How to characterize open sets in the unit circle

Question

At the end of the second chapter of Munkres' "Topology" there are some supplementary exercises concerning topological groups. One such group is $(S^1, \cdot)$. My problem is that the unit circle has not been dealt with so far in the theory and hence all I can assume is that it possesses the subspace topology it inherits from the metric topology of the complex plane. However, this does not seem to be enough to easily characterize the opens sets in $S^1$. Therefore I have turned to the site's help but I have only found answers relying on the notion of opennes of the map $x \mapsto e^{ix}$ which I only know how to be proved using compacteness preservation under a continuous function, which has not been yet introduced in Munkres at this stage.

My question is then if there is any other way around that does only use concepts and tools up to Munkres' "Topology" second chapter.

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As always, any comment or answer is much appreciated and let me know if I can explain myself clearer!

Answer 1

Let $p :\mathbb R \to S^1, p(x) = e^{ix} = \cos x + i \sin x$. This map is well-known to be a continuous surjective group homomorphism from $(\mathbb R,+)$ to $(S^1,\cdot)$.

We shall show that the set of circle segments $$\mathfrak S = \{ S(a,b) = p((a,b)) \mid (a,b) \in \mathfrak I \}$$ (where $\mathfrak I$ denotes the set of open intervals in $\mathbb R$) forms a basis for the topology on $S^1$.

Note that if $b -a > 2\pi$, then $S(a,b) = S^1$, and if $b - a \le 2\pi$, then $S(a,b)$ is the circle segment between $p(a)$ and $p(b)$ (where we travel counterclockwise; observe that $p(a) = p(b)$ in case $b -a = 2\pi$).

Trivially $S(a',b') \subset S(a,b)$ if $(a',b') \subset (a,b)$.

We have to show that the $S(a,b)$ are open in $S^1$ and form a basis.

1. The sets $S(-r,r)$ with $0 < r \le \pi/2$ are open in $S^1$ and form a neighborhood basis of the point $1 \in S^1$.

Consider the projection $\phi : S^1_+ = \{ z \in S^1 \mid \operatorname{Re} z > 0\} \to (-1,1), \phi(z) = \operatorname{Im} z$. This is a homeomorphism with inverse $\phi^{-1}(t) = \sqrt{1-t^2} + it$. We have $\phi(S(-r,r)) = \{\phi(p(x)) \mid x \in (-r,r) \} = \{\sin x \mid x \in (-r,r) \} = \sin((-r,r))$. Since $\sin$ maps $(-\pi/2,\pi/2)$ homeomorphically onto $(-1,1)$, we see that $\sin((-r,r))$ is open in $(-1,1)$, thus $S(-r,r)$ is open in $S^1_+$ and therefore open in $S^1$.

Now let $U \subset S^1$ be an open neighborhood of $1$ in $S^1$. Write $U = V \cap S^1$ with some open $V \subset \mathbb R^2 = \mathbb C$. Choose $s \in (0,1]$ such that $(1-s,1+s) \times (-s,s) \subset V$ and let $U_s = ((1-s,1+s) \times (-s,s)) \cap S^1$. This is an open neighborhood of $1$ in $S^1$ which is contained in $U$. Let $r \in (0,\pi/2]$ be the unique point such that $\sin r = s$. We have $\phi(U_S) = (-s,s) = \sin((-r,r)) = \phi(S(-r,r))$, thus $U_s =S(-r,r)$. This shows $0 \in S(-r,r) \subset U$.

2. The sets $S(c-r,c+r)$ with $r \in (0, \pi/2]$ are open in $S^1$ and form a neighborhood basis of the point $p(c) \in S^1$.

For each $z \in S^1$ the multiplication map $\mu_z : S^1 \to S^1, \mu_z(w) = z \cdot w$, is a homeomorphism (with inverse $\mu_z^{-1} = \mu_{z^{-1}})$. If $z = p(c)$, then $$\mu_z(S(a,b)) = \{p(c) \cdot p(x) \mid x \in(a,b)\} = \{p(c + x) \mid x \in (a,b)\} = p((c+a,c+b)) = S(c+a,c+b) .$$

For each $z \in S^1$ we have $\mu_z(1) = z$ and therefore the sets $\mu_z(S(-r,r))$ are open and form a neighborhood basis of the point $z$. With $z = p(c)$ we therefore see that the sets $S(c-r,c+r)$ with $r \in (0,\pi/2]$, are open and form a neighborhood basis of the point $z$.

3. All $S(a,b)$ are open (using 2. and the surjectivity of $p$ this shows that $\mathfrak S$ forms a basis for the topology on $S^1$).

Let $z \in S(a,b)$. Write $z = p(c)$ with $c \in (a,b)$ and choose $r \in (0,\pi/2]$ such that $(c-r,c+r) \subset (a,b)$. Then $z \in S(c-r,c+r) \subset S(a,b)$.