A positive number smaller than any positive infinitesimals I've been searching everywhere for an answer, but to no avail. I know that the hyperreals are a quotient of the space $\mathbb{R}^\mathbb{N}$ of sequences of real numbers modulo a fixed free ultrafilter on $\mathbb{N}$. Can this method be generalised to sequences of hyperreal numbers to yield numbers smaller than any positive infinitesimals but still greater than $0$?
If someone has already provided a semblance of an answer to this question, please give me the link.
 A: Quibble: it's not really right to say that the hyperreals are a quotient of $\mathbb{R}^\mathbb{N}$. Rather, there is a general notion of a hyperreal field, roughly any ordered field containing $\mathbb{R}$ satisfying an appropriate transfer property. Every nontrivial ultrapower $\mathbb{R}^\mathbb{N}/U$ is a hyperreal field, and this is the sort of hyperreal field most commonly considered, but not all hyperreal fields have this form.

Well, there's a bit of a misuse of the term "infinitesimal" here - by definition, a (positive) infinitesimal is a positive number smaller than all positive reals, so anything positive and smaller than some infinitesimal is again an infinitesimal - but I think the following addresses your intuition.
Suppose we have an embedding of ordered fields $F\subseteq K$. We can say that $K$ has $F$-infinitesimals iff there is some $\alpha\in K$ such that $0<\alpha$ but $\alpha<\beta$ for every $\beta\in F$ with $\beta>0$. (Keep in mind that since $F$ is a sub-ordered field of $K$ the ordering on $F$ is the restriction of the ordering on $K$, so there's no ambiguity with me using "$<$" above.) For example, $\mathbb{R}$ does not have $\mathbb{Q}$-infinitesimals, but every non-Archimedean field does have $\mathbb{Q}$-infinitesimals and if $F$ is any hyperreal field then $F$ has $\mathbb{R}$-infinitesimals.
The usual ultrapower construction of the hyperreals then lets us prove:

For every ordered field $F$ there is an ordered field $K\supseteq F$ which has $F$-infinitesimals.

This is a good exercise. Basically, take as our index set the set $I=\{\alpha\in F: \alpha>0\}$, and show that there is an ultrafilter $U$ on $I$ such that for each $\alpha\in F$ we have $\{\beta\in I: \beta<\alpha\}\in U$. Now think about what happens with the identity function $id:\alpha\mapsto\alpha$ in the power structure $F^I$ after we "mod out by $U$" ...
(Alternatively, just use compactness! Keep in mind that compactness can be proved via ultraproducts, so it's not so much an alternative to the above as it is a repackaging of the basic idea so that we don't have to do it over and over again in other contexts.)
So in a sense we can always find "more infinitesimal" elements - we just have to go to larger and larger fields in order to do so.
