# Why should $\mathcal{A}$ not have elements of size $0$ to form $\text{Seq}(\mathcal{A})$?

In Flajolet's symbolic method, given a combinatorial class $$\mathcal{A}$$, $$\text{Seq}(\mathcal{A})$$ is defined as $$\mathcal{E}+\mathcal{A}\times \mathcal{A}+ \mathcal{A}\times\mathcal{A}\times \mathcal{A}+...$$ and assuming that the generating function of $$\mathcal{A}$$ is $$A(z)$$ then the generating function of $$\text{Seq}(\mathcal{A})$$ is $$1+A(z)+A(z)^2+A(z)^3+...=\frac{1}{1-A(z)}$$.In the bibliography it is requested that $$\mathcal{A}$$ be a combinatorial structure without objects of size $$0$$, but it is not clear to me why, could someone explain it to me (I think the same thing happens with the $$\text{Cyc}( \mathcal{A})$$ operation).

If there is an object $$z\in \mathcal A$$ with weight zero, then there would be infinitely many sequences in $$\text{Seq}(\mathcal A)$$ with weight zero. Namely, $$(z),(z,z),(z,z,z),\dots,(z,z,\dots,z),\dots$$ all have weight zero. A combinatorial class is only allowed to have finitely many elements of each weight, so $$\text{Seq}(\mathcal A)$$ is not a well-defined combinatorial class in this case.
• @user19872448 Same is true for $\text{Cyc}$ and $\text{MSet}$. There might be others. Apr 13 at 16:12
• so, to define $\text{Seq}(\mathcal{A})$ $\mathcal{A}$ must not have a neutral element, but if I define $\mathcal{B}=\text{Seq}(\mathcal{A})$ then $\mathcal{B}$ does have a neutral element ? Apr 20 at 22:07