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Real Analysis: Bounded Subset of Rationals Has No Infimum

+1 vote
asked Aug 24, 2016 in Calculus and Pure Mathematics by Pandy (2,850 points)
edited Aug 24, 2016 by Pandy

The question is a True/False one and reads:

"Every subset of the rational numbers \(\mathbb{Q}\) which is bounded from below has a greatest lower bound \(c \in \mathbb{Q}\)."

and the answer given is false.

My understanding is that, since the Completeness Axiom says every non-empty set which is bounded from below has a greatest lower bound, the \(c\in \mathbb{Q}\) must be the part which makes the statement false but I can't think of why.

Thanks :)

1 Answer

+2 votes
answered Aug 24, 2016 by simon_rigby (4,220 points)
selected Aug 24, 2016 by Pandy
Best answer

Yes, you're right that the completeness axiom doesn't hold in \(\mathbb{Q}\)!

Here's a counterexample:

Consider the sequence \(\{a_n\}_{n=1}^\infty\) where \(a_n = (1+\frac{1}{n})^n\). It can be shown that this sequence is increasing,\(^1\) and bounded. By the monotone convergence theorem, \( \sup_{n \in \mathbb{N}}a_n  = \lim_{n \to \infty}a_n\) and it so happens that this limit is \(e = 2.718... \notin \mathbb{Q} \). Therefore, \(\{a_n\}_{n=1}^\infty\) has no supremum in \(\mathbb{Q}\), and \(\{-a_n\}_{n=1}^\infty\) has no infimum in \(\mathbb{Q}\).

\(^1\) Think of it as the accumulation of a unit under an effective interest rate \(i = 1\) compounded \(n\) times per year. The more frequent the compounding, the more it accumulates.

commented Aug 24, 2016 by Pandy (2,850 points)

This makes much more sense, thanks so much!