# Basis for infinite dimensional vector space definition

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?

• I don't know weher you confuse. Do you know definition of vector space? Feb 4 at 15:25
• It is poorly worded. You could save the definition by just omitting the first instance of the word "finite." Feb 4 at 15:31
• @AlexOrtiz exactly Feb 4 at 15:40

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.
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 $$X$$.