Let A and B be sets. If \(A ∩ X = B ∩ X = \phi\) and \(A ∪ X = B ∪ X\) for some set X, show that A = B. (Hints \(A = A ∩ (A ∪ X), B = B ∩ (B ∪ X)\) and use distributive law)
Solution and Explanation
Let A and B be two sets such that \(A ∩ X = B ∩ X = f\) and \(A ∪ X = B ∪ X\) for some set X.
To show: \(A = B \)
It can be seen that
\(A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X] \)
\(= (A ∩ B) ∪ (A ∩ X) \) \( [\)Distributive law] = \((A ∩ B) ∪ \phi [A ∩ X = \phi] \)
\(= A ∩ B\) …………………………………………………………….. (1)
Now, \(B = B ∩ (B ∪ X) \)
\(= B ∩ (A ∪ X) [A ∪ X = B ∪ X] \)
\(= (B ∩ A) ∪ (B ∩ X)\) [Distributive law]
\(= (B ∩ A) ∪ \phi [B ∩ X = \phi]\)
\(= B ∩ A \)
\(= A ∩ B\) …………………………………………………………… (2)
Hence, from (1) and (2), we obtain A = B.
Top Questions on Subsets
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?- In the following, all subsets of Euclidean spaces are considered with the respective subspace topologies. Define an equivalence relation \( \sim \) on the sphere \[ S = \left\{ (x_1, x_2, x_3) \in \mathbb{R}^3 : x_1^2 + x_2^2 + x_3^2 = 1 \right\} \] by \( (x_1, x_2, x_3) \sim (y_1, y_2, y_3) \) if \( x_3 = y_3 \), for \( (x_1, x_2, x_3), (y_1, y_2, y_3) \in S \). Let \( [x_1, x_2, x_3] \) denote the equivalence class of \( (x_1, x_2, x_3) \), and let \( X \) denote the set of all such equivalence classes. Let \( L : S \to X \) be given by \[ L\left( (x_1, x_2, x_3) \right) = [x_1, x_2, x_3]. \] If \( X \) is provided with the quotient topology induced by the map \( L \), then which one of the following is TRUE?
- Let \( X \) be an uncountable set. Let the topology on \( X \) be defined by declaring a subset \( U \subset X \) to be open if \( X - U \) is either empty or finite or countable, and the empty set to be open. Then, which of the following is/are TRUE?
- All rings considered below are assumed to be associative and commutative with \( 1 \neq 0 \). Further, all ring homomorphisms map 1 to 1.
Consider the following statements about such a ring \( R \):
P1: \( R \) is isomorphic to the product of two rings \( R_1 \) and \( R_2 \).
P2: \( \exists r_1, r_2 \in R \) such that \( r_1^2 = r_1 \neq 0 \), \( r_2^2 = 0 \), and \( r_1 + r_2 = 1 \).
P3: \( R \) has ideals \( I_1, I_2 \subset R \) with \( R \neq I_1 \), \( (0) \neq I_2 \), and \( R = I_1 + I_2 \) and \( I_1 \cap I_2 = (0) \).
P4: \( \exists a, b \in R \) with \( a \neq 0 \), \( b \neq 0 \) such that \( ab = 0 \).
Then, which of the following is/are TRUE?
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Concepts Used:
Operations on Sets
Some important operations on sets include union, intersection, difference, and the complement of a set, a brief explanation of operations on sets is as follows:
1. Union of Sets:
- The union of sets lists the elements in set A and set B or the elements in both set A and set B.
- For example, {3,4} ∪ {1, 4} = {1, 3, 4}
- It is denoted as “A U B”
2. Intersection of Sets:
- Intersection of sets lists the common elements in set A and B.
- For example, {3,4} ∪ {1, 4} = {4}
- It is denoted as “A ∩ B”
3.Set Difference:
- Set difference is the list of elements in set A which is not present in set B
- For example, {3,4} - {1, 4} = {3}
- It is denoted as “A - B”
4.Set Complement:
- The set complement is the list of all elements present in the Universal set except the elements present in set A
- It is denoted as “U-A”