 12:30 PM
13  Let $X$ be an infinite dimensional Banach space. Prove that every basis of $X$ is uncountable. Can anyone help how can I solve the above problem?

This was posted as a comment there:
Can't we prove it without Baire Category Theory in other words without axiom of dependent choice — Sushil 12 mins ago
However, I am not sure whether it makes sense to ask this without axiom of choice.
Or how should it be formulated so that it makes sense without Axiom of Choice.
The fact that every vector space has a Hamel basis needs full axiom of choice. I am not sure what happens if I replace "every vector space" with "every Banach space".
Should I perhaps read the question in the comment as: "Assuming that $X$ has a Hamel basis, can we show that it is not countable."?
@Sushil You have a much better chance of getting some answer if you post your question as a question, not just as a comment. However, before posting such question, some clarifications are needed in my opinion. See here for some comments. — Martin Sleziak 11 secs ago 12:53 PM
Oh I see. But I want some clarity. Cardinality of Hamel basis(if exist) are equal does it imply AC(or ADC). If this implication is wrong I may ask Let X be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable without Baire Category Theory. — Sushil 6 mins ago
Infact A vector space can't have countable basis and uncountable basis simultaneously. Do we really need AC here or can we prove it with Axiom of countable choice(if needed) — Sushil 2 mins ago
@Sushil: Without the axiom of choice it is consistent for a vector space to have two bases which have different cardinalities. I don't know if it consistent that one of them is well-ordered, though. — Asaf Karagila 1 min ago
I think that since Asaf noticed your question, you have a good chance of getting some reasonable answer. He knows a lot about AC and how things behave without AC.
@AsafKaragila Can we also prove cardinality of Hamel bases(if exist) are equal then it implies AC. — Sushil 8 mins ago
14  Assuming the axiom of choice does not hold we have that there is a vector space without a basis. The situation can be, in some sense, worse. It is consistent that there are vector spaces that have two bases with completely different cardinalities. Is anything known on when a vector space is spa... 1:18 PM
@MartinSleziak Ok but as I noticed from link mentioned by Asaf "vector space dimension is well-defined" is much weaker than DC. Hence now my question makes sense to ask: 'Prove that every Hamel basis in an infinite dimensional Banach Space of X is uncountable without Baire Category Theory assuming vector space dimension is well-defined". I think I can post it now. — Sushil 5 mins ago

4 hours later… 5:12 PM
-1  Prove that every Hamel basis in an infinite dimensional Banach Space is uncountable without using Baire Category Theory. We are assuming axiom that vector space dimension(if exist) is well-defined. Note: Axiom that vector space dimension(if exist) is well-defined is weaker than DC. http://mathov...