# Given an arbitrary ordered field (F,<), is there always a sequence of strictly positive elements of F that tends to zero?

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.

• I think not, in $\mathbb{R}^*$. Because for any sequence tending to zero, you can find a sequence that tends to zero faster. – Lior B-S Aug 12 at 8:28

$$\DeclareMathOperator{\cof}{cof}$$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.