Why does the contradiction in this R^∞ dimensionality proof work?

[JamesGold]

The proof goes as follows:

For contradiction, assume there exists |s|< ∞ such that s = {e₁, e₂, ... , e_n} and span(s} = ℝ^∞.

The above makes at least some sense to me. The proof goes on...

Let m > n and u = e_n+1 + e_n+2 + ... + e_m

u $\notin$ span(s), u $\in$ s

Because {e₁, e₂, ... , e_n} $\notin$ s, this implies a contradiction. Therefore ℝ^∞ is infinite-dimensional.

I don't see what we just did, and I don't see how what we just did proves that R^infinity is infinite-dimensional. Please help!

 

[mfb]

In words:
Assume that there is a finite base with a finite number of base vectors e₁ to e_n*.
Consider e_n+1 (+maybe more vectors) - where it is unclear in which basis that is supposed to be*. This vector cannot be represented by a sum of e₁ to e_n, therefore the initial assumption is wrong and a finite base cannot exist.

* I think the combination of those are a serious problem. While is possible to name (arbitrary) base vectors with e_i, this cannot be extended to other vectors. If e_i refer to the standard base, you cannot simply assume that a finite base uses those vectors.