# Proving that a quotient space is compact but not Hausdorff

Let ∼ be the equivalence relation on $\mathbb{R^2}$ defined by $(x, y) ∼ (x_0 , y_0 )$ if and only if there is a nonzero $t$ with $(x, y) = (tx_0 , ty_0$ ). Prove that the quotient space $\mathbb{R^2}/ ∼$ is compact but not Hausdorff.

My attempt : $\mathbb{R^2}/ ∼$ is homeomorphic to $\{e^{i\theta} \ | \ 0 \leq \theta < \pi\} \bigcup \{(0,0)\}$ which is homeomorphic to $[0,1]$ (if we consider the standard topology on $\mathbb{R^2}$ and $\mathbb{R^2}/ ∼$) by the homeomorphism $f:\mathbb{R^2}/ ∼ \to [0,1]$ defined by $$f(e^{i\theta})=1-\frac{\theta}{\pi} \ \ \ \ \ \ 0 \leq \theta < \pi$$ and $$f((0,0))=0$$ This function is bijective and continuous (since preimage of any open set is open in $\mathbb{R^2}/ ∼$) with a continuous inverse $$f^{-1}: [0,1] \to \mathbb{R^2}/ ∼$$ $$f^{-1}(x)=e^{\pi(1-x)} \ \ \ if \ \ 0 < x \leq 1$$ and $$f^{-1}(0)=(0,0)$$ Because of this homeomorphism $\mathbb{R^2}/ ∼$ is compact. But in the standard topology this space is not Hausdorff , since any neighborhood of $(0,0) \in \mathbb{R^2}/ ∼$ contains every other point in $\mathbb{R^2}/ ∼$ .

Is my argument about compactness and Hausdorffness of $\mathbb{R^2}/ ∼$ correct ?

But I was thinking : If we give the discrete topology to $\mathbb{R^2}$, this topology induces discrete topology on $\mathbb{R^2}/ ∼$ as the quotient topology, in that case $\mathbb{R^2}/ ∼$ is indeed Hausdorff but NOT compact since $\mathbb{R^2}/ ∼$ is infinite.

• $\Bbb R^2/\sim$ can't be homeomorphic to $\{e^{i\theta} \ | \ 0 \leq \theta < \pi\} \bigcup \{(0,0)\}$; one is Hausdorff, but the other (purportedly) is not – Omnomnomnom Aug 11 '14 at 0:19

Note that the image of $D = \{(x,y) : x^2 + y^2 \leq 1\} \subset \Bbb R^2$ under the quotient map $\Bbb R^2 \to \Bbb R^2 /\sim$ is the entirety of the target space. That is, there is a continuous map from $D$ onto $\Bbb R^2 /\sim$. Note that $D$ is compact.
Your idea about the space not being Hausdorff is ostensibly correct; the space is not Hausdorff because every (open) neighborhood of $(0,0)/\sim$ is the entire space.
• Thank you so much !Can you please shortly explain why there is a continuous map from $D$ onto $\mathbb{R^2} / \sim$ ? – the8thone Aug 11 '14 at 2:49
• The quotient map from $\Bbb R^2 \to \Bbb R^2/\sim$ is continuous. So is its restriction to $D \subset \Bbb R^2$. – Omnomnomnom Aug 11 '14 at 12:29