Proving connectedness of punctured disc without using path-connectedness Using path-connectedness it is easy to see that the punctured disc $$D:=\lbrace(x,y)\in\mathbb{R}^2:0<x^2+y^2<1\rbrace$$ is connected. I was wondering if there is a proof that $D$ is a connected set of $\mathbb{R}^2$ from the definition of connectedness, or without using path-connectedness ?
 A: The intervals $(0,1),[0,2\pi]$ are connected. Hence $(0,1)\times[0,2\pi)$ is connected. Consider the function $f:(0,1)\times[0,2\pi)\rightarrow D$ that sends $(r,\theta)$ to $(r\cos\theta,r\sin\theta)$. It is easy to verify that $f$ is continuous and surjective. Thus $D$ is the continuous image of $f$. Hence $D$ is connected.
A: Hint: Prove that it is the continuous image of a connected space.
A: *

*Identify the plane $ \mathbb{R}^{2} $ with $ \mathbb{C} $.

*The punctured plane is the image of the plane under the exponential function.

*As $ \mathbb{C} $ is connected and $ \exp $ is continuous, the punctured plane is thus connected.

*The punctured disk is homeomorphic to the punctured plane. Therefore, the punctured disk is connected as well.
Note: As $ \mathbb{R} $ is connected, we see that the plane $ \mathbb{R}^{2} $, and hence $ \mathbb{C} $, is connected in the first place.
A: Here I found another soloution based on complex exponentiation: Let's consider the set $S:=(-\infty ,0]\times[0,2\pi)$ which is connected and $f:S\to D$ is defined as $$f(x,y)=(e^x\cos y,e^x\sin y)$$ Then $f$ is continous and $f(S)=D$.
