What would a rigorous proof that the vector space of functions from $\mathbb{Z}$ to $\mathbb{R}$ is not finite-dimensional look like?

Question

So consider the vector space of functions from $\mathbb{Z}$ to $\mathbb{R}$. Intuitively, this vector space is not finite-dimensional. Indeed, $\mathbb{Z}$ and $\mathbb{N}$ are the same by "zig-zagging", and with regards to the vector space of functions from $\mathbb{N} \to \mathbb{Z}$, indeed I can exhibit via construction infinitely many linearly independent vectors via just taking vectors$$(0, 0, 0, \ldots, 0, 1, \ldots).$$My concern is, I feel as if this is not rigorous. What would a rigorous proof that the vector space of functions $\mathbb{Z}$ to $\mathbb{R}$ is not finite-dimensional look like? Thank you in advance!

Answer 1

You may prove it by contradiction.

Just assume the given space was finite dimensional with dimension $n$.

Then any $n+1$ vectors would be linearly dependent.

But, you can find $n+1$ linearly independent vectors $e_1, \ldots , e_{n+1}$ with $e_k(i) = \delta_{ik}$ for $k=1, \ldots , n+1$ and $i \in \mathbb{N}$.

This contradiction shows that the assumption was wrong.