We are given two sets \( A \) and \( B \) such that
\[ A \cap X = \varnothing,\quad B \cap X = \varnothing \] and \[ A \cup X = B \cup X, \] for some set \( X \).
We have to show that \[ A = B. \]
Using the given hint, we write
\[ A = A \cap (A \cup X). \]
Now apply the distributive law:
\[ A \cap (A \cup X) = (A \cap A) \cup (A \cap X). \]
Since \( A \cap A = A \) and \( A \cap X = \varnothing \), we get
\[ A = A \cup \varnothing = A. \]
Hence,
\[ A = (A \cap A) \cup (A \cap X). \]
Similarly, for set \( B \),
\[ B = B \cap (B \cup X). \]
Using distributive law,
\[ B \cap (B \cup X) = (B \cap B) \cup (B \cap X). \]
Since \( B \cap B = B \) and \( B \cap X = \varnothing \),
\[ B = B \cup \varnothing = B. \]
Given that \[ A \cup X = B \cup X, \] taking intersection with \( A \) on both sides, we get
\[ A \cap (A \cup X) = A \cap (B \cup X). \]
Using distributive law on the right side,
\[ A \cap (B \cup X) = (A \cap B) \cup (A \cap X). \]
Since \( A \cap X = \varnothing \),
\[ A = A \cap B. \]
Similarly, intersecting \( B \) with both sides of \( A \cup X = B \cup X \), we get
\[ B = A \cap B. \]
Therefore,
\[ A = A \cap B = B. \]
Hence, it is proved that
\[ \boxed{A = B}. \]
Consider the following subsets of the Euclidean space \( \mathbb{R}^4 \):
\( S = \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1^2 + x_2^2 + x_3^2 - x_4^2 = 0 \} \),
\( T = \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1^2 + x_2^2 + x_3^2 - x_4^2 = 1 \} \),
\( U = \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1^2 + x_2^2 + x_3^2 - x_4^2 = -1 \} \).
Then, which one of the following is TRUE?
Let the functions \( f: \mathbb{R} \to \mathbb{R} \) and \( g: \mathbb{R}^2 \to \mathbb{R} \) be given by \[ f(x_1, x_2) = x_1^2 + x_2^2 - 2x_1x_2, \quad g(x_1, x_2) = 2x_1^2 + 2x_2^2 - x_1x_2. \] Consider the following statements:
S1: For every compact subset \( K \) of \( \mathbb{R} \), \( f^{-1}(K) \) is compact.
S2: For every compact subset \( K \) of \( \mathbb{R} \), \( g^{-1}(K) \) is compact. Then, which one of the following is correct?