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- #1

- Jan 29, 2012

- 661

Show that ℜ^2 - (0,0) in the standard topology is homemorphic to S^1 x ℜ? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter Fernando Revilla
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- Thread starter
- #1

- Jan 29, 2012

- 661

Show that ℜ^2 - (0,0) in the standard topology is homemorphic to S^1 x ℜ? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter
- #2

- Jan 29, 2012

- 661

We can express $\mathbb R^2 \setminus \{(0,0)\}$ as disjoint union of circles: $$\mathbb R^2 \setminus \{(0,0)\}=\displaystyle\bigcup_{r\in (0,+\infty)}C_r\;,\qquad C_r=\{(x,y)\in\mathbb{R}^2:x^2+y^2=r^2\}$$ This is equivalent to say that $\mathbb R^2 \setminus \{(0,0)\}$ is homeomorphic to $S^1 \times (0,+\infty)$. Now, use that $(0,+\infty)$ is homeomorphic to $\mathbb{R}$.

If you have further questions, you can post them in the Topology and Advanced Geometry section.