Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are such transfinite numbers as $\aleph$ and $\omega$. Do they have infinitesimal analogues? I'm especially interested in numbers 'below' $0$ having one peculiarity: it should be non-negative. Is there chance there is/are infinitesimal number(s) which are non-negative, but still lower than zero? I'm not sure I should find what I need among infinitesimals, but your answer might help me to move forward with finding what I exactly need. So, are there such numbers that might be considered to be 'transfinitesimal'?

share|cite|improve this question

Yes, though these 'transfinitesimal' numbers aren't below zero, they are above zero but smaller than every positive real number. What you are looking for is called non-standard analysis,

Proving the existance of such things is usually rather easy, BTW, if you apply the completeness and compactness theorems. It then amounts to

  1. Choosing a set of axioms that describes the structure you're interested in, like the natural numbers or the reals. Note that you must do this in first-order predicate logic, to have the completeness and compactness theorem available.
  2. Define a (usually infinite) set of axioms which, if taken all together, force the existance of "strange" objects. In the case of the reals, you could add a constant $c_\omega$ and take the set of axioms $\left\{0 < c_\omega < 1, 0 < c_\omega < \frac{1}{2}, 0 < c_\omega < \frac{1}{3}, \ldots\right\}$.
  3. Then, by compactness, if every finite subset of those new axioms is consistent with the original theory, the whole set is also.
  4. And by completeness, if the whole set is consistent, it has a model. That model must then include a "strange" object, since it must supply some value for $c_\omega$, and that value must fullfill the whole set of newly added axioms.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.