Question

can u conclude that \(A = B\) if A, B and C are sets such that

a) \(A \cup C = B \cup C\) ? b) \(A \cap C = B \cap C\) ?

c) \(A \cup C = B \cup C\) and \(A \cap C = B \cap C\)?

Short answer

a) No

b) No

c) A=B

Step-by-step solution

Set Operations

Union\(A \cup B\): All elements that are either in A or in B.

Intersection\(A \cap B\): All elements that are both in A and in B.

Difference\(A - B\): All elements in A that are Not in B.

Complement\(\overline A \): All elements in the universal set U not in A.

Operation done using the union of two elements

(a) Given \(A \cup C = B \cup C\)

A and B then do not necessarily have to be the same set.

For eg.\(A = \left\{ 0 \right\}\)\(B = \left\{ 1 \right\}\)\(c = \left\{ {0,1} \right\}\)

By definition \(A \cup C = \left\{ {0,1} \right\}\)\(A \cup B = \left\{ {0,1} \right\}\)

We then note that the equality \(A \cup C = B \cup C\) holds, while the two sets A and B are unequal. Thus we found a case where A=B in not true.

Operation done using the intersection of two elements

(b) Given\(A \cap C = B \cap C\).

A and B then do not necessarily have to be the same set.

For eg.\(A = \left\{ 0 \right\}\)\(B = \left\{ {0,2} \right\}\)\(c = \left\{ 0 \right\}\)

By definition \(A \cap C = \left\{ 0 \right\}\)\(A \cap B = \left\{ 0 \right\}\)

We then note that the equality \(A \cap C = B \cap C\) holds, while the two sets A and B are unequal. Thus we found a case where A=B in not true

Operation done using the union and intersection of two elements

(c) Given \(A \cup C = B \cup C\) And\(A \cap C = B \cap C\)

To prove A=B

Let \(x \in A\) => \(x \in A \cup C\) then by a) \(x \in B \cup C\)

Let \(x \in C\)=> \(x \in A \cap C\)then by b) \(x \in B \cap C\)

\(x \in B\)

Therefore \(A \subseteq B\) .

Similarly

Let \(x \in B\) => \(x \in B \cup C\) then by a) \(x \in A \cup C\)

Let \(x \in C\)=> \(x \in B \cap C\)then by b) \(x \in A \cap C\)

\(x \in A\)

Therefore \(B \subseteq A\) .

Hence A=B.