Exercise · ZW-WELL-ORDERING-010 · ★★★ Advanced
Is \(\mathbb{N}\) well-ordered by the reverse order \(\ge\)? Choose the correct statement.
Choose one option.
A Yes, because the first natural number is \(1\). B No, because under \(\ge\), \(\mathbb{N}\) has no first element: there is no natural number greater than all the others. C Yes, because every finite subset has a maximum. D No, because \(\mathbb{N}\) is not totally ordered under \(\ge\). Show answer B. No, because under \(\ge\), \(\mathbb{N}\) has no first element: there is no natural number greater than all the others.← Back to the question
Explanation Under \(\ge\), the first element would be the largest natural number, which does not exist; well-ordering requires a first element in every nonempty subset , including \(\mathbb{N}\) itself.
A set \(A\) is well-ordered by \(<\) if \(<\) is a total order on \(A\) and every nonempty subset of \(A\) has a least element.
\[\forall S\subseteq A(S\ne\varnothing\Rightarrow\exists m\in S\forall s\in S\,m\le s)\]
More practice problems: well-ordering