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?
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 $I\cup\bigcup_{n\in\Bbb N_{\ge 0}}I_n$.