I just have a question about why some of the theorems involving dual space fails for infinite dimension.
A standard problem is below:
"Suppose V is finite-dimensional and v is an element of V with v does not equal 0. Prove that there exists an element in the dual space of V such that phi(v)=1"
The answer is to extend v to a basis of V, then we have a corresponding basis for V'(V' is the dual space of V) and by definition, phi1(v)=1 and 0 for all the other basis of V. So why does the question requires V to be finite dimension? Could we not do the same if V was infinite dimensional?
Best Answer
The argument works fine in infinite dimensions except that you need the axiom of choice to guarantee that infinite-dimensional vector spaces have bases, and in fact "every vector space has a basis" is equivalent to the axiom of choice; this is due to Blass.
In the finite-dimensional case this is true in ZF and it's not necessary to get into the set-theoretic weeds like this.
Here is a nice relatively explicit example: $\mathbb{R}$ is an infinite-dimensional vector space over $\mathbb{Q}$. Can you write down any nonzero linear functional $\mathbb{R} \to \mathbb{Q}$ whatsoever? In fact it's consistent with ZF that there don't exist any, and any such linear functional must be non-measurable.