Let $V$ be a vector space in which $f$ is a function defined on the natural numbers. $f(2x)$ $=$ $0$ for every $x$ greater than or equal to $0$ and there exists a number $n$ greater than or equal to $0$ such that $f(x)$ $=$ $0$ whenever $x$ is greater than or equal to $n$. Find a spanning set of $V$ other than the trivial set and $V$ itself.
So I take from this that every odd element in the space may not be $0$ up until the $n$ $-$ $1th$ element.
Therefore, I thought all vectors in the set could be written in the form a(0, 1, 0, 0,.....) + b(0, 0, 0, 1, 0, 0,......) + c(0, 0, 0, 0, 0, 1, 0, 0, 0,.......)+.......
This led me to believe the following could be a suitable spanning set:
$span${(0, 1, 0, 0,.....),(0, 0, 0, 1, 0, 0,......), (0, 0, 0, 0, 0, 1, 0, 0, 0,.......)}
The reason I thought this was suitable was that the odd indexes were the only ones that could contain numbers other than $0$ up till a certain point. And I see that for the first three possibilities, this set works, but if we keep on going, their are an infinite number of places 1 could be. So I am not sure how I would illustrate that as a spanning set other than something like this. Note that the natural numbers convention in this case starts from $0$.
Any help?