Prove that $A=${$(x,y) \in \mathbb{R}^2 | xy=0$} with the subspace topology of $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}$.
Prove that $A$ with the subspace topology of $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}$.
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general-topology
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0http://math.stackexchange.com/q/1232290/9464 – 2017-01-05
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0I think that $\mathbb{R}-{0}$ is connected and $A-{f(0,0)}$ is not connected, where $f$ is an application from $\mathbb{R}$ to $A$ @Jack – 2017-01-05
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0How many parts are if you remove $(0,0)$? What about the corresponding point in $\mathbb{R}$? – 2017-01-05
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0I restrict f in $\mathbb{R} -0$ @copper.hat – 2017-01-05
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0A homeomorphism will preserve connected components. – 2017-01-05
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0How would you do it? @copper.hat – 2017-01-05
1 Answers
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Suppose there is such a homeomorphism $f$.
Consider $A \setminus \{0\}$ and $\mathbb{R} \setminus \{f(0)\}$.
The first has 4 connected components, the second has two.
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0Ok. Thanks for help! – 2017-01-05