Let \( R\subset\mathbb{N}\times\mathbb{N} \). Which of the following statement is necessarily true?
Step-by-step Solution:
Let's analyze each option: Option A: Having cardinality at most 1 means some elements in the domain may have 0 mappings. This represents a partial function, not necessarily a total function \( f:\mathbb{N}\rightarrow\mathbb{N} \). Option B: This defines an injective property or partial inverse function, not a function from \( \mathbb{N} \) to \( \mathbb{N} \). Option C: If \( R_a \) is infinite, since \( R_a \subset \mathbb{N} \), it must be countably infinite (cardinality \( \aleph_0 \)). Since \( R \subset \mathbb{N} \times \mathbb{N} \), \( R \) is at most countably infinite. Because \( R_a \times \{a\} \subseteq R \), \( R \) must also be at least countably infinite. Thus, both \( R_a \) and \( R \) have the exact same cardinality (\( \aleph_0 \)). This is necessarily true. Option D: Both would be countably infinite, meaning their cardinalities are equal, not larger.