Given $f\colon X \times Y \to \mathbb{R}$, prove that \begin{align*} \min_{(x,y) \in X \times Y} f(x,y) = \min_{x \in X}\left( \min_{y \in Y} f(x,y) \right). \end{align*}
Intuitively, I see that the left hand side is the minimum value over all possible pairs $(x,y)$ in $X \times Y$. Say that the minimiser is $(x^o,y^o)$.
For the right hand side, we pick $x = x_1$, then find a $y=y_1$ which minimises $f(x_1,y)$, then repeat this process for all $x \in X$. Now in this process when $x = x^o$, $y$ will take value $y^o$, since that is the minimum value the function can take for any possible pair. Hence they equality seems true.
How do I show this rigorously?
