Exercise · ZW-COMPLETENESS-AXIOM-002 · ★☆☆ Basic
Which of these sets of rationals has no supremum in \(\mathbb Q\)?
Choose one option.
Show answer
- C. \(\{x\in\mathbb Q: x^2<2\}\)
Explanation
Its rational upper bounds have no least element, since the supremum would be \(\sqrt2\), which is irrational. This is why \(\mathbb Q\) is not complete.
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-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?
- ZW-COMPLETENESS-AXIOM-009 Let S be nonempty and bounded above, and M=sup S. Which statement is true?