Exercise · ZW-COMPLETENESS-AXIOM-010 · ★★★ Advanced
Why does the completeness axiom require the set to be nonempty?
Choose one option.
Show answer
- B. Every real is an upper bound of the empty set, so there would be no least one
Explanation
Every number is an upper bound of \(\varnothing\); the set of bounds is all of \(\mathbb R\), which has no least element. The empty set is bounded and is a subset of \(\mathbb R\).
Review the concept: Completeness axiom
The property that sets \(\mathbb R\) apart from \(\mathbb Q\): every nonempty set \(S\) of reals with an upper bound \(M\) has a supremum.
\[S\ne\varnothing,\ \exists M:\ s\le M\ \forall s\in S\ \Longrightarrow\ \sup S\in\mathbb R\]
More practice problems: completeness axiom
- ZW-COMPLETENESS-AXIOM-001 What does the completeness axiom state?
- ZW-COMPLETENESS-AXIOM-002 Which of these sets of rationals has no supremum in mathbb Q?
- ZW-COMPLETENESS-AXIOM-003 For which set does the hypothesis of the completeness axiom NOT hold?
- ZW-COMPLETENESS-AXIOM-004 If S is nonempty and bounded below, what does the completeness axiom guarantee?
- ZW-COMPLETENESS-AXIOM-005 Compute sup\x∈mathbb R: x^3<8\.
- ZW-COMPLETENESS-AXIOM-006 Let S=\x∈mathbb R: x^2+x<6\. What is inf S+sup S?
- ZW-COMPLETENESS-AXIOM-007 Compute sup\(2n)/(n+1): n∈mathbb N\.
- ZW-COMPLETENESS-AXIOM-008 Which structure satisfies the ordered field axioms but not completeness?