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 $[(1,0)]$.

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

More details are hidden below.

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 $[(1,0)]$.

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