I'm trying to prove that the time order is invariant in a lightlike interval, I have this
$$P_1=(X_1^0, X_1^1,0,0)$$ $$P_2=(X_2^0, X_2^1,0,0)$$
Two lightlike events in a inertial frame where $t_2>t_1$. Under Lorentz transformations from a $\Sigma '$ frame I have
$$\Delta t' = \gamma(v)(\Delta t-\frac{v}{c^2}\Delta x)$$
Since the events are lightlike $c^2\Delta t^2=\Delta x^2$, where $\Delta t>0$, previously I got that, since $\Delta t>0$, $\Delta x>0$ as well, but, back in the equation above, it doesn't guarantee $\Delta t'>0$, I realized that if $\Delta x<0$ then $\Delta t'>0$, should it be like that? How do I get $\Delta x<0$ knowing the events are lightlike?
EDIT: I did this
Since the events are likelight and $c^2\Delta t^2=\Delta x^2$, we can just get $\Delta x$ and substitute this into the Lorentz transform expression above, having
$$\Delta t' = \gamma(v)\left(\Delta t-\frac{v}{c^2}(c\Delta t)\right)$$
Then
$$\Delta t' = \gamma(v)\Delta t\left(1-\frac{v}{c}\right)$$
Now, from the velocity-addition formula, $v<c$, so $\frac{v}{c}<1$ and $\left(1-\frac{v}{c}\right)$ is always greater than zero, thus, since the Lorentz gamma factor is also grater than zero as well as the interval $\Delta t$ (from the problem)
$$\Delta t'>0$$
Finally, the time-order is invariant under Lorentz transformation, thus time-order is a relativistic invariant in lightlike events as well as timelike events
