# Rational power series over $\mathbb N \cup \{\infty\}$, rationality of singular part

Let $\Sigma$ be a finite alphabet, and consider the formel power series over $\Sigma$ considered as non-commuting variables with coefficients in the semiring $\mathcal N := \mathbb N \cup \{\infty\}$ with $$n \cdot \infty = \infty\cdot n = \infty \mbox{ for all n \ne 0, } \qquad \infty\cdot \infty = \infty \qquad n + \infty = \infty + n = \infty \mbox{ for all n, } \qquad 0\cdot \infty = \infty\cdot 0 = 0 \qquad \infty + \infty = \infty.$$

A formal power series is a polynomial if only a finite number of coefficients are non-zero. A formel power series is called rational if it could be built up from polynomials by a finite application of addition (pointwise), multiplication (convolutional product), and the star operatation, which for $f = \sum_{w \in \Sigma^{\ast}} c_w w$ is defined by $$f^{\ast} = 1 + f + f^2 + f^3 + \ldots$$ where the sum taken over an infinite set $\{ k_i \in \mathcal N \mid i \in I \}$ is defined as $\sup\{ \sum_{j\in J} k_j \mid J\subseteq I \mbox{ finite } \}$, i.e. the supremum of all finite sums, hence the above expression makes sense.

These notions go back to Schützenberger and Eilenburg, where the last one calls these $\mathcal N$-subsets instead of formal power series. Now given $f$ we define $$xf_s := \left\{ \begin{array}{ll} \infty & xf = \infty \\ 0 & \mbox{ otherwise.} \end{array}\right.$$ the singular part, i.e. all entries whose coefficient is $\infty$ are retained, and $$xf_n := \left\{ \begin{array}{ll} xf & xf \ne \infty \\ 0 & \mbox{ otherwise } \end{array} \right.$$ the non-singular part, where we retain those entries that do not evaluate to $\infty$. Surely $$f = f_s + f_n$$ and we have \begin{align*} (f + g)_s & = f_s + g_s \\ (f g)_s & = f_s g + g_s f. \end{align*}

Now let $\mathcal C$ be the class of formal power series $f$ such that $f_s$ is rational, then why does the above formula imply that $\mathcal C$ is closed under multiplication?

The argument is taken from Samuel Eilenberg, Automata, Machines, Languages, Volume A to show that for rational $f$, also $f_s$ and $f_n$ are rational, and he uses exactly the above argument. Buf if $f_s$ is rational, and $g$ such that $g_s$ is rational, why should then $f_s g + f g_s$ be rational?

• Isn't it the case that if $f$ and $f_s$ are rational then so is $f_n$? – Michaël Cadilhac Jul 5 '18 at 16:13
• I do not think so, rational languages/power series are in general not closed under complement, but even if this would hold, then how is it related to my question? – StefanH Jul 5 '18 at 22:25