To prove a combinatorial identity I'm building an ad hoc vector space that I will define below.

First, I define the set of Pseudo Power Series. A Pseudo Power Series (pps) $\alpha$ is an infinite sequence of real numbers $a_0, a_1, a_2, \dots$ denoted by $\alpha = (a_0, a_1, a_2, \dots)$, in which the order of the elements $a_0, a_1, a_2, \dots$ matters. We will denote by PPS the set of all pps.

Before going any further, let us set $f_0 = (1, 0, \dots, 0, \dots)$ and, in general, \begin{align*} f_i = (0, 0, \dots, a_{i-1} = 0, a_i = 1, a_{i+1} = 0, \dots). \end{align*} Moreover, we set $(0, 0, \dots, 0, \dots) = \boldsymbol{0}$ and $(1, 1, \dots, 1, \dots) = \boldsymbol{1}$. Finally, it's immediate to see that for any $f_i$, $f_j$ it holds \begin{align*} f_i * f_j =\begin{cases} f_i & \text{if } i = j\\ \boldsymbol{0} & \text{if } i \neq j.\\ \end{cases} \end{align*}

We also define the scalar multiplication \begin{align*} \cdot : \mathbb{R} \times (\text{PPS}, +, \boldsymbol{0}) \rightarrow (\text{PPS}, +, \boldsymbol{0}) \end{align*} for $r \in \mathbb{R}$, and $\alpha \in$ PPS as \begin{align*} r \cdot \alpha = (ra_0, ra_1, \dots, ra_i, \dots). \end{align*}

Now we come to my question: I thought that the set $\{f_0, f_1, \dots, f_k, \dots\}$ could be considered as a basis for the vector space $(\text{PPS}, +, \boldsymbol{0})$, but it is not true. To be a basis it has to span every element of PPS, but it is not possible: no finite linear combination of scalars and element of that basis can span every element of PPS.

So my question is: how can I extend the concept of basis for this infinite dimensional vector space? Do you know a way? Maybe if I use modules or other structures I can use infinite linear combinations? Thank you a lot.

  • 2
    $\begingroup$ Welcome to stackexchange. Your vector space does have a basis, but not one that's easy to come by. You can read about it if you search for Hamel basis and follow a link that matches your level of mathematical sophistication. Consider math.lsa.umich.edu/~kesmith/infinite.pdf $\endgroup$ – Ethan Bolker Jun 25 '18 at 12:24
  • $\begingroup$ Thank you. I will read the pdf you have posted. In particular "Example 2" seems to represent my case. $\endgroup$ – Matteo Bodini Jun 26 '18 at 7:37

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