Set Operations

Key Ideas:

  • Introducing basic operations on sets: Union, Intersection, Substraction

Manipulating sets:

The same way we can combine numbers through addition, substractions, etc. There are a number of questions we may want to ask about multiple sets.

Union:

The Union of two sets $A$ and $B$ represent all the elements that belong to either of the two sets. The union operation is represented by the symbol $\cup$

Formally then, $A \cup B = \{x \in U: x \in A \lor x \in B\}$ Given this definition, how would you evaluate $A \cup \emptyset$ ? Try it out.

This evaluates to $A$. The empty set has no elements to contribute to the union

How about $A \cup U$?

This evaluates to $U$. The universal set already contains all the elements of $A$

Intersection

The Intersection of two sets $A$ and $B$ represent all the elements that belong to both of the two sets. The intersection operation is represented by the symbol $\cap$

Formally then, $A \cap B = \{x \in U: x \in A \land x \in B\}$ Given this definition, how would you evaluate $A \cap \emptyset$?

This evaluates to $\emptyset$. The empty set has no elements, so there are no elements in common between it and $A$

How about $A \cap U$?

This evaluates to $A$. The universal set already contains all the elements of $A$, so those are the elements the two sets have in common

Substraction:

The substraction of two sets $A$ and $B$ represent all the elements that belong to $A$ but not to $B$ of the two sets. The intersection operation is represented by the symbol $\setminus$

Formally then, $A \setminus B = \{x \in U: x \in A \land x \notin B\}$

Notice that while order did not matter for union and intersection, $A \setminus B$ and $B \setminus A$ will produce different results.

Given this definition, how would you evaluate the following $A \setminus \emptyset$?

This evaluates to $A$. The empty set has no elements, so there are no elements to remove $A$

How about $A \setminus U$ and $U \setminus A$? We should expect these to be different.

$A \setminus U$ evaluates to $\emptyset$. The universal set already contains all the elements of $A$, so we would 'take away' all the elements, leaving the empty set.

$U \setminus A$ evaluates to $\bar A$. By our definition, this would be all the elements that do not belong to $A$, in other words its complement.

Section Video

References:

For further details, you can read through the following chapter: