We know that for an infinite dimensional Banach Space the Hamel Basis needs to be uncountable. Why is it difficult from the point of view of constructive analysis to obtain an Uncountable Hamel Basis for an infinite dimensional Banach Space ?
I have found no resources online regarding the construction of a uncountable hamel basis for any infinite dimensional Banach space
Think about how you might create such a basis. Let's say for $\ell^1$. You start with the standard unit vectors, whose span merely gives you vectors which have only finitely many non-zero entries. What do you add next? Maybe $(\frac1n)$? Then $(2^{-n})$? You keep adding things in. Then you lose track of what is a linear span of what you have previously added.
It is similar to creating a Hamel basis of $\mathbb R$ over $\mathbb Q$. Start with $1$, $\sqrt 2$, $\sqrt 3$, $\pi$. Can we add $e$, or is that already in the linear span of the previous elements?
If you want to make it rigorous, you would want to show that there is a model of set theory in which $\ell^1$ does not have a Hamel basis. I found this, but it doesn't precisely answer this question. https://mathoverflow.net/questions/233359/linear-space-with-hamel-basis-and-the-axiom-of-choice
This seems to answer your question rigorously: A Hamel basis for $\ell^p$?
