# Hyperreals as a vector space?

1. Aug 31, 2010

### Rasalhague

Can the hyperreal numbers be described as a vector space over the reals with a basis (0,...,e2,e1,e0,e-1,e-2...), where e1 is a first order infinitesimal number, e0 = 1, and e1 a first order infinite number?

2. Aug 31, 2010

### Hurkyl

Staff Emeritus
Nope!

Omitting zero, that is a basis for a real sub-vector space of the hyperreals. However, it certainly does not span the hyperreals -- your basis has countably many vectors, but the hyperreals are a real vector space of uncountable dimension.

A specific hyperreal that is not in the span is $$\sqrt{e^1}$$.

Also, I think you have some misconceptions about infinitessimals. In particular, for any nonzero infinitessimal e, there are other infinitessimals (such as $\sqrt{e}$) that are infinitely larger in magnitude. I don't think "first order infinitessimal" makes any sense -- unless you have previously chosen a reference infinitessimal to compare things to.

After pondering it for a bit, I think (but have not proven) any basis must actually contain two hyperreals whose ratio is finite, but not infinitessimal.

3. Aug 31, 2010

### Rasalhague

Ah, okay, I didn't realise there wasn't a natural choice of "reference infinitesimal". Would it be possible to somehow arbitrarily define a particular infinitesimal to play this role, perhaps allowing the exponents to be real numbers?

4. Aug 31, 2010

### Rasalhague

For an infinitesimal e, unlimited numbers H and K, and an appreciable number b, according to Goldblatt: Lectures on the hyperreals, p. 51, e/b, e/H, b/H are infinitesimal, b/e, H/e, H/b are unlimited, for e, b not equal to 0, while e/b, H/K, eH are undetermined. So if the ratio x/y is finite (limited), wouldn't that imply x is real, and y real and nonzero? But then x/y would be a multiple of 1 = e0, or whichever other nonzero real number we had in our basis.

5. Aug 31, 2010

### Hurkyl

Staff Emeritus
You can choose a reference infinitessimal. Arguments I've seen often start off with something like "choose an infinitessimal e" or "choose an infinite hyperinteger H".

Let e be a nonzero infinitessimal. The ratio
$$\frac{e+1}{e}$$​
is finite and not infinitessimal, but not standard, and not a real multiple of 1.

Oh, also of interest may be that, assuming we chose e positive, ee is an infinitessimal bigger than er for all positive standard numbers.

Last edited: Aug 31, 2010
6. Sep 1, 2010

### Rasalhague

Another snag: I just remembered, Goldblatt also calls H + K undetermined where H and K are both unlimited.