# A Vector space (no topology) basis

#### mathman

For a topic that is specified to be not topological, there is sure a lot of topology involved To be quite frank, we don't seem to understand what your objective is, mathman.

The definition of Cauchy sequence (assuming at least some kind of metric present) states roughly that $d(a_n,a_m)$ eventually becomes arbitrarily small. Therefore, whether a sequence is Cauchy depends on given topology (is a given $\varepsilon$-ball contained in a certain open set or not). A basis is just a way to express the sequence elements in terms of basis elements.

Furthermore, we have no control how the Schauder basis operates. The definition states that every element can be expressed as a uniquely determined infinite combination of basis elements, that's it. (For instance, you could have almost all scalars equal to zero). We're not allowed to make arbitrary combinations and declare them elements in the space.

On the other hand the problem seems kind of trivial. There is a representation $\sum _{n\in\mathbb N} s_ne_n = (0,1,0,1,\ldots)$. By definition the series converges so the partial sums give us a Cauchy sequence.
I am not sure what the terms in the sequence are. If they are (0,1,0,1,...) for n terms, the sup norm for the difference is always 1, so they don't converge. If you drop a requirement that only finite combinations of basis elements are allowed, then I have no problem with the Schauder basis.

#### nuuskur

Certain infinite combinations are allowed (the ones provided by definition). That's the best you can do and the whole point of a Schauder basis. If you want to make arbitrary combinations, then you are restricted to finite combinations, because of potential well-definedness problems.

The last example I gave is sloppy for I didn't specify where - if we are talking about $\ell_\infty$ (sup-norm), it doesn't have a Schauder basis, because it isn't separable. Indeed, if $E$ is a Schauder basis (wo.l.o.g normalised), then $X := \{\sum _{j=1}^m x_je_j \mid e_j\in E, x_j\in \mathbb Q, m\in\mathbb N\}$ is a countable dense subset (a good exercise for FA1 students).

To be completely fair, the example is rotten to the core. The sequence $(0,1,0,1,\ldots )$ is not in any $\ell _p$ for $1\leq p<\infty$. Replace with appropriate sequence and my argument about a Cauchy sequence applies.

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#### mathman

When all s said and done, does $l_\infty$ have a basis by any definition?

#### nuuskur

Yes, it is a nonzero space, thus it has a basis. It is known that the axiom of choice holds if and only if every spanning subset contains a minimal spanning subset.

It is also known that the axiom of choice is independent of ZF. Thus, if you don't accept AC, then you will have to relinquish the existence of bases for arbitrary nonzero vectorspaces.

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#### mathman

The term basis is still confusing to me. Is the set of vectors of the form (0,0,0,...0,0,1,0,..) where the 1 is at any position a 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 im pretty sure theres 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

Gold Member
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.

"Vector space (no topology) basis"

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