Exercise · ZW-EQUIPOTENCE-009 · ★★★ Advanced
The inclusion of \(\mathbb{N}\) in \(\mathbb{R}\) is an injective function . Why does it not prove that \(\mathbb{N}\sim\mathbb{R}\)?
Choose one option.
A Because the inclusion is injective but not surjective, and the non-natural reals are never reached B Because the inclusion shows that \(\mathbb{N}\) and \(\mathbb{R}\) have the same number of elements C Because \(\mathbb{R}\) is not an infinite set D Because \(\mathbb{N}\) is not infinite Show answer A. Because the inclusion is injective but not surjective, and the non-natural reals are never reached← Back to the question
Explanation The inclusion never reaches the reals that are not natural numbers. In fact \(\mathbb{N}\) and \(\mathbb{R}\) are not equipotent, as Cantor's diagonal argument shows.
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-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?