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This problem is asking to prove that if there is a surjective (onto) function from set \( A \) to set \( B \), then the sets \( A \) and \( B \) are equinumerous (have the same cardinality).

To prove this:

1. If a surjection \( f: A \to B \) exists, every element in \( B \) must be mapped to by at least one element in \( A \).
2. Since there is no element in \( B \) that isn't mapped from \( A \), this shows that the size of \( A \) is at least as large as \( B \).
3. If \( f \) is surjective, a bijection (one-to-one correspondence) can be formed between subsets, implying that \( |A| = |B| \).

Do you need a deeper explanation or examples? Here are five related questions to explore:

1. What is the difference between surjective, injective, and bijective functions?
2. How can a bijection be constructed from a surjection and an injection?
3. What are the cardinalities of common infinite sets, like \( \mathbb{N} \) and \( \mathbb{R} \)?
4. Can a set have more elements than another and still map surjectively onto it?
5. How do you prove two infinite sets have the same cardinality?

**Tip**: For finite sets, a surjection automatically implies that the sets have the same number of elements.

Prove that if there exists a surjection from set A to set B and a surjection from set B to set A, then sets A and B are equinumerous.

Learn how to prove that two sets A and B are equinumerous using surjective functions. This problem involves key concepts in set theory, specifically surjections and cardinality, and applies the Cantor-Bernstein Theorem. Suitable for university-level students.

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