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Take $\mathbb{H}=\mathbb{R}^\mathbb{N}/\mathcal{U}$, where $\mathcal{U}$ is some ultrafilter. Questions:

  • Are there more than one independent infinitessimal in this field. This means $\epsilon_1 > 0$ and $\epsilon_2 > 0$ such that $0 < \epsilon_1 < \epsilon_2 < r$ for all $0 < r \in \mathbb{R}$, and there is no relation (lineair or algebraic) between them ?
  • If there are, are there examples ?


  1. This is like asking for an "hyperlarge number comparable to the ordinal $\omega_n$ with $n>1$", I only wish to know if they can be constructed in $\mathbb{H}$.
  2. I suspect you can proof the existence of more infinitessimals in $\mathbb{H}$ by roughly the same diagonal argument which is used to proof the uncountability of $\mathbb{R}$, but I can't lay my finger on it, and it doesn't lead to a construction (I am not one of the constructionalist. but in this case I wish to know if you can do more than an existence result).


I thought of introducing "a second infinitessimal" to $\mathbb{H}=\mathbb{R}^\mathbb{N}/\mathcal{U}$ by looking at $(\mathbb{R}^\mathbb{N}/\mathcal{U})^\mathbb{N}/\mathcal{V}$, with a second ultrafilter, but (aside from the seond ultrafilter), I don't know in wich swamp I am moving then. Any thoughts on this would be appreciated.

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"No relation" is a slippery concept. We could try to specify it as "there is no polynomial $p(x,y)$ with standard coefficients such that $p(\epsilon_1,\epsilon_2)=0$. But even then something like $\epsilon_1 = e^{-1/\epsilon_2}$ would probably count as "independent" under that definition. – Henning Makholm Jul 5 '12 at 22:38
Willem, I am not sure how this is "like asking for an ordinal $\omega_n$ with $n>1$". – Asaf Karagila Jul 5 '12 at 22:42
If you are a constructivist, you will have trouble with the existence of a non-principal ultrafilter $\mathcal{U}$ needed to produce the non-standard model. – André Nicolas Jul 5 '12 at 23:01
One should also be careful when saying "the hyperreals" since unlike the real numbers there are several non-isomorphic hyperreal fields. – Asaf Karagila Jul 5 '12 at 23:02
@Asaf 1) The $\omega_n$ stuff is to be seen with two large quotes around it, of course you cannot expect that (for example) $\omega_1 \in \mathbb{H}$. This, and the fact that ordinals are not the same as hyperlarge numbers. I did put some quotes around that remark. – Willem Noorduin Jul 6 '12 at 7:25
up vote 5 down vote accepted

Let $$\epsilon_2=\left\langle\frac1{n+1}:n\in\omega\right\rangle^{\mathscr{U}}$$ and $$\epsilon_1=\left\langle\frac1{n!}:n\in\omega\right\rangle^{\mathscr{U}}\;;$$ clearly $a\epsilon_1^b<\epsilon_2$ for any $a,b\in\Bbb R^+$. Is that sufficient independence?

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One might or might not consider $\epsilon_1 = \frac{1}{\Gamma(1/\epsilon_2)}$ a relation... – Henning Makholm Jul 7 '12 at 11:05

Here is a partial answer for linearly independence:

Let $V=\{\pm\epsilon\in\Bbb H\mid\epsilon\text{ is infinitesimal}\}\cup\{0\}$. Note that the sum of two infinitesimals is either zero or an infinitesimal. Furthermore, if $a\in\mathbb R$ then $a\cdot\epsilon$ is also an infinitesimal.

This is a vector space over $\mathbb R$. Its dimension cannot be $1$ since $\epsilon^2\neq\epsilon$, but if $\epsilon^2=a\cdot\epsilon$ then $\epsilon=a\in\mathbb R$ but that could only be if $\epsilon=0$ to begin with.

So linearly independent over $\mathbb R$ there has to be more than one infinitesimal. I suspect that a similar argument will show that there cannot be finitely many.

We can also observe that $V$ is closed under products and forms an integral domain. I think we can squeeze this argument down to show that as an algebra it is also not finitely generated, so the independence is not only linear but also polynomial.

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