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Let $R$ be any ordered field. Define $R((x)):=\{\sum\limits_{k=-n}^\infty a_kx^k:a_k \in R \ , \ n \in\mathbb N\}$

Addition: $\sum_\limits{k=-n}a_kx^k+\sum_\limits{k=-m}b_kx^k:= \sum_\limits{k=-max\{n,m\}} (a_n+b_n)x^n$

Multiplication: $\sum_\limits{k=-n}a_kx^k \times\sum_\limits{k=-m}b_kx^k:= \sum_\limits{k=-(n+m)} \left(\sum_\limits{i+j=k}(a_ib_j)\right)x^k$

Order Relation: $\left(\sum_\limits{k=-n}^\infty a_kx^k\right) \geq 0$ iff the first nonzero coefficient is greater than or equal to $0$. Using this we can define $\alpha \leq \beta$ iff $0 \leq \beta-\alpha.$

Then $\left(R((x)),+,\times, \leq\right)$ is an ordered field.

Archimedean Property: By the order relation defined above, we can see that there exists $x^{-1} \in R((x))$ with $1<2<...<n<...<x^{-1}$ $\implies$ the field is non-archimedean.

Defining modulus function: $\left|\sum_\limits{k=-n}a_kx^k\right|_{R((x))}:=\begin{cases} \sum_\limits{k=-n}a_kx^k & \text{if first nonzero coefficient postive} \\ -\sum_\limits{k=-n}a_kx^k & \text{if first nonzero coefficient negetive} \end{cases}$

Cauchy Sequences: $(\alpha_n)$ is said to be cauchy, if for every $(0<)\varepsilon \in R((x))$ there exists $k \in\mathbb N$ such that for all $n,m \geq k$ we have $|\alpha_n-\alpha_m|_{R((x))}<\varepsilon$.

Now I want to see if $R((x))$ is Cauchy Complete, (Do I need to assume that $R$ is Cauchy Complete?)

Let $(\alpha_n)$ be a Cauchy Sequence in $R((x))$ that is for any $(0<)\varepsilon \in R((x))$ there exists an $k \in\mathbb N$ such that for all $n,m \geq k$ we have $|\alpha_n-\alpha_m|_{R((x))}<\varepsilon$.

Taking the first nonzero coefficient from each $\alpha_n$ we can obtain a Cauchy sequence $(a_n)$ in $R$.

If I assume $R$ is Cauchy Complete then $(a_n) \to a(\in R)$.

But I don't know what to do next.

I am not sure whether the above definitions are meaningful. Any insight will be of great help.

Saikat
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    Your claim that $(a_n)_{n \in \mathbb{N}}$ is false in general. Take $\varepsilon_k:=x^k$. Given $n,m\geq k$ with $|\alpha_n - \alpha_m|< \varepsilon_k$. What can you say about the partial sums of $\alpha_n$ and $\alpha_m$ up to $x^{k-1}$ if you get what I mean? – nombre Jan 08 '21 at 19:00
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    If you really meant for all $\varepsilon>0 \in R((x))$ then $\alpha_n$ is Cauchy iff the coefficients of $x^k,k\le K$ are constant for $n$ large enough, so $R((x))$ is complete independently of $R$. – reuns Jan 08 '21 at 19:05
  • @reuns: Is K fixed for all $\varepsilon$? – Saikat Jan 09 '21 at 14:58
  • @nombre: The way I have defined modulus function and Cauchy sequences, are they ok? – Saikat Jan 09 '21 at 16:14
  • @reuns : Can you please explain why "Cauchyy iff the coefficients ..." and how that makes $R((x))$ complete? – Saikat Jan 09 '21 at 16:16
  • @Saikat Yes, the notion of Cauchy sequence and modulus function are the usual ones. – nombre Jan 10 '21 at 17:45
  • @nombre: I have added a question regarding this field. If possible please have a look at it here – Saikat Jan 11 '21 at 20:06

1 Answers1

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Given $(a_n)$ Cauchy,

For all $K$ there is $N_K$ such that for $n\ge m\ge N_K$ either $0\le a_n-a_m < x^K$ or $0\le a_m-a_n < x^K$ which means that $a_n-a_m\in x^K R[[x]]$.

Writing $a_n = \sum_{j\ge J_n} A_{n,j} x^j$ we get that $(A_{n,j})_{n\ge 1}$ is constant for $n$ large enough and $$\lim_{n\to \infty} a_n = \sum_{j\ge \min(0,J_{N_1})} (\lim_{n\to \infty} A_{n,j}) x^j\in R((x))$$

(the topology is generated by the open sets $\{ f\in R((x)), b-\epsilon<f<b+\epsilon\}$)

reuns
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