Exercise · ZW-EQUIPOTENCE-007 · ★★☆ Intermediate
Is \(\{1,2\}\) equipotent to \(\{1,2,3\}\)?
Choose one option.
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- C. No, because there is no bijection between finite sets of different sizes
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Explanation
A finite set with \(n\) elements can only be put in bijection with another with \(n\) elements, so \(\{1,2\}\) and \(\{1,2,3\}\) are not equipotent.
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
- ZW-EQUIPOTENCE-001 What is needed for two sets A and B to be equipotent?
- ZW-EQUIPOTENCE-002 Is \a,b,c\ equipotent to \1,2,3\?
- ZW-EQUIPOTENCE-003 Which of these properties does equipotence satisfy?
- ZW-EQUIPOTENCE-004 Let ℕ=\0,1,2,dots\. Is ℕ equipotent to \0,2,4,dots\?
- ZW-EQUIPOTENCE-005 Which statement about ℕ and ℤ is correct?
- ZW-EQUIPOTENCE-006 How is it written that the sets A and B are equipotent?
- ZW-EQUIPOTENCE-008 What do you get when composing two bijections fcolon A→ B and gcolon B→ C?
- ZW-EQUIPOTENCE-009 The inclusion of ℕ in ℝ is an injective function. Why does it not prove that ℕsimℝ?