Maximum likelihood estimator of Cauchy distribution but with a catch I have an exercise to solve that states that we need to find the Maximum likelihood estimator of location parameter of the Cauchy distribution given a set X={x1,x2} , |x1-x2|<2
Now I worked using this approach: Maximum likelihood estimator of location parameter of Cauchy distribution
but I can't get to grasp how " given a set X={x1,x2} , |x1-x2|<2" changes the problem.
Thank you

From symmetry principle location can only be $\frac{x_1+x_2} 2$, when there is a unique solution.
this is regardless of the estimation approach. For instance, a particular estimator, such as MLE, may not even exist in every case. However, if it does exist and it is unique, then the symmetry dictates what should it be.
non MLE estimation
If we don't constrain ourselves with MLE, then we could get the parameters differently. Obviously, the peak (center) is still the same half-way between $x_1$ and $x_2$. So, we only need to figure the other parameter.
In physics we used to characterize Breit-Wigner function with half-width at half-height parameter, which is equal to a shape parameter $\gamma$ of Cauchy distribution definition in statistics: PDF reaches its half height when you step left or right by half-width from the peak. So, naturally, the "width" is $2\gamma$ and intuitively it should be estimated as $2\hat\gamma=|x_1-x_2|$.
The "height" of PDF is at its peak $x_0$:$$H=\frac 1 {\pi\gamma}$$ Solve the PDF for $x$ to equal $H/2$: $$\frac H 2=\frac 1 {2\pi\gamma}=\frac 1 {\pi\gamma\left[1+\left(\frac{x-x_0} \gamma\right)^2\right]}$$ to get $$\frac{x-x_0}\gamma=1$$ So when the variable is $x_0\pm\gamma$ the PDF reaches its half-height
