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Simple boundary point definition at Planet Math

Rudin gives the following as an example of a boundary point that is not simple:

If $\Omega = U - \{x : 0 < x \le 1\}$ then $\Omega$ is simply-connected. If $0 < \beta \le 1$, $\beta$ is a boundary point that is not simple.

No matter what sequences I try, I can not find a sequence that converges to $\beta$ but a path cannot connect its points. Would you please tell me how this example works?

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up vote 5 down vote accepted

Try $$x_n=\beta+\frac{(-1)^n}{n} i$$

Is these were connected by a path $\gamma:[0,1)\to\Omega$, then the real part of $\gamma$ would have to attain negative values along a sequence converging to $1$.

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I tried something like this, but why can not the path go infinitely many times up and down around the origin across the negative real axis? – PeterM Feb 22 '13 at 21:40
@PeterM Because having $\operatorname{Re}\gamma(t_n)<0$ for some sequence $t_n\to 1$ is not consistent with $\lim_{t\to 1}\gamma(t)=\beta$. You are not looking just for a path connecting the points, but for a path that also converges to $\beta$. – user53153 Feb 22 '13 at 21:48
But $t_n$ is only a subsequence. $\gamma(t_n) = x_n$ by definition so it always has a positive real part. I'm trying to formulate an argument that a compact path cannot make infinite roundtrips like this, but I'm unsuccessful so far. – PeterM Feb 22 '13 at 22:04
@PeterM It can make all kinds of roundtrips if you do not insist on it having a limit as $t\to 1$. – user53153 Feb 22 '13 at 22:16
Ohhh. I was too fixated on the subsequence $t_n$ that I forgot that $\gamma$ must be continuous (and hence have a limit at $t \to 1$ too. Got it now. Thanks! – PeterM Feb 22 '13 at 23:34

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