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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 ?

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4 Answers 4

up vote 2 down vote accepted

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.

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You probably want $(0,1)\times[0,2\pi]$ instead. – T. Eskin Jan 1 '13 at 15:19
Yes. you are right. Tthanks – Amr Jan 1 '13 at 15:20
Woudn't it be $f:(0,1)\times [0,2\pi)\to D$ ? – pritam Jan 1 '13 at 15:32
There is no need to ensure that $f$ is injective – Amr Jan 1 '13 at 15:36
@pritam, yes I thought I wrote $[0,2\pi]$ – Amr Jan 1 '13 at 15:37
  • 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.

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C=R*R R is connected hence C – Koushik Jan 1 '13 at 15:23
@Haskell Curry, did part of your answer get chopped off? – Amzoti Jan 1 '13 at 15:24
Yes. Edited. Thanks! – Haskell Curry Jan 1 '13 at 15:24
@Azmoti: Having chopped-off answers is the consequence of writing math with the touchscreen keypad of an iPad. :( – Haskell Curry Jan 1 '13 at 15:58

Hint: Prove that it is the continuous image of a connected space.

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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 $f(S)=D$.

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