Proving connectedness of punctured disk using "disconnectedness" definition I was wondering if there is a proof that $H = \{z \in \mathbb{C} : 0 < |z| < 1\}$ is a connected set of $\mathbb{C}$ from the "topological" definition of connectedness?
This is what I have so far: Suppose BWOC that $H$ is not connected. Then there exist open sets $U$ and $V$ such that $U \cap H \neq \varnothing, V \cap H \neq \varnothing, \ H \subseteq U \cup V, \ \mathrm{and} \ U \cap V = \varnothing.$
Now I have no idea how to contradict the existence of such sets $U$ and $V$!!
 A: You won't manage to do it using only topological properties. In general, it is false that if $(X, \tau)$ is a connected topological space and $p \in X$, then $(X \setminus\{p\}, \tau_{X \setminus\{p\}})$ is connected. The catch here is that in $\Bbb C$, connectedness is equivalent to path-connectedness, for open sets. Once you have this result, it is easy to prove that your set there is path-connected, hence connected.

A subset $H \subset \Bbb C$ is path-connected if given $p,q \in H$, there exists $\gamma: [a,b] \to H$ continuous such that $\gamma(a) = p$ and $\gamma(b) = q$. Let $H$ be the set in your problem. Take $p$ and $q$ in $H$. If $p,q$ and $0$ are not collinear, you can take $\gamma:[0,1]\to H$ given by $\gamma(t) = p + t(q-p)$, that is, join the points by a straight line. If said line would contain $0$, which we don't want, take a little detour: the point $1/2$ is in $H$. Then consider $\gamma:[0,2] \to H$ given by: $$\gamma(t) = \begin{cases} p + t\left(\frac{1}{2}-t\right),\, &\text{if } 0 \leq t \leq 1 \\ \frac{1}{2}+(t-1)\left(q-\frac{1}{2}\right) ,&\text{if } 1 \leq t \leq 2\end{cases}$$
Then $\gamma$ is well-defined, it is continuous, and $\gamma(0)=p$, $\gamma(2) = q$. I leave to you the verification that all the points in $\gamma(t)$ really are inside $H$. This ensures that $H$ is path-connected.
A: Since this has a complex analysis tag, it might be worth pointing out that $\{0<|z|<1\}= f(U),$ where $f(z) = e^z$ and $U$ is the open left half plane. Hence $\{0<|z|<1\}$ is connected.
