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Let $(F,<)$ be an ordered field. Is there always a sequence $(a_n)_{n\in\mathbb N}$ of strictly positive elements of $F$ such that $\lim_{n\to\infty}a_n=0$? (To define limits, I'm using the topology induced by the total ordering of $F$).

If the field is archimedean, than setting $a_n=\frac{1}{n+1}~\forall n\in\mathbb N$ will work just fine. However, if the field is non-archimedean, then there is a strictly positive $\epsilon\in F$ such that $\forall n\in\mathbb N:\epsilon<\frac{1}{n+1}$, thus the sequence defined above is not eventually in the interval $(-\epsilon,\epsilon)$, and it does not approach zero.

So the problem seems to be the existence of infinitesimal elements.

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  • $\begingroup$ I think not, in $\mathbb{R}^*$. Because for any sequence tending to zero, you can find a sequence that tends to zero faster. $\endgroup$ – Lior B-S Aug 12 at 8:28
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$\DeclareMathOperator{\cof}{cof}$The answer for a given ordered field $F$ depends on its cofinality $\cof(F)$. This is the least cardinality of an unbounded subset of $F$, or equivalently, the least order type of a well-ordered subset of $F$ without upper bound.

If $\cof(F)=\aleph_0$, then picking a strictly increasing sequence $(u_n)_{n \in \mathbb{N}}$ of strictly positive elements without upper bound, then the sequence $(\frac{1}{u_n})_{n \in \mathbb{N}}$ tends to zero.

If $\cof(F)>\aleph_0$, then for every sequence $(a_n)_{n\in \mathbb{N}}$ of strictly positive elements, the sequence $(\frac{1}{a_n})_{n \in \mathbb{N}}$ has an upper bound $A$, so $(a_n)_{n\in \mathbb{N}}$ lies above $\frac{1}{A}$ and thus does not converge to zero.

For instance, fields of hyperreal numbers have uncountable cofinality and thus have no strictly positive sequences which tend to zero.

On the other hand, many ordered fields have countable cofinality: archimedean fields, the field of rational fractions / functions over an archimedean field, Laurent and Puiseux series, Levi-Civita series, logarithmic-eponential transseries...

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