Consider A={1} and B={1,2,3}, and the following relation R⊂A×B:
R={⟨1,1⟩, ⟨1,2⟩, ⟨1,3⟩}

Check all that apply:
R
is injective
R
is surjective
R
is a function
None of the other options

We are given the relation \( R \subset A \times B \), where \( A = \{1\} \) and \( B = \{1, 2, 3\} \), and the relation \( R = \{(1,1), (1,2), (1,3)\} \).

Let's check the conditions:

### 1. **Is \( R \) a function?**

A relation \( R \) is a function if each element of the domain \( A \) is related to **exactly one** element of the codomain \( B \).

- Here, \( A = \{1\} \), and the element \( 1 \in A \) is mapped to multiple elements in \( B \): \( 1 \), \( 2 \), and \( 3 \). 
- Since the element \( 1 \in A \) is mapped to **more than one** element in \( B \), \( R \) **is not a function**.

### 2. **Is \( R \) injective?**

A relation (or function) is injective (one-to-one) if different elements in the domain map to different elements in the codomain. However, for injectivity, we require that the relation is first a function.

- Since \( R \) is not a function, we do not check for injectivity. Hence, \( R \) **is not injective**.

### 3. **Is \( R \) surjective?**

A relation (or function) is surjective (onto) if every element in the codomain \( B \) has at least one preimage in the domain \( A \).

- Here, the codomain \( B = \{1, 2, 3\} \), and every element in \( B \) is mapped from the element \( 1 \in A \) (i.e., \( 1 \to 1 \), \( 1 \to 2 \), and \( 1 \to 3 \)).
- Therefore, \( R \) **is surjective**, as every element of \( B \) has a preimage in \( A \).

### Conclusion:
- \( R \) **is surjective**.
- \( R \) **is not a function**.
- \( R \) **is not injective**.

The correct options are:
- \( R \) is surjective
- \( R \) is not a function

Would you like more details on any part of this explanation?

Here are 5 related questions:
1. What makes a relation a function in set theory?
2. How is injectivity different from surjectivity in functions?
3. Can a relation be surjective without being a function?
4. How does the Cartesian product \( A \times B \) influence the nature of a relation?
5. In what cases can a relation be both injective and surjective?

**Tip:** For a relation to be a function, each input (from the domain) must correspond to exactly one output (in the codomain).

Consider A={1} and B={1,2,3}, and the following relation R⊂A×B: R={⟨1,1⟩, ⟨1,2⟩, ⟨1,3⟩} Check all that apply: R is injective, R is surjective, R is a function, None of the other options

Explore the relation R⊂A×B where A={1} and B={1,2,3}, with R={⟨1,1⟩, ⟨1,2⟩, ⟨1,3⟩}. Learn whether this relation is a function, injective, or surjective, and understand the underlying concepts of set theory. This detailed solution is ideal for students in Grades 10-12.

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