# Why is this quotient of the punctured plane not Hausdorff (Hatcher 1.3.25)?

This is from question 1.3.25 of Hatcher's Algebraic Topology:

Let $$\phi : \mathbb{R}^2 \to \mathbb{R}^2$$ be the linear transformation $$\phi(x,y) = (2x, y/2)$$. This generates an action of $$\mathbb{Z}$$ on $$X = \mathbb{R}^2 - \{0\}$$. [...] Show that the orbit space $$X/\mathbb{Z}$$ is non-Hausdorff [...].

I think the idea is to note that $$(1,1)$$ and $$(1,0)$$ are in distinct orbits but that their orbits contain all points of the form $$(2^n,2^{-n})$$ and $$(2^n,0)$$, respectively, for all $$n \in \mathbb{Z}$$. Using the fact that the distance in $$\mathbb{R}^2$$ between $$(2^n,2^{-n})$$ and $$(2^n,0)$$ tends to $$0$$, we should be able to conclude that the quotient is non-Hausdorff.

However, I'm having trouble turning this into a rigorous proof. I don't see why this implies that there can't be neighbourhoods in the quotient that separate $$[(1,1)]$$ and $$[(1,0)]$$.

• Is the action the one that sends $n \in \mathbb{Z}$ and $p \in \mathbb{R}^2$ to $\phi^n(p)$? – preferred_anon Sep 27 at 15:11
• Yep, $p \in \mathbb{R}^2 - \{0\}$ to be exact – SFeesh Sep 27 at 15:13

There's a reason you're having trouble turning your idea into a rigorous proof: there actually are neighborhoods that separate $$[(1,1)]$$ and $$[(1,0)]$$! For instance, let $$U=\{(x,y):xy>1/2\}$$ and $$V=\{(x,y):xy<1/2\}$$. Then $$U$$ and $$V$$ are $$\mathbb{Z}$$-invariant open sets and their images in the quotient are disjoint neighborhoods of $$[(1,1)]$$ and $$[(1,0)]$$.
You can get a better intuitive idea for what's going on by thinking about an action of $$\mathbb{R}$$ rather than just an action of $$\mathbb{Z}$$: let $$t\in\mathbb{R}$$ act on $$(x,y)$$ by $$t\cdot (x,y)=(e^t x,e^{-t}y)$$. Then the orbits of this action are (almost) just branches of hyperbolas of the form $$xy=c$$. Intuitively, this would (almost) make the quotient space Hausdorff: such branches of hyperbolas form four continuous families (one in each quadrant), and each family looks like a copy of $$\mathbb{R}_+$$ (parametrized by the value of $$c$$). So, the quotient looks like a disjoint union of four copies of $$\mathbb{R}_+$$, which is Hausdorff.
But, why did I keep saying "(almost)"? There are four special orbits that are not branches of hyperbolas: the halves of the coordinate axes, which are the "branches" of the degenerate hyperbola $$xy=0$$. Intuitively, the positive branches of hyperbolas $$xy=c$$ as $$c\to 0^+$$ should be approaching both the positive $$x$$-axis and the positive $$y$$-axis. So, this would make the quotient non-Hausdorff, since the positive $$x$$-axis and positive $$y$$-axis could not have disjoint neighborhoods.
OK, now let's return back to the actual quotient in the problem where we have $$\mathbb{Z}$$ instead of $$\mathbb{R}$$, and make all this rigorous. Based on the discussion above, we should expect the problem to arise with points on the axes. So, for instance, can you show that $$[(0,1)]$$ and $$[(1,0)]$$ do not have disjoint neighborhoods? Instead of showing this directly, you might find it easier to exhibit a sequence that converges to both of them.
Consider the sequence of points $$(1/2^n,1)$$, which converge to $$(0,1)$$. But $$(1/2^n,1)$$ is in the same orbit as $$(1,1/2^n)$$ (apply $$\phi$$ $$n$$ times), and $$(1,1/2^n)\to (1,0)$$. So in the quotient, the sequence $$[(1/2^n,1)]$$ converges to both $$[(0,1)]$$ and $$[(1,0)]$$.