We are required to prove the following identities using properties of sets:
(i) \[ A \cup (A \cap B) = A \] (ii) \[ A \cap (A \cup B) = A. \]
Proof of (i):
Starting with the left-hand side,
\[ A \cup (A \cap B). \]
Using the distributive law of union over intersection,
\[ A \cup (A \cap B) = (A \cup A) \cap (A \cup B). \]
Using the idempotent law,
\[ A \cup A = A. \]
Therefore,
\[ A \cup (A \cap B) = A \cap (A \cup B). \]
Using the absorption law,
\[ A \cap (A \cup B) = A. \]
Hence,
\[ A \cup (A \cap B) = A. \]
Proof of (ii):
Starting with the left-hand side,
\[ A \cap (A \cup B). \]
Using the distributive law of intersection over union,
\[ A \cap (A \cup B) = (A \cap A) \cup (A \cap B). \]
Using the idempotent law,
\[ A \cap A = A. \]
Therefore,
\[ A \cap (A \cup B) = A \cup (A \cap B). \]
Using the absorption law,
\[ A \cup (A \cap B) = A. \]
Hence,
\[ A \cap (A \cup B) = A. \]
Thus, both the identities are proved.
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?