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Consider $$ S_1:=\left\{z\in\mathbb{C}: \lvert z\rvert =1\right\},\\E:=\left\{0\right\}\cup\bigcup_{n\in\mathbb{N}}\left\{(1-2^{-n})e^{\pi i k/2^n}: k\in\left\{0,1,\ldots,2^{n+1}-1\right\}\right\}. $$ Show that $E\cup S_1$ is compact with the induced topology on $\mathbb{C}$.

Can you please give me some help to show this?

Which characterization of compactness do you recommend?

Every open cover has finite open cover? Every sequence has a convergent subsequence? Or which one?

Edit

Isn't $(E\cup S_1)^C=E^C\cap S_1^C$ open, because $E^C$ is open, $S_1^C$ is open and finite intersections of open sets are open? Thus $E\cup S_1$ is closed and since $E\cup S_1\subset D, D:=\left\{z\in\mathbb{C}: \lvert z\rvert\leqslant 1\right\}$ and $D$ is compact it follows that $E\cup S_1$ is compact?

1 Answers1

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Here are three approaches to showing that $E\cup S^1$ is compact. For all of them it will be handy to let

$$E_n=\left\{(1-2^{-n})e^{\pi i k/2^n}: k\in\left\{0,1,\ldots,2^{n+1}-1\right\}\right\}$$

for each $n\in\Bbb Z^+$; $E_n$ is a set of $2^{n+1}$ points spaced uniformly around the circle of radius $1-2^{-n}$ centred at $0$. I’ll denote that circle by $C_n$.

  • Show that $E\cup S^1$ is closed and bounded. It’s obviously bounded, so we need only show that it’s closed. First show that $S^1\cup\{0\}\cup\bigcup_{n\in\Bbb Z^+}C_n$ is closed by showing that every point of its complement is either outside the closed unit disk, in the open annular region between $C_n$ and $C_{n+1}$ for some $n\in\Bbb Z^+$, or between $0$ and $C_1$. Then show that each point of $C_n\setminus E_n$ has an open nbhd that misses $E\cup S^1$.

  • Show that every sequence in $E\cup S^1$ has a convergent subsequence. Let $\langle z_n:n\in\Bbb N\rangle$ be a sequence in $E\cup S^1$. If infinitely many terms of the sequence lie in $S^1$, we’re done, so without loss of generality we may assume that the sequence lies entirely in $E$. Choose $k_1\in\{0,1,2,3\}$ so that infinitely many terms of the sequence lie in the sector determined by $0,e^{\pi ik_1/2}$, and $e^{\pi i(k_1+1)/2}$ (where $k_1+1$ is computed mod $4$). Given a sector determined by $0,e^{\pi ik_n/2^n}$, and $e^{\pi i(k_n+1)/2^n}$ (where $k_n+1$ is computed mod $2^{n+1}$) that contains infinitely many terms of the sequence, let $k_{n+1}=2k_n$ if the sector determined by $0,e^{\pi ik_n/2^n}$, and $e^{\pi i(2k_n+1)/2^{n+1}}$ contains infinitely many terms of the sequence; otherwise let $k_{n+1}=2k_n+1$. (This is just the familiar bisection procedure.) Show that $\langle e^{\pi ik_n/2^n}:n\in\Bbb Z^+\rangle$ is a convergent sequence in $S^1$, say with limit $w$, and that $\langle z_n:n\in\Bbb N\rangle$ has a subsequence converging to $w$.

  • Show that every open cover of $E\cup S^1$ has a finite subcover. Let $\mathscr{U}$ be an open cover of $E\cup S^1$. $S^1$ is compact, so some finite $\mathscr{U}_0\subseteq\mathscr{U}$ covers $S^1$. Let $U=\bigcup\mathscr{U}_0$. Use compactness of $S^1$ to show that there is an $\epsilon>0$ such that the $\epsilon$-ball around every point of $S^1$ is contained in $U$. Conclude that $U$ contains all but finitely many of the circles $C_n$, so that $(E\cup S^1)\setminus U$ is finite.

Added: It may be helpful to regard $E\cup S^1$ as a circular version of the following subset of $\Bbb R^2$. Let $I=[0,1]\times\{0\}$. For $n\in\Bbb N_{\ge 0}$ let

$$I_n=\left\{\left\langle\frac{k}{2^n},\frac1{2^n}\right\rangle:k\in\{0,\ldots,2^n-1\}\right\}\;.$$

The desired subset of $\Bbb R^2$ is then $I\cup\bigcup_{n\in\Bbb N_{\ge 0}}I_n$.

Brian M. Scott
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  • Thank you. To your first way: Do not understand what you mean with "each point of $C_n\setminus E_n$ has an open neighborhood that misses $E\cup S^1$ and why from this it follows that $E\cup S^1$ is closed. –  Jan 14 '15 at 15:52
  • @math12: I’m suggesting showing that the complement of $E\cup S^1$ is open by showing that each point of the complement is the centre of an $\epsilon$-ball that is disjoint from $E\cup S^1$. For the points that are outside $S^1$ or between the circles $C_n$ this is easy. For a point $z$ on $C_n$ that is not in $E_n$, you have to work a little harder, but not much: it must lie between two adjacent points of $E_n$, so you just need to choose $\epsilon$ small enough to exclude those points, and also small enough to avoid $C_{n+1}$, the next circle out. – Brian M. Scott Jan 14 '15 at 15:56
  • Now I got it, thanks very much. –  Jan 14 '15 at 15:59
  • @math12: You’re very welcome. – Brian M. Scott Jan 14 '15 at 16:00
  • Is it the same argument to say that $E$ is discrete in $E\cup S^1$? Just choose small enough neighbourhoods around each point in $E$? –  Jan 14 '15 at 16:01
  • @math12: Yes, $E$ is a discrete set for that reason. Note, though, that it’s not a closed discrete set. In fact, $\operatorname{cl}E=E\cup S^1$: every point of $S^1$ is the limit of a sequence in $E$. – Brian M. Scott Jan 14 '15 at 16:02
  • If $y\in S^1$, then $y$ is the limit of which sequence in $E$? –  Jan 14 '15 at 16:08
  • @math12: Use the sector idea in my second approach. If $y=e^{i\theta}$, for each $n\in\Bbb Z^+$ choose $k_n\in{0,\ldots,2^{n+1}-1}$ so that $\pi k_n/2^n\le\theta<\pi(k_n+1)/2^n$ (where $k_n+1$ is computed mod $2^{n+1}$). The sequence $\langle (1-2^{-n})e^{\pi ik_n/2^n}:n\in\Bbb Z^+\rangle$ converges to $y$. – Brian M. Scott Jan 14 '15 at 16:15