# Completeness axiom as having no holes in the set

1. Oct 31, 2012

### Bipolarity

Completeness axiom as having "no holes" in the set

My textbook describes the completeness axiom as essential to showing that there are no "holes or gaps" in the real numbers. That is, for any two reals A and B, there exists a real C such that A<C<B.

Of course, we all know that the actual statement of the completeness axiom is that any bounded set of real numbers has a least upper bound.

I was wondering, how can we use the explicit statement of the completeness axiom to show that there are no "holes or gaps" in the reals? Is it possible or did my textbook just use this as an intuitive explanation for the completeness axiom?

Thanks all!

BiP

Last edited: Oct 31, 2012
2. Oct 31, 2012

### micromass

Staff Emeritus
Re: Completeness axiom as having "no holes" in the set

The intuitive textbook explanation is rubbish. Even the rational numbers satisfy that between any two points A and B, there exists a rational C such that A<C<B. And of course, the rational numbers are not complete. So the completeness axiom is not needed at all to show this fact.

Of course, the intuition that complete space has "no holes" is a good intuition. But you should be very careful with specifying what exactly is a "hole".

3. Oct 31, 2012

### Bipolarity

Re: Completeness axiom as having "no holes" in the set

It must then be the case that I did not comprehend what my textbook meant to be a "hole". So what is the "hole" ? My textbook doesn't explain it, perhaps you might know.

Also, if a set of numbers is complete, then does it necessarily mean that it satisfies the A<C<B property that I outlined above?

BiP

4. Oct 31, 2012

### micromass

Staff Emeritus
Re: Completeness axiom as having "no holes" in the set

It's just intuition. I don't think it is meant to be very formal. Maybe you could rigorously define what a "hole" is though.

The idea is the following: consider the following sequence

$$3,~3.1,~3.14,~3.141,~3.1415,...$$

This is a sequence of numbers and it appears that as you go through the sequence, you get closer to a certain number. However, if we work in the rationals, then the limit of this sequence (i.e. the number $\pi$) is not contained in the rationals. So you can get very close to $\pi$, but you can never actually be $\pi$. Thus we say that the number $\pi$ is a missing number (or hole). So a missing number is a number that is not in the set but where you can get close to. The completeness axiom states exactly that there are no missing numbers.

No: consider $[0,1]\cup [2,3]$. This is complete, but it doesn't satisfy the above property.

5. Oct 31, 2012

### lurflurf

Re: Completeness axiom as having "no holes" in the set

Yes the holes the real numbers do not have are equivalently cuts, or limits, or least upper bounds. Just having numbers everywhere is being dense, not complete.