Basis for infinite dimensional vector space definition

Question

I'm reading Luenberger's Optimization by Vector Space Methods which has the following definition:

A finite set $S$ of linearly independent vectors is said to be a basis for the space $X$ if $S$ generates $X$. A vector space having a finite basis is said to be finite dimensional. All other vector spaces are said to be infinite dimensional.

Am I going crazy for thinking that this definition does not really allow for infinite dimensional spaces since $S$ is defined as a finite set? Maybe there's some subtlety I'm missing here?

Answer 1

That definition makes sense. Take, for instance, the vector space $\Bbb R^{\Bbb N}$ of all sequences of real numbers. There is no finite set $S\subset\Bbb R^{\Bbb N}$ which spans $\Bbb R^{\Bbb N}$. Therefore, $\Bbb R^{\Bbb N}$ is infinite-dimensional.

There is a problem with that definition however: it doesn't allow the existence of bases of infinite-dimensional vector spaces.

Answer 2

The definition is: $X$ is infinite dimensional iff there does not exist any finite subset $S\subseteq X$ such that $S$ is a basis for $X$.

For instance $X = {\Bbb R}^{\Bbb N}$ (the vector space of all real sequences) is infinite dimensional because whatever finite subset $S \subset X$ you pick, it isn’t a basis for $X$.

Answer 3

Notice that by the Axiom of Choice, a basis will always exist. Notice too, that in the finite-dimensional case, the basis will be a Hamel Basis. While in infinite dimensional spaces it may be infinite in cardinality, the sums $$ \Sigma_{i=1}^n b_iv_i $$ , will be finite sums; sums over finite subsets of our infinite basis. In some of these infinite-dimensional vector spaces, when they're normed, there may be Schauder Bases , where we have infinite sums, which require a notion of convergence.