Vector space (no topology) basis

In summary: Yes, this vector space has a basis, but it is uncountable. The basis is not easily writable as it requires the axiom of choice. However, a countable basis can be constructed using a Schauder basis, where the basis elements are sequences with only one non-zero entry.
  • #36
nuuskur said:
Which vector space do you have in mind, exactly? As long as it's non-zero, there is a basis.
Vector space elements are infinite sequences of zeros and ones, with arithmetic mod 2 for the scalars..
 
Physics news on Phys.org
  • #37
mathwonk said:
but i do not know at all how to write down a vector basis for the direct product ("hamel basis"), and presumably no one else does either. so this is a nice explicit example of something whose existence is guaranteed by zorn's lemma, but apparently no one has ever seen or explicitly described one.
yes I am pretty sure there's a theorem saying existence of hamel basis is equivalent to AC so if someone managed to write down explicit basis then that can only mean trouble...

this is like the issue with ultrafilters and well ordering of reals and many other things.
 
  • #38
mathman said:
Vector space elements are infinite sequences of zeros and ones, with arithmetic mod 2 for the scalars..
Then your scalars are essentially just 0 and 1 then, I guess.
 
  • #39
WWGD said:
Then your scalars are essentially just 0 and 1 then, I guess.
Yes. Scalar arithmetic is mod 2.
 

Similar threads

  • Linear and Abstract Algebra
Replies
9
Views
192
  • Linear and Abstract Algebra
Replies
6
Views
869
  • Linear and Abstract Algebra
Replies
9
Views
567
  • Linear and Abstract Algebra
Replies
4
Views
2K
  • Linear and Abstract Algebra
Replies
8
Views
873
  • Linear and Abstract Algebra
Replies
3
Views
293
Replies
26
Views
2K
Replies
12
Views
3K
  • Linear and Abstract Algebra
2
Replies
43
Views
5K
  • Linear and Abstract Algebra
Replies
17
Views
9K
Back
Top