Sets - Union of Two Sets-advanced

Let $A = \{x : x \in \mathbb{N}, x < 6\}$ and $B = \{x : x \text{ is a prime number } < 10\}$. Find $A \cup B$.

First, list the elements of each set in roster form: $A = \{1, 2, 3, 4, 5\}$ $B = \{2, 3, 5, 7\}$ Now, find the union by combining all elements, ensuring no duplicates: $A \cup B = \{1, 2, 3, 4, 5, 7\}$

To find the union, we include every element that appears in at least one of the sets. The elements $2, 3,$ and $5$ are common to both but are listed only once.

In a class of $50$ students, $30$ students like Mathematics and $25$ students like Science. If $10$ students like both subjects, find the number of students who like either Mathematics or Science.

Let $M$ be the set of students who like Mathematics and $S$ be the set of students who like Science. Given: $n(M) = 30$ $n(S) = 25$ $n(M \cap S) = 10$ Using the formula: $n(M \cup S) = n(M) + n(S) - n(M \cap S)$ $n(M \cup S) = 30 + 25 - 10$ $n(M \cup S) = 55 - 10 = 45$

We use the Principle of Inclusion-Exclusion. Adding $n(M)$ and $n(S)$ counts the $10$ students who like both subjects twice, so we subtract $n(M \cap S)$ once to get the correct count of unique students.

If $A \subset B$, find $A \cup B$.

By definition, $A \subset B$ means every element of $A$ is also an element of $B$. Since $A \cup B$ is the set of all elements in $A$ or $B$, and all elements of $A$ are already in $B$, the resulting set is simply $B$. Therefore, $A \cup B = B$.

This is a property of subsets. When one set is entirely contained within another, their union is the larger set (superset).