# 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?
– Yos
Commented Feb 4, 2023 at 15:25
• It is poorly worded. You could save the definition by just omitting the first instance of the word "finite." Commented Feb 4, 2023 at 15:31
• @AlexOrtiz exactly Commented Feb 4, 2023 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.

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

• Thank you for confirming. This is what I got hung up on (not allowing bases to exist for infinite dimensional vector spaces). Commented Feb 4, 2023 at 15:39
• I'm glad I could help. Commented Feb 4, 2023 at 15:40

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 $$X$$.

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.

• Ah, another anonymous down voter. Please tell the reason for the doenvote.
– MSIS
Commented Dec 25, 2023 at 12:51