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This question came up in a test from the previous year and i have no idea how to begin.

We define the following equivalence relation on $\mathbb{R}^2$: $(x_1,y_1) \approx(x_2,y_2) \iff \exists t>0$ so that $x_2 = tx_1$ and $y_2=y_1/t$.

Let Y = $\mathbb{R}^2\mod\approx$

Prove / Disprove: Y is Hausdorff.

I cant seem to visualize what this quotient would even look like. Any hints, tips will be greatly appreciated, as well as straight up solutions. Thank you!

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    Hint: Generic equivalence classes are graphs of functions $y=f(x)=c/x$. – 2017-02-08

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Let's get some ideas of which points get glued together. Take $(x_1, y_1) = (1,1)$. We will glue to this point all pairs $(x,y)$ such that $x = t$ and $y = 1/t$. That is to say, we are gluing together along hyperbolas of the form $f(x) = k/x$ for an arbitrary real number $k$. On the other hand, we can also look at points on the axes, like $(0,1)$. This gets glued to points of the form $x = 0, y = 1/t$, where $t$ is real, so in other words, the $y$ axis gets identified, except for the origin. Similarly for the $x$ axis.

Now we know that hyperbolas can be separated, since the plane is normal, so this is not an issue. The question remains to deal with the axes. Let one of the points be the origin - this is its own class. Can we separate it from the image of the axes with disjoint open sets?

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    No! We cant! Because i cant think of a way how! – 2017-02-08
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    A set is open in the quotient topology if and only if it is open in the preimage. So we need an open set containing the punctured axis that doesn't contain the origin. This is clearly impossible. So the set is not Hausdorff. – 2017-02-08