Main Answer:
An infinite-dimensional Banach space cannot have countably infinite algebraic dimension.
Step-by-step explanation:
In mathematics, the algebraic dimension of a vector space refers to the size of its basis. For a Banach space, an infinite-dimensional space, if it were to have a countably infinite algebraic dimension, it would imply the existence of a countable basis. However, this contradicts the Baire Category Theorem, which states that the intersection of any countable collection of dense open sets in a complete metric space is non-empty. In the context of a Banach space, this implies that the space itself cannot be expressed as the countable union of nowhere-dense sets.
Assuming the contrary, where an infinite-dimensional Banach space has a countably infinite algebraic dimension, we can construct such nowhere-dense sets using finite-dimensional subspaces spanned by subsets of the basis. The union of these sets, being countable, should cover the entire space. This leads to a contradiction with the Baire Category Theorem.
Consequently, an infinite-dimensional Banach space cannot possess a countably infinite algebraic dimension, emphasizing the profound implications of functional analysis in understanding the structure of infinite-dimensional spaces.