This is slightly different question. First I need to mention that I am neither a mathematician nor a researcher. As an ordinary student the separation " with and without Axiom of choice " trouble me a lot. But I am really interested to know further.
Every ideal in a ring contained in some maximal ideal.
Every vector space has a basis.
Every linear space can be made into a Normed space.
Product of any collection of compact topological spaces is compact with respect to the product topology.
There exists a subset of $\Bbb{R}$ which is not measurable (Lebesgue Measurable) for an example the famous Vitali Set .
All the above theorem can be proved using Axiom of choice.
The natural question is , what happened if we are not allowed to use AC ?
Without AC, can we get -
An ideal which is not contained in any maximal ideal in some ring.
A linear space without basis.
A linear space in which no norm can be defined.
Product of any collection of compact topological spaces which is not compact with respect to the product topology.
A non measurable subset of $\mathbb{R}$ .
If the answer of above questions are "Yes", then the study will be more complicated . [except for 5), it will be fun to have all subsets of $\mathbb{R} $ are measurable]. I don't know a linear space without any basis will be useful or not ?
Why the proof of such kind of theorem involves " if we assume the Axiom of choice, then... " ?
Why not we take it as granted as an extra axiom and we forget about the situation happening without the AC ?
As we know given any two distinct points we can draw a unique straight line passes through them. But no one ask the question if the Euclid's axiom is not listed as an axiom , can we draw more lines using two distinct points.
My question is very clear, why not we avoid the phrase "if we assume the AC " and take it as granted ?