5. For each pair of real numbers $(x, y)(0 \leqslant x \leqslant y \leqslant 1)$, consider the set
$$
A=\{x y, x y-x-y+1, x+y-2 x y\} .
$$

Let the maximum value of the elements in set $A$ be $M(x, y)$. Find the minimum value of $M(x, y)$.
5. Let $q_{0}=x y$,
$$
\begin{array}{l}
q_{1}=x y-x-y+1=(1-x)(1-y), \\
q_{2}=x+y-2 x y .
\end{array}
$$

Then $q_{0}, q_{1}, q_{2}$ are the areas of three figures within the unit square in Figure 1 determined by the point $(x, y)$ ($q_{2}$ is the sum of the areas of two rectangles).

From Figure 2, we know that when the point $(x, y)$ moves in a direction perpendicular to the diagonal $\{y=x\}$, the area of the shaded region decreases. Thus, $q_{2}$ decreases, while the areas of $q_{0}$ and $q_{1}$ increase. Therefore, we can assume that the point that minimizes $M(x, y)$ satisfies $q_{0}=\max A$ (or $q_{1}=\max A$) or lies on the diagonal $\{y=x\}$.

If $q_{0}=\max A$, fix $q_{0}$ as $q$, and let the point $\left(x_{0}, y_{0}\right)$ maximize $q_{1}$, with $y_{0}=x_{0}$. Geometrically, this means that the family of hyperbolas $\{x y=a \mid a>0\}$ and $\{(1-x)(1-y)=a \mid a>0\}$ (determined by $x, y>0$) have branches that face each other, and are perpendicular to the diagonal $\{y=x\}$.

Thus, the branch of the hyperbola $\{(1-x)(1-y)=a\}$ that intersects the hyperbola $\{x y=q\}$ at the maximum value $a$ intersects the diagonal $\{y=x\}$ at the point $(\sqrt{q}, \sqrt{q})$. This indicates that moving along the hyperbola $\{x y=q\}$ towards the diagonal $\{y=x\}$, $q_{0}$ remains $q$, while $q_{1}$ increases and $q_{2}$ decreases.
Since $q_{0} \geqslant \frac{1}{3}$, and $q_{1} \leqslant q_{0}$, when $x=y$,
$$
q_{1}<\frac{1}{3} \text{. }
$$

Therefore, the minimum value of $M(x, y)$ is obtained on the diagonal $\{y=x\}$.
Finally, consider the case when $x=y$.
If $x \leqslant \frac{1}{3}$, then the maximum value of $A$ is $q_{1}$. If $\frac{1}{3} \leqslant x \leqslant \frac{2}{3}$, then the maximum value of $A$ is $q_{2}$. If $x \geqslant \frac{2}{3}$, then the maximum value of $A$ is $q_{0}$. Hence, when $(x, y)=\left(\frac{1}{3}, \frac{1}{3}\right),\left(\frac{2}{3}, \frac{2}{3}\right)$, the minimum value of $M(x, y)$ is $\frac{4}{9}$.