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I am working on the old preliminary exam from my university. I found trouble to solve the following problem. Could you please help me on it. Let $X = \mathbb R^2/{\sim}$ be quotient space where $x\sim y$ iff $x = cy$ for real positive $c$.

a) prove that any closed subset of $X$ must contain the origin.

b) $X$ is not Hausdorff

c) $X$ is normal

To prove the first part I tried with contradiction: Let $A$ be nonempty closed subset of $X$ not containing the $[0]$, then $X-A$ is open in $X$ and inverse image of quotient map contains the nbhd of origin, that is it contains the point of every equivalence class. hence it is contradiction as it also contains the point of $A$. But I am not sure about this. Could you please help me?

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  • $\begingroup$ In part a), the restriction that the closed subset be non-empty is missing. $\endgroup$ – Daniel Fischer Aug 7 '14 at 22:18
  • $\begingroup$ @Daniel coul you please share me your idea? $\endgroup$ – Ganga Aug 7 '14 at 22:30
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In the list of properties to prove, you or the poser of the exam forgot the restriction that the closed subset be non-empty in part a). You have stated that demand in your proof of part a), so it could be that you just forgot to type that word in the question.

Your proof for part a) is correct, but in my opinion not formulated clearly. It took me a bit to parse

then $X−A$ is open in $X$ and inverse image of quotient map contains the nbhd of origin, that is it contains the point of every equivalence class.

I would suggest that you name the canonical projection $\pi \colon \mathbb{R}^2 \to X$ and $U = X\setminus A$.

Let $A\subset X$ a closed set not containing $[0]$. Then $U = X\setminus A$ is an open neighbourhood of $[0]$, and hence $V = \pi^{-1}(U)$ is an open neighbourhood of $\pi^{-1}([0]) = 0$. Thus there is a $\delta > 0$ such that $B_\delta(0) \subset V$, and since $V$ is saturated, $B_{t\cdot \delta}(0) = t\cdot B_\delta(0) \subset V$ for all $t > 0$. But $$\bigcup_{t>0} B_{t\delta}(0) = \mathbb{R}^2,$$

so $U = \pi(V) = \pi(\mathbb{R}^2) = X$, and $A = X\setminus U = \varnothing$.

This result also directly implies the parts b) and c), you just have to explain why it does.

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