If \[A = \left\{ {0,2,4} \right\}\] find \[A \cap \phi \] and \[A \cap A\]

Hint: First, we understand the definition of the intersection of two sets. Now use the properties of intersection. Then give set \[A\] to the intersection with an empty set. Then set \[A\] to the intersection with itself. Finally, we get an answer.

Complete step-by-step answer:
The given set \[A\] is nonempty. Set \[A = \left\{ {0,2,4} \right\}\]
First, we understand the intersection meaning \[A\] and \[B\] are sets. The common elements or objects of the two sets form the set. That is called \[A\] intersection \[B\]
\[A \cap B = \left\{ {x:x \in A\,and\,x \in B} \right\}\]
This is the definition of the intersection of two sets.
Now first we find \[A \cap \phi \]
Set \[A\] is \[\left\{ {0,2,4} \right\}\]. \[\phi \] means an empty set or null set. The set has no elements.
So that \[\phi = \left\{ {} \right\}\]
Now we find the intersection of \[A\] an empty set.
\[A \cap \phi = \left\{ {} \right\}\]
Because the common element of both sets is none. So that answer of \[A\] intersection empty set is a null set.
Next, we find \[A \cap A\] value,
We know that \[A\] value is \[\left\{ {0,2,4} \right\}\]
Set \[A\] is the intersection to itself. That means \[A \cap A = \left\{ {0,2,4} \right\}\]
Because the common element of both sets is whole elements. It means the intersection value \[A\] and \[A\] is \[A\].
So that,
\[
  A \cap A = \left\{ {0,2,4} \right\} \\
   = A \\
 \]
The intersection of the empty set to non-empty is always an empty set. And the intersection of a set with itself is always getting the same set. These are important properties of the intersection of two sets.
Finally, we get the answers
\[A \cap \phi = \left\{ {} \right\},\,A \cap A = A\]

Note: First given a question to understand the definition of intersection or union of two sets. The intersection of the empty set to non-empty is always an empty set. And the intersection of a set with itself is always getting the same set. These properties are used carefully. Don't confuse intersection and union properties. Because union property is changeable.