To show that \[ A \cap B = A \cap C \] need not imply \[ B = C, \] we give a counterexample.
Let
\[ A = \{1\}, \quad B = \{1, 2\}, \quad C = \{1, 3\}. \]
Now, find the intersection of \( A \) with \( B \):
\[ A \cap B = \{1\} \cap \{1, 2\} = \{1\}. \]
Next, find the intersection of \( A \) with \( C \):
\[ A \cap C = \{1\} \cap \{1, 3\} = \{1\}. \]
Hence,
\[ A \cap B = A \cap C. \]
But clearly,
\[ B = \{1, 2\} \neq \{1, 3\} = C. \]
Therefore,
\[ A \cap B = A \cap C \;\text{need not imply}\; B = C. \]
Hence, the statement is proved by counterexample.
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?