A Vector space (no topology) basis

[nuuskur]

It is a basis in the "usual sense" if the space contains only those sequences for which almost all coordinates are zero. That's because you are allowed finite combinations.

As for Schauder bases - the sequences you describe give a Schauder basis in any $\ell _p$, where $1\leq p<\infty$.

Important thing to note: a Schauder basis is not necessarily a basis.

 

[mathman]

So far as my understanding: (Hamel) basis allows only a finite linear combination - no topology. Schauder basis allows countable combinations with limiting such. as $l_p,\ 1\le p\lt \infty$. My question: can a concept of basis (any name) be used in $l_\infty$?

 

[nuuskur]

My question: can a concept of basis (any name) be used in $\ell _\infty$?

Standard math accepts the axiom of choice. Yes, the space does have a basis, but it's difficult to describe explicitly and it is uncountable.

 

[mathman]

What about the vector space using integers mod2 as the field?

 

[nuuskur]

Which vector space do you have in mind, exactly? As long as it's non-zero, there is a basis.

 

[mathman]

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..

 

[black hole 123]

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.

 

[WWGD]

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.

 

[mathman]

Then your scalars are essentially just 0 and 1 then, I guess.

Yes. Scalar arithmetic is mod 2.