To prove that a relation $R$ on a set $A$ is complete if and only if $R \cup R^{-1} = A \times A$, we proceed as follows:

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Definitions:

1. A relation $R$ on $A$ is complete if for every pair $a, b \in A$, either $(a, b) \in R$ or $(b, a) \in R$.
2. The inverse of $R$, denoted $R^{-1}$, is defined as $R^{-1} = \{(b, a) \mid (a, b) \in R \}$.
3. The Cartesian product $A \times A$ contains all ordered pairs $(a, b)$ where $a, b \in A$.

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Proof:

(⇒) If $R$ is complete, then $R \cup R^{-1} = A \times A$.

• Assume $R$ is complete. By definition, for any pair $a, b \in A$:
  • Either $(a, b) \in R$ or $(b, a) \in R$.
• The set $R \cup R^{-1}$ includes all pairs in $R$ and all pairs in $R^{-1}$, so:
  • If $(a, b) \in R$, then $(a, b) \in R \cup R^{-1}$.
  • If $(b, a) \in R$, then $(b, a) \in R^{-1}$, which implies $(b, a) \in R \cup R^{-1}$.
• Since for every pair $(a, b) \in A \times A$, either $(a, b) \in R$ or $(b, a) \in R$, it follows that $R \cup R^{-1} = A \times A$.

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(⇐) If $R \cup R^{-1} = A \times A$, then $R$ is complete.

• Assume $R \cup R^{-1} = A \times A$.
• Take any pair $a, b \in A$. Since $(a, b) \in A \times A$, it must be true that:
  • $(a, b) \in R$, or
  • $(a, b) \notin R$, in which case $(b, a) \in R^{-1}$ because $R \cup R^{-1} = A \times A$.
• If $(b, a) \in R^{-1}$, by the definition of $R^{-1}$, $(b, a) \in R$.
• Therefore, for every pair $a, b \in A$, either $(a, b) \in R$ or $(b, a) \in R$, which means $R$ is complete.

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Conclusion:

We have shown both directions:

1. If $R$ is complete, then $R \cup R^{-1} = A \times A$.
2. If $R \cup R^{-1} = A \times A$, then $R$ is complete.

Thus, $R$ is complete if and only if $R \cup R^{-1} = A \times A$. $\blacksquare$

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Questions for Further Exploration:

1. What is an example of a complete relation on a finite set?
2. How does the concept of completeness differ for partial orders and equivalence relations?
3. Can a relation $R$ be complete but not transitive? Why or why not?
4. How does the symmetry of $R$ affect $R \cup R^{-1}$?
5. What happens if $R$ is not defined on all pairs of $A \times A$?

Tip:

When working with relations, always write down the definitions of key terms like completeness, symmetry, and transitivity to avoid confusion.