Miscellaneous Exercise on Chapter 1.php
Maths - Exercise
Q1. Decide, among the following sets, which sets are subsets of one and another:
\(\mathrm{A = \{ x : x \in \mathbb{R}\text{ and x satisfy }x^2 – 8x + 12 = 0 \}}\),
B = { 2, 4, 6 },
C = { 2, 4, 6, 8, . . . }, D = { 6 }.
Solution
We first determine the elements of each set explicitly in order to compare them.
\[ \begin{aligned} A &= \{x : x \in \mathbb{R} \text{ and } x^2 - 8x + 12 = 0\} \end{aligned} \]
Solving the quadratic equation, \[ \begin{aligned} x^2 - 8x + 12 &= 0 \\ (x-2)(x-6) &= 0 \end{aligned} \] we obtain \(x = 2\) or \(x = 6\). Hence, \[ A = \{2,6\} \]
The remaining sets are given as \[ \begin{aligned} B &= \{2,4,6\}, \\ C &= \{2,4,6,8,\ldots\}, \\ D &= \{6\} \end{aligned} \]
Since every element of \(A\), namely \(2\) and \(6\), is contained in \(B\) and also in \(C\), we conclude that \[ A \subset B \quad \text{and} \quad A \subset C \]
The set \(D = \{6\}\) contains a single element, which is present in all the sets \(A\), \(B\), and \(C\). Therefore, \[ D \subset A, \quad D \subset B, \quad \text{and} \quad D \subset C \]
However, the element \(4\) belongs to \(B\) and \(C\) but not to \(A\), so \(B\) is not a subset of \(A\), and similarly \(C\) is not a subset of \(A\). Also, since \(8 \in C\) but \(8 \notin B\), \(C\) is not a subset of \(B\).
Thus, the subset relationships are \(D \subset A \subset B \subset C\), with no reverse containment among these sets.
Q2. In each of the following, determine whether the statement is true or false. If it is
true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B , then x ∈ B
(ii) If A ⊂ B and B ∈ C , then A ∈ C
(iii) If A ⊂ B and B ⊂ C , then A ⊂ C
(iv) If A ⊄ B and B ⊄ C , then A ⊄ C
(v) If x ∈ A and A ⊄ B , then x ∈ B
(vi) If A ⊂ B and x ∉ B , then x ∉ A
Solution
We examine each statement carefully using the definitions of membership and subset relations.
\[ \begin{aligned} \text{(i)}\quad \text{If } x \in A \text{ and } A \in B, \text{ then } x \in B \end{aligned} \]
This statement is false. Membership of \(x\) in \(A\) does not imply membership of \(x\) in \(B\) merely because \(A\) itself is an element of \(B\). For example, let \(A=\{1,2\}\), \(B=\{\{1,2\},3\}\), and \(x=1\). Then \(x \in A\) and \(A \in B\), but \(x \notin B\).
\[ \begin{aligned} \text{(ii)}\quad \text{If } A \subset B \text{ and } B \in C, \text{ then } A \in C \end{aligned} \]
This statement is false. A subset of an element of a set need not itself be an element of that set. For instance, let \(A=\{1\}\), \(B=\{1,2\}\), and \(C=\{\{1,2\}\}\). Then \(A \subset B\) and \(B \in C\), but \(A \notin C\).
\[ \begin{aligned} \text{(iii)}\quad \text{If } A \subset B \text{ and } B \subset C, \text{ then } A \subset C \end{aligned} \]
This statement is true. Since every element of \(A\) belongs to \(B\) and every element of \(B\) belongs to \(C\), it follows directly that every element of \(A\) also belongs to \(C\). Hence, \(A \subset C\).
\[ \begin{aligned} \text{(iv)}\quad \text{If } A \nsubseteq B \text{ and } B \nsubseteq C, \text{ then } A \nsubseteq C \end{aligned} \]
This statement is false. Let \(A=\{1,2\}\), \(B=\{2,3\}\), and \(C=\{1,2,3\}\). Then \(A \nsubseteq B\) and \(B \nsubseteq C\) is false here, but modifying slightly with \(C=\{1,2\}\), we get \(B \nsubseteq C\) and yet \(A \subset C\). Hence the conclusion does not always follow.
\[ \begin{aligned} \text{(v)}\quad \text{If } x \in A \text{ and } A \nsubseteq B, \text{ then } x \in B \end{aligned} \]
This statement is false. Let \(A=\{1,2\}\), \(B=\{2\}\), and take \(x=1\). Then \(x \in A\) and \(A \nsubseteq B\), but \(x \notin B\).
\[ \begin{aligned} \text{(vi)}\quad \text{If } A \subset B \text{ and } x \notin B, \text{ then } x \notin A \end{aligned} \]
This statement is true. If \(x\) were an element of \(A\), then since \(A \subset B\), it would follow that \(x \in B\). This contradicts the given condition. Therefore, \(x \notin A\).
Q3. Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.
Solution
We are given that \[ A \cup B = A \cup C \quad \text{and} \quad A \cap B = A \cap C. \] Our aim is to show that \(B = C\).
Let \(x\) be an arbitrary element of \(B\). We consider two possible cases.
First, suppose \(x \in A\). Then \(x \in A \cap B\). Since \(A \cap B = A \cap C\), it follows that \(x \in A \cap C\), and hence \(x \in C\).
Next, suppose \(x \notin A\). Since \(x \in B\), we must have \(x \in A \cup B\). Using the equality \(A \cup B = A \cup C\), this implies that \(x \in A \cup C\). As \(x \notin A\), it follows that \(x \in C\).
Thus, in both cases, every element of \(B\) belongs to \(C\), and hence \(B \subset C\).
By applying the same argument with the roles of \(B\) and \(C\) interchanged, we similarly obtain \(C \subset B\).
Since \(B \subset C\) and \(C \subset B\), we conclude that \[ B = C. \]
Q4. Show that the following four conditions are equivalent :
(i) A ⊂ B
(ii) A – B = φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Solution
We show that each condition implies the others, thereby establishing their equivalence.
Assume first that \[ \begin{aligned} \text{(i)}\quad A \subset B. \end{aligned} \] If every element of \(A\) belongs to \(B\), then there is no element of \(A\) that lies outside \(B\). Hence, \[ \begin{aligned} A - B = \varnothing, \end{aligned} \] which proves condition (ii).
Now assume \[ \begin{aligned} \text{(ii)}\quad A - B = \varnothing. \end{aligned} \] This means there is no element of \(A\) that is not in \(B\). Therefore, every element of \(A\) is already contained in \(B\), and adding elements of \(A\) to \(B\) does not enlarge \(B\). Thus, \[ \begin{aligned} A \cup B = B, \end{aligned} \] which gives condition (iii).
Next, assume \[ \begin{aligned} \text{(iii)}\quad A \cup B = B. \end{aligned} \] Since the union does not introduce any new elements beyond those already in \(B\), all elements of \(A\) must be contained in \(B\). Consequently, the common elements of \(A\) and \(B\) are exactly the elements of \(A\), and hence, \[ \begin{aligned} A \cap B = A, \end{aligned} \] which establishes condition (iv).
Finally, assume \[ \begin{aligned} \text{(iv)}\quad A \cap B = A. \end{aligned} \] This equality shows that every element of \(A\) is also an element of \(B\). Therefore, \[ \begin{aligned} A \subset B, \end{aligned} \] which is condition (i).
Since \((i) \Rightarrow (ii) \Rightarrow (iii) \Rightarrow (iv) \Rightarrow (i)\), all four conditions are equivalent.
Q5. Show that if A ⊂ B, then C – B ⊂ C – A.
Solution
We are given that \(A \subset B\). We show that every element of \(C - B\) also belongs to \(C - A\).
Let \(x\) be an arbitrary element of \(C - B\). Then, by definition of set difference, \[ \begin{aligned} x &\in C \\ x &\notin B \end{aligned} \]
Since \(A \subset B\), every element of \(A\) is also an element of \(B\). Therefore, if \(x \notin B\), it is impossible for \(x\) to belong to \(A\). Hence, \[ \begin{aligned} x \notin A. \end{aligned} \]
Combining the results, we have \[ \begin{aligned} x \in C \quad \text{and} \quad x \notin A, \end{aligned} \] which implies \[ \begin{aligned} x \in C - A. \end{aligned} \]
Since every element of \(C - B\) is also an element of \(C - A\), we conclude that \[ \begin{aligned} C - B \subset C - A. \end{aligned} \]
Q6. Show that for any sets A and B, A = ( A ∩ B ) ∪ ( A – B ) and A ∪ ( B – A ) = ( A ∪ B )
Solution
We prove each identity separately using the basic definitions of intersection, union, and set difference.
First, we show that \[ A = (A \cap B) \cup (A - B). \]
Let \(x \in A\). Then either \(x \in B\) or \(x \notin B\). If \(x \in B\), then \(x \in A \cap B\). If \(x \notin B\), then \(x \in A - B\). Hence, every element of \(A\) belongs to \((A \cap B) \cup (A - B)\).
Conversely, let \(x \in (A \cap B) \cup (A - B)\). Then \(x \in A \cap B\) or \(x \in A - B\). In both cases, \(x \in A\). Therefore, \[ \begin{aligned} (A \cap B) \cup (A - B) = A. \end{aligned} \]
Next, we prove that \[ A \cup (B - A) = A \cup B. \]
Let \(x \in A \cup (B - A)\). Then either \(x \in A\) or \(x \in B - A\). In the second case, \(x \in B\). Thus, in both situations, \(x \in A \cup B\), which shows that \[ A \cup (B - A) \subset A \cup B. \]
Conversely, let \(x \in A \cup B\). If \(x \in A\), then clearly \(x \in A \cup (B - A)\). If \(x \in B\) and \(x \notin A\), then \(x \in B - A\), and hence \(x \in A \cup (B - A)\). Thus, \[ A \cup B \subset A \cup (B - A). \]
Since both inclusions hold, we conclude that \[ \begin{aligned} A \cup (B - A) = A \cup B. \end{aligned} \]
Q7. Using properties of sets, show that
(i) A ∪ ( A ∩ B ) = A
(ii) A ∩ ( A ∪ B ) = A.
Solution
We verify each identity by applying basic properties of sets and logical reasoning.
First, we show that \[ A \cup (A \cap B) = A. \]
Let \(x \in A \cup (A \cap B)\). Then either \(x \in A\) or \(x \in A \cap B\). In both cases, \(x \in A\). Hence, \[ A \cup (A \cap B) \subset A. \]
Conversely, let \(x \in A\). Clearly, \(x \in A \cup (A \cap B)\). Therefore, \[ A \subset A \cup (A \cap B). \]
Since both inclusions hold, we conclude that \[ \begin{aligned} A \cup (A \cap B) = A. \end{aligned} \]
Next, we prove that \[ A \cap (A \cup B) = A. \]
Let \(x \in A \cap (A \cup B)\). Then \(x \in A\) and \(x \in A \cup B\). The first condition already ensures that \(x \in A\). Hence, \[ A \cap (A \cup B) \subset A. \]
Conversely, let \(x \in A\). Since \(A \subset A \cup B\), it follows that \(x \in A \cup B\). Thus, \[ x \in A \cap (A \cup B), \] which implies \[ A \subset A \cap (A \cup B). \]
Combining both results, we obtain \[ \begin{aligned} A \cap (A \cup B) = A. \end{aligned} \]
Q8. Show that A ∩ B = A ∩ C need not imply B = C.
Solution
To show that the condition \(A \cap B = A \cap C\) does not necessarily imply \(B = C\), it is sufficient to provide a suitable counterexample.
Consider the sets \[ \begin{aligned} A &= \{1\}, \\ B &= \{1,2\}, \\ C &= \{1,3\}. \end{aligned} \]
Now, we evaluate the intersections with \(A\).
\[ \begin{aligned} A \cap B &= \{1\} \cap \{1,2\} = \{1\}, \\ A \cap C &= \{1\} \cap \{1,3\} = \{1\}. \end{aligned} \]
Thus, \[ A \cap B = A \cap C. \]
However, the sets \(B\) and \(C\) are clearly not equal since \(2 \in B\) but \(2 \notin C\), and \(3 \in C\) but \(3 \notin B\). Hence, \[ B \ne C. \]
This example shows that even if \(A \cap B = A \cap C\), it does not follow that \(B = C\).
Q9. Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B
Solution
We are given that \[ A \cap X = \varnothing,\quad B \cap X = \varnothing \quad \text{and} \quad A \cup X = B \cup X. \] We show that \(A = B\).
Let \(x\) be an arbitrary element of \(A\). Then \(x \in A \cup X\). Since \(A \cup X = B \cup X\), it follows that \[ \begin{aligned} x \in B \cup X. \end{aligned} \]
Thus, either \(x \in B\) or \(x \in X\). But \(A \cap X = \varnothing\), so no element of \(A\) can belong to \(X\). Hence, \[ \begin{aligned} x \notin X, \end{aligned} \] and therefore \(x \in B\).
This shows that every element of \(A\) belongs to \(B\), and hence \[ \begin{aligned} A \subset B. \end{aligned} \]
By repeating the same argument with the roles of \(A\) and \(B\) interchanged, and using the fact that \(B \cap X = \varnothing\), we obtain \[ \begin{aligned} B \subset A. \end{aligned} \]
Since \(A \subset B\) and \(B \subset A\), we conclude that \[ \begin{aligned} A = B. \end{aligned} \]
Q10. Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ.
Solution
To satisfy the given conditions, we must choose sets \(A\), \(B\), and \(C\) such that each pairwise intersection is non-empty, while the intersection of all three sets together is empty.
Consider the following sets: \[ \begin{aligned} A &= \{1,2\}, \\ B &= \{2,3\}, \\ C &= \{1,3\}. \end{aligned} \]
Now, we evaluate the pairwise intersections.
\[ \begin{aligned} A \cap B &= \{2\}, \\ B \cap C &= \{3\}, \\ A \cap C &= \{1\}. \end{aligned} \]
Each of these intersections contains at least one element and is therefore non-empty.
Next, we find the intersection of all three sets.
\[ \begin{aligned} A \cap B \cap C &= \{1,2\} \cap \{2,3\} \cap \{1,3\} = \varnothing. \end{aligned} \]
Thus, the sets \(A\), \(B\), and \(C\) satisfy the required conditions: all pairwise intersections are non-empty, but the intersection of all three sets is empty.
