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 sets \( A = \{1\} \) and \( B = \{1, 2, 3\} \), along with the relation \( R \subseteq A \times B \) defined as:

\[
R = \{(1, 1), (1, 2), (1, 3)\}
\]

Let's evaluate the given options.

1. **Is \( R \) injective?**

   A relation is **injective** (one-to-one) if no two distinct elements in \( A \) are mapped to the same element in \( B \). Here, we only have one element in \( A \) (which is \( 1 \)), and it is related to three different elements in \( B \) (namely \( 1, 2, \) and \( 3 \)). Since the same element in \( A \) is mapped to multiple elements in \( B \), the relation is **not injective**.

2. **Is \( R \) surjective?**

   A relation is **surjective** (onto) if every element of \( B \) has a preimage in \( A \). In this case, all elements of \( B \) (1, 2, and 3) are mapped from the element \( 1 \in A \). Therefore, the relation is **surjective** because every element in \( B \) is covered.

3. **Is \( R \) a function?**

   A relation is a **function** if every element of \( A \) is related to exactly one element in \( B \). However, in this case, the element \( 1 \in A \) is related to multiple elements in \( B \) (1, 2, and 3). This violates the definition of a function, so \( R \) is **not a function**.

4. **None of the other options?**

   Since the relation is surjective but neither injective nor a function, this option does **not** apply.

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

Would you like further clarification on any of these concepts?

### Related questions:
1. Can a relation be injective but not surjective?
2. What conditions must a relation satisfy to be a function?
3. How can we visualize injective and surjective relations using diagrams?
4. Can a function be neither injective nor surjective?
5. How would the answer change if \( A \) had more elements?

### Tip:
For a relation to be a function, each input (from set \( A \)) must correspond to **exactly one** output (from set \( B \)).

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 how to determine whether a relation between sets A={1} and B={1,2,3} is injective, surjective, or a function. This step-by-step analysis covers the core concepts of relations, injectivity, surjectivity, and functions. Suitable for Grades 10-12 students studying advanced set theory.

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