Is there more than one infinitessimal among the hyperreal numbers Take $\mathbb{H}=\mathbb{R}^\mathbb{N}/\mathcal{U}$, where $\mathcal{U}$ is some ultrafilter. Questions:


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*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 ?


Notes:


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*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 $\mathbb{H}$.

*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).


Edit:
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.
 A: 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?
A: 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.
