Exercise · ZW-BOLZANO-WEIERSTRASS-THEOREM-009 · ★★★ Advanced
Why does the Bolzano-Weierstrass theorem fail if only rational numbers are allowed?
Choose one option.
A Because \(\mathbb Q\) is unbounded B Because \(\mathbb Q\) has no order C Because \(\mathbb Q\) is not countable D The decimal truncations of \(\sqrt2\) are bounded and no subsequence converges in \(\mathbb Q\) Show answer D. The decimal truncations of \(\sqrt2\) are bounded and no subsequence converges in \(\mathbb Q\)← Back to the question
Explanation The proof uses the completeness of \(\mathbb R\). In \(\mathbb Q\) those truncations form a bounded sequence whose subsequences all converge to \(\sqrt2\), which is not rational. The other reasons are false: \(\mathbb Q\) is ordered and countable.
Every bounded real sequence \((a_n)\) has at least one subsequence \((a_{n_k})\) that converges to a number in the set \(\mathbb R\) of reals.
\[\exists\,(n_k):\ \lim_{k\to\infty}a_{n_k}\in\mathbb R\]
More practice problems: bolzano-weierstrass theorem