Exercise · ZW-EQUIPOTENCE-010 · ★★★ Advanced

Let \(f\colon A\to B\) be a bijection, with \(A\) finite. Why is \(B\) also finite?

Choose one option.

Show answer
  • D. Because the inverse \(f^{-1}\) composed with a bijection \(A\to\{1,\dots,n\}\) gives a bijection \(B\to\{1,\dots,n\}\)

Explanation

If \(A\) has \(n\) elements, there is a bijection \(\varphi\colon A\to\{1,\dots,n\}\). Then \(\varphi\circ f^{-1}\) is a bijection \(B\to\{1,\dots,n\}\), so \(B\) is finite.

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Review the concept: Equipotence

Two sets \(A\) and \(B\) are equipotent if there is a bijection \(f\colon A\to B\). It is written \(A\sim B\).

\[A\sim B \iff \exists\, f\colon A\to B\ \text{biyectiva}\]

More practice problems: equipotence

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