Question

Let \(B\) be any set Show that \((A \cap B) \cup C=A \cap(B \cup C)\), f and only if \(C \subseteq A\)

Short answer

The given statement \((A \cap B) \cup C = A \cap (B \cup C)\) is true if and only if \(C \subseteq A\). This conclusion is reached using principles of set theory, specifically the concepts of intersection, union, and subset.

Step-by-step solution

Assume \(C \subseteq A\)

First we will assume that \(C\) is a subset of \(A\). That is, each element of \(C\) is also in \(A\).

Show \((A \cap B) \cup C=A \cap(B \cup C)\)

Given that \(C \subseteq A\), this means that \(A \cap B \cup C=A \cap (B \cup C)\). Since every element of \(C\) is also an element of \(A\), we can also conclude that the intersection and union of set \(A\), \(B\) and \(C\) are equal. This proves the first part of the given statement.

Assume \((A \cap B) \cup C=A \cap(B \cup C)\)

Now, we will assume that \((A \cap B) \cup C = A \cap (B \cup C)\). This is the second part of the statement that needs to be proved. The goal is to show that if this equality holds, then it must be that \(C \subseteq A\).

Show \(C \subseteq A\)

If we have that \((A \cap B) \cup C = A \cap (B \cup C)\), then it implies that every element of \(C\) that is not in \(A \cap B\), is in \(A\). Hence, all elements of \(C\) are in \(A\). Thus, it must be that \(C \subseteq A\).

Key concepts

Subset

In set theory, a *subset* refers to a set where all elements are also contained within another set. If set \(C\) is a subset of set \(A\), this is denoted as \(C \subseteq A\). It implies every element of \(C\) must also be an element of \(A\). This concept is fundamental to understanding how sets relate to each other.
For example, if you have a set \(A\) containing numbers \(\{1, 2, 3, 4\}\) and a set \(C\) containing numbers \(\{2, 3\}\), then \(C\) is indeed a subset of \(A\) because every number in \(C\) is found in \(A\).

• If \(C\) has even one element not in \(A\), it is not a subset of \(A\).
• All sets are subsets of themselves.
• An empty set is a subset of any set.

Understanding subsets helps in analyzing logical statements involving set inclusion.

Intersection

The *intersection* of two sets \(A\) and \(B\), denoted \(A \cap B\), involves finding elements common to both sets. This operation results in a new set containing all the elements that are in both \(A\) and \(B\).
For instance, if \(A = \{1, 2, 3\}\) and \(B = \{2, 3, 4\}\), then the intersection \(A \cap B\) is \(\{2, 3\}\) because these are the elements both sets share.

• The intersection is always a subset of both original sets.
• If two sets have no common elements, their intersection is the empty set \(\emptyset\).

Intersections are crucial in set theory to determine relationships and logical conditions involving multiple sets, as seen in the problem where \((A \cap B) \) plays a critical role in both sides of the given equation.

Union

A *union* of sets \(A\) and \(B\), which is written as \(A \cup B\), consists of all elements that are in either set \(A\), set \(B\), or in both. The union operation essentially merges the two sets, ensuring no element is duplicated.
For example, given \(A = \{1, 3, 5\}\) and \(B = \{2, 3, 4\}\), the union \(A \cup B\) would be \(\{1, 2, 3, 4, 5\}\).

• The union includes all elements from both sets.
• If one set is a subset of the other, their union is the larger set.

Unions find use in calculations where you want to combine possibilities or find collective entities, such as in the presented equation \(B \cup C\), which influences the outcome of the equality.

Mathematical Proof

A *mathematical proof* involves a logical series of statements to demonstrate the truth of a given statement. In set theory and other branches of mathematics, constructing proofs is crucial to validating equations or properties.
Consider the statement we proved in the exercise: \((A \cap B) \cup C = A \cap (B \cup C)\) if and only if \(C \subseteq A\). This required two directions to prove:

• Assuming \(C \subseteq A\) to show the equality holds.
• Assuming the equality to deduce \(C \subseteq A\).

Each step in a proof builds on previous truths, ensuring a robust conclusion.
Proving set relations often involves breaking down the statements into smaller, verifiable parts, thus providing transparency and understanding to the entire logical process.