How does completeness axiom provide the existence of irrational numbers?

Question

So I've been reflecting on the Completeness Axiom. It is said that this is what provides continuity to the real numbers. An example is the set $S = {x: x^2 <2}$. This will have no least upper bound in $\Bbb Q$, but will in $\Bbb R$. Now if it is to be said that it is the Completeness Axiom that provides this continuity, that has to mean that he existence of $\sqrt 2$ is provided axiomatically by the completeness axiom. If $\sqrt 2$ is to exist based on some other reason/principle/axiom, then that will be what provides continuity, and the completeness axiom will just be a restatement of something it has no part in creating, and thus won't even be an axiom anymore.

I am on this philosophical query because it seems from everywhere I read that it is in fact the completeness axiom that provides continuity. Then it must be that the existence of $\sqrt 2$ is provided axiomatically by the completeness axiom. Am I correct?

Answer 1

I suspect (you have to tell!) that what you mean by continuity is the fact that the real line has the linear continuum property, namely that:

1. It has the least upper bound property
2. and for each $x,y$ satisfying $x \lt y$, there exists $z$ such that $x \lt z \lt y$.

From there, I suggest that you have a look at the various way to build the real numbers. This is instructive of the way the completeness of $\mathbb R$ is deduced.

Answer 2

There are a few different things I could say here. I’m going to ramble a bit, and hopefully some proper subset of the things I say will answer your questions.

First, I think we should be careful to separate the order properties of $\mathbb{R}$ from the algebraic properties of $\mathbb{R}$. The existence of irrational numbers is an algebraic property of $\mathbb{R}$, while completeness is an order property.

Let’s first examine some properties of linear orders.

Density: an order $(L, \leq)$ is dense if and only if for every pair of points $x,y \in L$ with $x \leq y$, there is a third point $z \in L$ such that $x \leq z \leq y$. We can think of this as “if you give me any two points, we can find a point in between them”.

Completeness: an order $(L, \leq)$ is complete if and only if every bounded subset $A \subseteq L$ has a least upper bound and a greatest lower bound.

In some sense, both of these correspond to the order feeling “continuous”. I would caution against thinking of completeness as “continuity”. Perhaps an example will illustrate why.

We may take obvious order-preserving injections $\mathbb{Q} \: \hookrightarrow \: \mathbb{R} \: \hookrightarrow \:^*\mathbb{R}$. If you’re unfamiliar, $^*\mathbb{R}$ denotes the “hyperreal numbers”. The hyperreals are an ordered field which is non-Archimedean. In particular, there are “infinite” and “infinitesimal” hyperreal numbers.

All three orders $\mathbb{Q}$, $\mathbb{R}$, and $^* \mathbb{R}$ are dense, but only $\mathbb{R}$ is complete. When we extend to the hyperreals, we add an infinitesimal $\epsilon$ such that for every $r \in \mathbb{R}$, we have $0 < \epsilon < r$. This sure “feels” like plugging a hole in the real numbers, doesn’t it? But the real numbers were complete. So what hole could we be plugging if the real numbers have no holes to plug? This is why I would advocate for thinking of completeness in a different way than continuity.

If you want an interpretation of completeness that makes sense, we might pass to the world of topology. I don’t know if you’ve learned any topology yet, but if you give me an ordered set $(L, \leq)$, we may put a topology on $L$ generated from half-open intervals. One may show the following theorem:

Theorem: Let $(L, \leq)$ be endowed with the order topology. $L$ is connected if and only if the order is dense and complete.

This “connected” notion might be more useful than “continuity”.

Bringing this back to the world of algebra, it is a minor miracle to me that taking the completion of $\mathbb{Q}$ to get $\mathbb{R}$ preserves the field properties of $\mathbb{Q}$. This is not always the case. There are ordered fields that cannot be completed to get another field. The alignment of “irrational numbers” and “Dedekind cuts” is a consequence of this miracle.