$\{q\in \mathbb{Q}:q\ge 0,q^2<2\}$ is closed and bounded in $\mathbb{Q}$ and is not compact

Refuted claim: in every ordered field a closed bounded set is compact, so the completeness hypothesis of the Heine-Borel characterisation is unnecessary (FALSE: in every ordered field a closed bounded set is compact, so Heine-Borel needs no completeness).

The witness is the ordered field $\mathbb{Q}$ (The rationals as equivalence classes of pairs of integers, The rationals form a totally ordered field) together with

$S:=\{q\in \mathbb{Q}:q\ge 0\text{ and }{q}^2<2\}\mathrm{.}$

The set $S$ is bounded, is closed in $\mathbb{Q}$, and is not compact in $\mathbb{Q}$, all with respect to the vocabulary of Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset transposed from $\mathbb{R}$ to $\mathbb{Q}$ exactly as set out in FALSE: in every ordered field a closed bounded set is compact, so Heine-Borel needs no completeness, where the refutation is carried out in full. This item records the witness and says what makes it work.

• What is closed in $\mathbb{Q}$ is not closed in $\mathbb{R}$. Read inside $\mathbb{R}$, the same collection of numbers is bounded and fails to be closed: the real $\sqrt{2}$, which exists by Square roots exist: a unique $\sqrt{a}\ge 0$ with $(\sqrt{a}{)}^2=a$; the positives are $\{x^2:x\mathrm{\ne }0\}$ and is not rational by FALSE: some rational number squares to 2, is adherent to it and absent from it. The set is closed in $\mathbb{Q}$ precisely because the point that would have to be adjoined to close it does not lie in $\mathbb{Q}$. Closedness is a statement about a set inside an ambient field, not about the set alone.

• Only one half of Heine-Borel fails here. That a compact set is closed and bounded (A compact subset of $\mathbb{R}$ is closed and bounded) uses no completeness at all, only the Archimedean property and the existence of maxima of finite sets; the converse (Heine-Borel by bisection: every closed bounded interval $[a,b]$ is compact and A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded) is the half that spends completeness, and it is the half refuted here.

• Why this witness rather than $\mathbb{Q}\cap [0,1]$. Both work: a set with rational endpoints is also closed and bounded in $\mathbb{Q}$ and also fails to be compact there, but its non-compactness has to be produced by splitting it at some irrational chosen for the purpose. In $S$ the irrational is already built in, and the same fact, the irrationality of $\sqrt{2}$ (FALSE: some rational number squares to 2), delivers both closedness in $\mathbb{Q}$ and the absence of a finite subcover. The witness therefore runs on exactly the mechanism of the false statement it refutes.