# Real Analysis: Bounded Subset of Rationals Has No Infimum

+1 vote
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edited Aug 24, 2016

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 :)

answered Aug 24, 2016 by (4,220 points)
selected Aug 24, 2016 by Pandy

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 (2,850 points)

This makes much more sense, thanks so much!