Path-connected space So I have this following question: Find a compact path-connected space $X \subset \mathbb{C}$ such that $X \backslash \{ O \}$ has an infinity of arc components.
I though about using an infinity of line segments (of length 1 for instance) which all have O as an extremity. However, I cannot get myself convinced that this set is compact.
So is this example fine, and if so how to prove it ? And else, do you have another example ?
Thank you for your help.
 A: The following is a standard example: let $q_n$ be an enumeration of the rational numbers in $[0,2\pi)$. Define $S_n$ as the segment with endpoints $(0,\frac{e^{iq_n}}{2^n})$. Define $S=\cup S_n$.
It is easy to see that $S$ is path connected and bounded. We claim that it is closed (compactness then follows from Heine-Borel):
To prove it, let $\{x_n\}\subset S$ be a converging sequence. Now, either $\{\arg(x_n)\}$ is a finite set (and we are done, since $S_n$ is closed and so $S_{n_1}\cup \dots \cup S_{n_k}$ is closed and so $\lim_n x_n\in S$), or it is not. In this case, however, $\lim_n |x_n|=0$, and so $x_n\to 0\in S$. Thus $S$ is closed.
We now claim that $S-\{0\}$ has a countable number of connected components. Indeed, $S_n-\{0\}$ is a connected subset of $S-\{0\}$, and it is maximal: to see this, let $X$ be the connected component of $S-\{0\}$ containing $S_n-\{0\}$. Its image through $\arg$ must be connected and contained in $\mathbb{Q}\cap [0,2\pi)$, and so it must be constant, which implies $X=S_n-\{0\}$.
One can also construct a compact set $X$ such that $X-\{(0,0)\}$ has an uncountable number of components: let $C$ be the cantor set, and $C_x$ the segment joining $(x_k,1)\in C\times \{1\}$ and the origin. Then $K=\cup_{x\in C} C_x$ is compact and path connected, but removing the origin gives us $K-\{(0,0)\}=\sqcup_{x\in C}C_x-\{(0,0)\}$, a set with $|C|=2^{\aleph_0}$ connected components
