77. Find the set $X$, if $X \cap A = X \cup A$, where $A$ is a given set.
$\triangle$ The given equality is an example of an equation with a set. In the future, we will encounter systems of such equations.

Based on property 2 of operations over sets (p. 29)

$$
X \cap A \subset X \subset X \cup A
$$

But since the extreme sets $X \cap A$ and $X \cup A$ coincide here, all three sets coincide:

$$
X \cap A = X = X \cup A
$$

When does the equality $X \cap A = X$ hold? Since $X \cap A \subset A$, substituting $X \cap A$ with the set $X$, we get that $X \subset A$.

![](https://cdn.mathpix.com/cropped/2024_05_21_0aa0e6ac168e80891953g-036.jpg?height=576&width=1334&top_left_y=1825&top_left_x=361)

35

When does the equality $X \cup A = X$ hold? Since $A \subset X \cup A$, then $A \subset X$.

But the inclusions $X \subset A$ and $A \subset X$ hold simultaneously only when $X = A$.

Strictly speaking, we need to verify the obtained answer. When $X = A$, based on property 1

$$
X \cap A = A, \quad X \cup A = A,
$$

and then the given equation becomes an identity.

Answer: $X = A$.