Let $\Omega(t)=\Bbb{R}(t)$ denote all rational functions with real coefficients. Supposing $x(t),y(t) \in \Omega(t)$, a total strict order $<$ is defined where $x(t)<y(t) \iff \exists \, T \in \Bbb{R}^+,\forall \, t>T,x(t)<y(t)$. $\Omega(t)$ is made into an ordered field in this sense.
Since the order is dense, a natural definition of sequence convergence arises where $x_n \rightarrow x \iff \forall \, \varepsilon>0,\exists \, N\in \Bbb{Z}^+,\forall \, n>N,-\varepsilon<x_n-x<\varepsilon$. Well-defined Cauchy series are also allowed where a sequence $\{x_k\}$ is Cauchy $\iff \forall \, \varepsilon>0,\exists \, K\in \Bbb{Z}^+,\forall \, p,q>K,-\varepsilon<x_p-x_q<\varepsilon$.
Q1: Is $\Omega(t)$ itself Cauchy complete in the above sense? If not, then does its Cauchy completeness contain a natural embedding of the hyperreal numbers? An explicit embedding is welcome.
Q2: Replace each $\Bbb{R}$ in the definition with $\Bbb{Q}$. The Cauchy completeness of $\Omega'(t)$ does not canonically contain $\Bbb{R}$ since usual convergent number series in $\Bbb{Q}$ does not converge under the existence of infinitesimal elements. However it is indeed Cauchy complete, so can $\Bbb{R}$ be embedded into the Cauchy completeness of $\Omega'(t)$? It is also better with an explicit answer.
Best Answer
The field $\mathbb{R}(t)$ itself is not Cauchy complete and neither is $\mathbb{Q}(t)$. Indeed, writing $t$ for the identity function, the sequence $(\sum \limits_{k=0}^n \frac{1}{k! \ t^k})_{n \in \mathbb{N}}$ is Cauchy but does not converge.
The Cauchy completion of $\mathbb{R}(t)$ is the field $\mathbb{R}((t))$ of formal Laurent series, i.e. formal sums $\sum \limits_{k>n} a_k t^{-k}$ where $n \in \mathbb{Z}$ and $(a_k)_{k >n}$ is a family of real numbers. Fields of hyperreal numbers don't embed into $\mathbb{R}((t))$. Indeed, on the one hand, in such a field $^*\mathbb{R}$, for each positive infinite element $H$, there is a hyperreal $\exp(H)$ with $\exp(H)>H^n$ for all $n\in \mathbb{N}$.
On the other hand, for any positive infinite $y \in \mathbb{R}((t))$, the sequence $(y^n)_{n \in \mathbb{N}}$ has no upper bound.
Likewise, the Cauchy completion of $\mathbb{Q}(t)$ is the field $\mathbb{Q}((t))$ of formal Laurent series with rational coefficients. This field does not contain an isomorphic copy of $\mathbb{R}$, for it contains no square root of $2$.