# Is min max equal to max min for this special problem?

Assume we have three finite sets $A_1, A_2, A_3$. It seems to me that two following optimization problems are equivalent. Am I right? if I am, how can we show that?

$\min_{x_1 \in A_1, x_2 \in A_2, x_3 \in A_3} \max_{i \in \{1,2,3\}} x_i = \max_{i \in \{1,2,3\}} \min_{x_i \in A_i} x_i$

What I have tried:

for any triple $(x_1,x_2,x_3) \in (A_1, A_2, A_3)$ it is true that $\max_{i \in \{1,2,3\}} x_i \geq x_j \quad \forall j \in \{1,2,3\}$. So it still holds when we minimize over $A_1, A_2, A_3$, that is, $\min_{x_1 \in A_1, x_2 \in A_2, x_3 \in A_3} \max_{i \in \{1,2,3\}} x_i \geq \min_{x_j \in A_j} x_j \quad \forall j \in \{1,2,3\}$. Finally, since the last inequality is true for any $j \in \{1,2,3\}$, it is true when me maximize over the right hand side of inequality: $\min_{x_1 \in A_1, x_2 \in A_2, x_3 \in A_3} \max_{i \in \{1,2,3\}} x_i \geq \max_{j \in \{1,2,3\}} \min_{x_j \in A_j} x_j$.

If the above proof is correct, this shows one side. What can we do to show the other side, that is $\min_{x_1 \in A_1, x_2 \in A_2, x_3 \in A_3} \max_{i \in \{1,2,3\}} x_i \leq \max_{j \in \{1,2,3\}} \min_{x_j \in A_j} x_j$?

Since the $\max_{i\in {1,2,3}} x_i$ is non decreasing in each of $x_i$, it is minimized when each of $x_i$ is minimized individually. That is, $$\min_{x_1\in A_1, x_2\in A_2, x_3\in A_3} \max_{i\in {1,2,3}} x_i = \max_{i\in\{1,2,3\}} \min(A_i)\\=\max_{i\in\{1,2,3\}}\min_{x_i \in A_i} x_i.$$