# How to show that $S^1$ with one point removed is still connected?

$S^1:= \{x \in \mathbb{R^2}: \|x\| = 1 \}$

Suppose $y_0 \in S^1$. Prove $(S^1-\{y_0\}, \mathcal{T}_{S^1- \{y_0\}})$ is connected. where $\mathcal{T}_{S^1- \{y_0\}}$ is the subspace topology coming from the topology on $S^1$.

My thought:

I show $$f: \mathbb{R} \to S^1, f(x) = (\cos(x),\sin(x))$$ continuous. The I restrict the codomain $S^1$ to $S^1- \{y_0\}$. I meant to use the fact, continuous function $f: X \to Y$, $f(X)$ connected if $X$ connected.

The issue here is that, I also have to restrict the domain of $\mathbb{R}$ to $\mathbb{R} - \{x_0\}$, since for all $y_0 \in S^1, \exists x_0 : y_0 = (\cos(x_0), \sin(x_0))$. This restriction of domain makes the new domain no longer connected. Hence, I can't use the theorem I want to.

Any help?

Hint: try showing it is still path connected.

• So I try to construct a particular function $f$ that $f(0) = x$ and $f(1) = y, \forall x,y \in S^1$ ? Commented Jul 19, 2015 at 4:00
• yep, and you do that by just tracing the arc between them, which always exists
– Alan
Commented Jul 19, 2015 at 4:09

Your idea is a good one, but it needs to be modified.

You say you need to restrict the map $f: \mathbb{R} \to S^1$, $x \mapsto (\cos x, \sin x)$ to $\mathbb{R}\setminus\{x_0\}$ to ensure that the image is $S^1\setminus\{y_0\}$. But this is not true, for example $x_0 + 2\pi \in \mathbb{R}\setminus\{x_0\}$ and $f(x_0 + 2\pi) = y_0$.

What you suggested was to remove one of the preimages of $y_0$ from the domain, but you need to remove all of them to ensure that $y_0$ is no longer in the image. That is, the map $f|_{\mathbb{R}\setminus f^{-1}(y_0)}$ has image $S^1\setminus\{y_0\}$. However, as $\mathbb{R}\setminus f^{-1}(y_0) = \bigcup_{k \in \mathbb{Z}}(x_0 + 2k\pi, x_0 + 2(k+1)\pi)$ is not connected, it does not immediately follow that $S^1\setminus\{y_0\}$ is connected.

What happens if you restrict the map $f$ to one of the connected components of $\mathbb{R}\setminus f^{-1}(y_0)$, say $(x_0, x_0 + 2\pi)$?

• thanks for the reply. Once restrict the domain to $(x_0, x_0+2\pi)$. The rest is to show that new restricted function $g: (x_0, x_0+2\pi) \to S^1-\{y_0\}$ is subjective. Then I'm done ? Commented Jul 19, 2015 at 4:27
• That's correct. Then $S^1 - \{y_0\}$ will be the continuous image of a connected set, and therefore connected. Commented Jul 19, 2015 at 4:30

It is homeomorphic to the corresponding Euclidean space, so...?

• f homeomorphism, I can say $S^1$ connected,but how do I conclude $S^1 - \{y_0\}$ connected as well? Commented Jul 19, 2015 at 4:13
• Pedro means that $S^1$ without a point is homeomorphic to $\mathbb{R}$ (via stereographic projection), there is connected, since the latter is connected. Commented Jul 19, 2015 at 4:21

I think this might work: consider $S^1 \subset \Bbb C$ as the unit circle, that is, the set of unimodular complex numbers $e^{i\theta}$, where $0 \le \theta < 2\pi$; note that even though the value $\theta = 2\pi$ is not allowed, every open set in $S^1$ which contains $1 = e^{0i}$ also contains all open arcs of the form

$\{ e^{i\alpha} \mid \alpha \in (2\pi - \epsilon, 0) \cup [0, \epsilon) \} \tag{1}$

for real $\epsilon > 0$ sufficiently small.

Pick $\theta_0 \in [0, 2\pi)$ and consider the space $S^1$ with $e^{i\theta_0}$ omitted; that is, $S^1 - \{ e^{i\theta_0} \}$. Further pick $\theta_0 \ne \theta_1, \theta_2 \in [0, 2\pi)$; without loss of generality, assume $\theta_1 < \theta_2$.

If $\theta _0 < \theta_1$ or $\theta_2 < \theta_0$, then for $t \in [0, 1$,

$(1 - t) \theta_1 + t \theta_2 = \theta_1 + t(\theta_2 - \theta_1) \in [\theta_1, \theta_2], \tag{2}$

so

$\theta_0 \ne (1 - t)\theta_1 + t \theta_2 \tag{3}$

for $t \in [0, 1]$. It follows that $e^{i((1 - t)\theta_1 + t\theta_2)}$ is a smooth path ´twixt $e^{i\theta_1}$ and $e^{i\theta_2}$ in $S^1 - \{ e^{i\theta_0} \}$.

In the event that $\theta_1 < \theta_0 < \theta_2$, we consider the three angles $\theta_1 - \theta_0 < 0 < \theta_2 - \theta_0$ obtained by subtracting $\theta_0$ from $\theta_1$, $\theta_0$, $\theta_2$ or equivalently rotating the configuration of angles on $S^1$ clockwise by an amount $\theta_0$. Since $\theta_1 - \theta_0 < 0$, it corresponds to the $\theta$-coordinate value $2\pi + (\theta_1 - \theta_0) \in [0, 2\pi)$, and since $\theta_2 < 2\pi$ and $0 \le \theta_1$ we have $\theta_2 < 2\pi + \theta_1$ implying

$0 < \theta_2 - \theta_0 < 2\pi + (\theta_1 - \theta_0) < 2\pi. \tag{4}$

It now follows by the previous remarks that

$e^{i((1 - t) (\theta_2 - \theta_0) + t(2\pi + (\theta_1 - \theta_0)))}, t \in [0, 1], \tag{5}$

is a path in $S^1 - \{ 1 \} = S^1 - \{ e^{i 0} \}$ which joins $e^{ i(\theta_2 - \theta_0)}$ with $e^{i(2\pi + (\theta_1 - \theta_0))}$. Finally, rotating everything counter-clockwise by an amount $\theta_0$, or equivalently multiplying (5) by $e^{i \theta_0}$, yields a path joining $e^{i \theta_2}$ with $e^{i \theta_1}$ in $S^1 - \{ e^{i \theta_0} \}$.

For all possible configurations of $\theta_0$, $\theta_1$, $\theta_2$, we see that $S^1 - \{ e^{i \theta_0} \}$ is path connected; and being path connected, it is in fact connected.

And that's how it may be shown!

QED!!!