Can one prove, using the completeness axiom, that the rational numbers are not open?

Question

It's clear to me that $\mathbb{Q}$ is not open in $\mathbb{R}$ just by taking open balls, however a homework problem that I'm looking at makes the following statement "We know that $\mathbb{Q} \subset \mathbb{R}$ is not open, by the completeness axiom." This is where I am stuck. Is this statement true in that this property is by the completeness axiom? I'm struggling to see how the completeness axiom by itself can imply anything about openness.

Answer 1

If your definition of the reals is "a complete ordered field", then of course in some sense the completeness axiom is what guarantees that $\mathbb{Q}$ is not an open subset of $\mathbb{R}$: if you only knew that $\mathbb{R}$ was an ordered field, it could be that $\mathbb{R}=\mathbb{Q}$. So, any proof that $\mathbb{Q}$ is not open in $\mathbb{R}$ will have to use the completeness axiom somewhere. But you're right that it's not in any way an immediate consequence of the completeness axiom.

In contrast, the fact that $\mathbb{Q}$ is not closed in $\mathbb{R}$ is more or less an immediate consequence of the completeness axiom, as indicated in BrianO's comment. I'm guessing that whoever wrote your homework problem actually had this fact in mind, and was just attempting to provide some (kind of feeble) motivation behind the homework problem without realizing that what they were saying didn't really make sense.

Answer 2

A thought: (possibly invalid, depending on the definitions used)

This is equivalent to showing that the irrationals form a closed set. This is false because the set $$ \mathbb{R} \setminus \mathbb{Q} \supset A = \{ -2^{1/2-n} \mid n \in \mathbb{N} \} $$ has least upper bound $0$, and $A$ ordered using the real total order forms a Cauchy sequence that approaches $0$, so there is a Cauchy sequence of irrationals converging to a rational, so the rationals aren't closed.