1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hyperreals as a vector space?

  1. Aug 31, 2010 #1
    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. jcsd
  3. Aug 31, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member


    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 [tex]\sqrt{e^1}[/tex].

    Also, I think you have some misconceptions about infinitessimals. In particular, for any nonzero infinitessimal e, there are other infinitessimals (such as [itex]\sqrt{e}[/itex]) 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.
  4. Aug 31, 2010 #3
    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?
  5. Aug 31, 2010 #4
    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.
  6. Aug 31, 2010 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
    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
  7. Sep 1, 2010 #6
    Another snag: I just remembered, Goldblatt also calls H + K undetermined where H and K are both unlimited.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook