Exercise · ZW-EQUIPOTENCE-005 · ★★☆ Intermediate
Which statement about \(\mathbb{N}\) and \(\mathbb{Z}\) is correct?
Choose one option.
A They are equipotent, with a bijection that alternates signs B They are not equipotent, because \(\mathbb{Z}\) is larger than \(\mathbb{N}\) C They are not equipotent, because \(\mathbb{Z}\) has negative numbers D \(\mathbb{N}\subseteq\mathbb{Z}\), so the inclusion is a bijection
Show answer A. They are equipotent, with a bijection that alternates signs
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Explanation The map \(0\mapsto0,\ 1\mapsto1,\ 2\mapsto-1,\ 3\mapsto2,\ 4\mapsto-2,\dots\) is a bijection. The inclusion \(\mathbb{N}\subseteq\mathbb{Z}\) is injective but not surjective, so it is not enough.
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-006 How is it written that the sets A and B are equipotent?ZW-EQUIPOTENCE-007 Is \1,2\ equipotent to \1,2,3\?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?