(a) Is every open connected subset of the circle path-connected?
(b) What about connected nonopen subsets of the circle?
For (a), the circle is a subset of $\mathbb{R}^{2}$ and if $U$ is an open subset of the circle, $U = N\cap C$ where $C$ is the circle and $N$ is an open set of $\mathbb{R}^{2}$ which is path-connected. Since $C$ is path-connected, then $U$ is path-connected.
Is corrected?
For (b), I dont know. Intuitively seemes true, but Im not sure. Can someone help me?
On
For $r\in \Bbb R$ let $P(r)=(\cos r, \sin r).$ Let $C=\{P(r):r\in \Bbb R\}.$ Suppose $S$ is a connected subset of $C,$ and $p_1, p_2$ are distinct members of $S.$ There exist $r_1,r_2$ such that $p_1=P(r_1), p_2=P(r_2)$ and $r_1<r_2<r_1+2\pi.$
(i). If $\{P(r): r\in (r_1,r_2)\}\subset S,$ let $f(x)=P(r_1+x(r_2-r_1))$ for $x\in [0,1].$
(ii). If $r_3\in (r_1,r_2)$ and $P(r_3)\not \in S$ then $\{P(r): r\in (r_2, r_1+2\pi)\}\subset S.$ Otherwise let $r_4\in (r_2,r_1+2\pi)$ with $P(r_4)\not \in S.$ But then $S\cap \{P(r): r\in (r_4-2\pi,r_3)\}$ and $S\cap \{P(r): r\in (r_3,r_4)\}$ are two non-empty disjoint open subsets of the space $S,$ and their union is $S,$ a contradiction.
$\quad$ So let $f(x)=P(r_2+x(r_1-r_2+2\pi)$ for $x\in [0,1].$
Intersections of connected sets are not, in general, connected. For example, consider sets in the shapes $\cup$ and $\cap$, overlaid on each other. The intersection in that case will have two components. That line of thought isn't going to work.
Well, no, it isn't. The circle minus a point is homeomorphic to the line. So then, we split into cases.
Case 1: $U$ is the whole circle. Then $U=C$ is path-connected.
Case 2: There is some point $x$ on the circle but not in $U$. Then $C\setminus x$ is homeomorphic to the line. Project to that line; the image $U'$ of $U$ is connected, so it's an interval*, and it's path-connected. Wrap back to the circle, and $U$ is path-connected.
* Empty intervals count.
There - that's the comment's argument patched up.