$ disconnects $D(0, \delta)$. QED.

(NB: Of course, for $\delta$ huge, e.g. $\delta > 2$, $D(0,\delta) \cap \Omega$ is connected, so the initial silly problem (without "arbitrarily small") is not contradicted. But we can cook something up similarly to contradict that too. If you insist I can tell you how, although I don't find that very important)


Now, if we wanted to fix all of this and try to say something true, I would use the "improved" definition of simple boundary point and maybe claim that

$\omega$ is a simple boundary point of $\Omega$ iff $\Omega$ is locally connected at $\omega$, meaning that there exists a basis of neighborhoods $\{U_n\}$ of $\omega$ such that $U_n \cap \Omega$ is connected for all $n$.

although for the moment it is not clear to me how to prove "$\Rightarrow$". I'll think about it if you're interested.

I hope this time I did not say too much nonsense!

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  • $\begingroup$ Thank you so much! Your lemmas simplify much of the problem. Let me write down some details for other persons who may be interested in this exercise. $\endgroup$ – Du Phan Dec 17 '13 at 6:12
  • $\begingroup$ I think that your lemma 3 is not true. For example, look at the open square $0$, $1$, $1+i$, $i$, then for each $n\geq 2$, remove the intervals $[\frac{1}{n},\frac{1}{n}+\frac{1}{2}i]$, the the remaining is connected, open, but the points in $\{iy:0<y<\frac{1}{2}\}$ give us counterexamples. $\endgroup$ – Du Phan Dec 17 '13 at 7:48
  • $\begingroup$ Also, I think that Lemma 2 is a corollary of the question, rather than a lemma; because the $(\Rightarrow)$ side requires (tacitly) $\Omega \cap D(w;\delta)$ is connected. $\endgroup$ – Du Phan Dec 17 '13 at 8:03
  • $\begingroup$ Oops, you're right about lemma 3, sorry about that. As for lemma 2, I agree that in the first version, it is also not correct. In the second version, I meant to say "for $\delta$ sufficiently small" and I accidentally deleted it (I just realized that). Do you agree with the second version? Anyway, I need to rework my answer. I'll do that tonight (Europe time) $\endgroup$ – Seub Dec 17 '13 at 10:23
  • $\begingroup$ The second version of Lemma 2 is true, if we know the question is true. I don't know how to prove its $(\Rightarrow)$ side if I don't have $\Omega \cap D(w;\delta)$ connected before. By the way, I appreciate any time you spend on this question. Don't have to rush. Regards :) $\endgroup$ – Du Phan Dec 17 '13 at 10:46

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