The question here asks for a counterexample to the statement:

$A \subseteq B \iff A \cup B = A$

To find a counterexample, we need to show that this statement does not hold universally. The statement essentially implies that if $A$ is a subset of $B$, then the union of $A$ and $B$ should equal $A$, and vice versa.

Explanation

1. If $A \subseteq B$, then every element in $A$ is also in $B$. However, this does not imply that $A \cup B = A$; in fact, $A \cup B$ would typically be equal to $B$ if $A \subseteq B$ and $B \neq A$.

2. The statement would be true only if $A = B$ because only then would $A \cup B = A$.

Counterexample

Let’s choose sets such that $A \subseteq B$ but $A \cup B \neq A$. Consider:

• $A = \{1\}$
• $B = \{1, 2\}$

Here:

• $A \subseteq B$ is true since $\{1\}$ is indeed a subset of $\{1, 2\}$.
• However, $A \cup B = \{1, 2\}$, which is not equal to $A = \{1\}$.

Thus, $A \subseteq B$ does not imply that $A \cup B = A$ in this case, providing a counterexample to the statement.

Would you like further details on this solution, or have any other questions?

Here are 5 related questions for deeper exploration:

1. What is the difference between a subset and a proper subset?
2. How does the union of two sets behave when one set is a subset of another?
3. Can you provide an example where $A \cup B = A$ and $A = B$?
4. What conditions are necessary for $A \cup B = A$?
5. How does the intersection of sets relate to their union?

Tip: When dealing with set relationships, visualize with simple examples or Venn diagrams to better understand subset and union properties.