Connected set in $\mathbb{S}^1$

Let $A$ be a connected set in $\mathbb{S}^1$. Show that $A^c$ is also connected.

My idea: suppose $A^c$ is not connected. Then, there exist $U,V \subset \mathbb{S}^1$ open and disjoint such that $U \cup V =A^c$. Therefore $\mathbb{S}^1 = A \cup U \cup V$. If I had that $A\cup U$ was open I could arrive at contradiction, since $\mathbb{S}^1$ is connected, but I can't say that...

• $A^c$ is disconnected iff there are disjoint non-empty subsets $U,V$ of the space $A^c$ with $U\cup V=A^c.$ If such $U,V$ exist then $U=A^c\cap U'$ and $V=A^c\cap V'$ for some $U',V'$ that are open subsets of the space $S^1.$ But $U',V'$ are not necessarily disjoint. – DanielWainfleet Sep 22 '17 at 22:16

If $A=S^1$, then $A^\complement=\emptyset$, which is connected.
Otherwise, $A^\complement$ has at least on point. We can assume, withou loss og generality, that that point is $(-1,0)$. Consider the map $f\colon(-\pi,\pi)\longrightarrow S^1$ defined by $f(x)=(\cos x,\sin x)$. Then $f^{-1}(A)$ is a connected subset of $(-\pi,\pi)$ in therefore it is an interval $I$. Therefore, its complement is either an interval or the union of two intervals and, if it turns out to be the union of two intervals, one of them begins at $-\pi$ and the other one ends at $\pi$. In each case, $f(I^\complement)\cup\{(-1,0)\}$ (which is $A^\complement$) is connected.
• Why is $f^{-1}\left( A \right)$ connected? – Figurac Sep 22 '17 at 18:54
• Because $f^{-1}$ is continuous and therefore maps connected sets onto connected sets. – José Carlos Santos Sep 22 '17 at 19:18
• Why can you say $f(I^\complement)\cup\{(-1,0)\}$ is connected? – Figurac Sep 25 '17 at 18:52
• @Figurac Because if $a$ and $b$ are the extreme points of $I$, then $f(I^\complement)\cup\{(-1,0\}$ is the arc of the unit circle going from $f(a)$ to $f(b)$ passing through $(-1,0)$. (one of even both extreme points of the arc may be missing, but that doesn't destroy connectedness). – José Carlos Santos Sep 25 '17 at 21:06