Chronology condition and metric perturbations 
Let $(M,g)$ be the quotient of the 2-dimensional Minkowski space-time by the discrete group of isometries generated by the map $f(t, x) = (t + 1, x + 1)$. Show that $(M, g)$ satisfies the chronology condition, but there exist arbitrarily small perturbations of $(M, g)$ of the form $g_{\epsilon}=g+\epsilon h$ (where $h$ is an arbitrary symmetric (2, 0)-tensor field with compact support),  which do not.

I was trying to do this proof and I can understand that indeed $(M,g)$ is chronological, but I don't get how to perturb it in order to eliminate this condition.
 A: Using the definition from Wu-Sachs:

By definition $(M,g)$ obeys the chronology condition iff no points in $M$ chronologically precedes itself (Rougthly: ancestorcide is impossible)

Now $M$ is diffeomorphic to a strip, say $(t,x)$ with $0\le t <1$ and $x\in \mathbf{R}$. The lines $t=0$ and $t=1$ are identified by $(t=0,x) \sim (t=1,x+1)$.
You can put a point in its past then by spreading the lightcone. Let us see how.
Consider a constant perturbation of the metric, with $h = \operatorname{diag}(0,-1)$. As a remark, I'm here using the convention of the Minkowski metric $\eta = \operatorname{diag}(-1,+1)$. With the metric $g_\epsilon = \eta + \epsilon h = \operatorname{diag}(-1,1-\epsilon)$, the lightrays are described by curves $x = x_0 \pm \frac{1}{\sqrt{1-\epsilon}}t$, with $0\le t<1$.
Now start with the origin $(t=0,x=0)$. With the metric $g_\epsilon$, it is then chronologically in the past of the cone $(t,x)$ with 
$|x| \le \frac{1}{\sqrt{1-\epsilon}}t \cong t [1 + \frac{\epsilon}{2}+\ldots]$ for $0<t<1$.
Since $\epsilon$ is positive, that means the origin is in the chronological past to a point slightly at the right of $(t=1,x=1)$, say a point $(t=1, x= 1+\delta)$ with $0<\delta < \frac{1}{\sqrt{1-\epsilon}}-1$. Now  $(t=1, x= 1+\delta) \sim (t=0, x= \delta)$, which is in the chronological past of the point $(t=1,x=1) \sim (t=0,x=0)$. Hence the origin is in the chronological past of itself, for any $\epsilon >0$.
