Why isn't every Hamel basis a Schauder basis?

Question

I seem to have tripped on the common Hamel/Schauder confusion.

If $X$ is any vector space (not necessarily finite dimension) and $B$ is a linearly independent subset that spans $X$, then $B$ is a Hamel basis for $X$.

If there exists a sequence $(e_n)$ such that for every $x \in X$ there exists a unique sequence of scalars $(\alpha_n)$ for which $\lim_{n \to \infty} || x - (\alpha_1e_1 + \cdots + \alpha_ne_n)|| = 0$, then $(e_n)$ is a Schauder basis for $X$.

So I'm tempted to think that every Hamel basis is also a Schauder basis; just extened the finite linear combination into an infinite one by adding zero coeeficients. I know I'm wrong, but what am I missing?

@Batominovski: Please stop flooding the front page with Hamel basis questions. — , Aug 12 '16 at 21:54
@Rahul Sorry, if my effort at helping organize topics into proper tags offends you. I know too much energy is spent to click onto the second page. — Batominovski, Aug 12 '16 at 21:56
@Batominovski But you're not really helping to organize topics into proper tags. The tags you've added - "Hamel basis" and "Schauder basis" - are each very rare (with less than thirty questions tagged with each, a significant number of which are ones you've recently tagged). Not every term needs to be a tag . . . — Noah Schweber, Aug 12 '16 at 22:13
@Batominovski: Relevant meta discussions: Retagging old questions that get bumped. How much bumping is too much? Tag editing etiquette on old questions — , Aug 12 '16 at 22:17
I know that not every term needs to be a tag. I just put an already existing tag on some relevant topics, and I didn't do to all of them, just some 10-ish (and I decided that it was enough before Rahul became offended). I can't imagine that mere 10-ish questions that got bumped would be such an issue. If I put tags on 20-30 old questions, then I would agree that I was being too obsessive. — Batominovski, Aug 12 '16 at 22:23
@Batominovski Re: "already-existing tag," note that every question tagged "Hamel-basis" was by you - as of now (after Asaf's rollback), there are zero questions with that tag. — Noah Schweber, Aug 12 '16 at 23:04
Somebody created it, not me, although I added information to the tag. I didn't notice that there was no topic with this tag. — Batominovski, Aug 12 '16 at 23:06

Noah Schweber · Answer 1 · 2016-02-13T21:26:59.473

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The issue is ". . . unique sequence of scalars . . .". If $B$ is a Hamel basis, then there will be at least one such sequence of scalars, but there may be more.

edited Feb 13 '16 at 21:26

answered Feb 13 '16 at 21:24

Noah Schweber

245,398

In more detail, a Schauder basis of an infinite dimensional Banach space will not be countable. – PhoemueX Feb 13 '16 at 21:26
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@PhoemueX I don't really see what that has to do with my answer? – Noah Schweber Feb 13 '16 at 21:27
Ok, now (after the edit), I see what you mean. I thought that you were refering to the fact that the definition assumes (among other things) that a Hamel basis is countable. – PhoemueX Feb 13 '16 at 21:34
A Hamel basis of Banach space can never be countably infinite, so the OP's $(e_n)$ could never be a Hamel basis. PhoemueX has a valid point. – Rob Arthan Feb 13 '16 at 21:52
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@RobArthan I wasn't saying it wasn't relevant to the question, just that it wasn't relevant to my answer. – Noah Schweber Feb 13 '16 at 22:17
Actually, I think the countable vs. uncountable clears things up a bit for me. A Hamel basis $B$ could be an uncountable set of linearly independent vectors. My way of extending this to a Schauder basis would then break down since Schauder combinations can only use countably many vectors? Is this right? Then is it fair to say that every countable Hamel basis is also a Schauder basis? – Fequish Feb 13 '16 at 22:18
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A normed vector space with a countably infinite Hamel basis cannot be a Banach space. A Hamel basis for a Banach space is either finite or uncountable. – Rob Arthan Feb 13 '16 at 22:28
I'm a bit confused as to why Banach spaces are being brought in. Don't my basis definitions above work for normed spaces in general? – Fequish Feb 13 '16 at 22:39
Your definitions "work", but Schauder bases are normally only considered in the context of Banach spaces. (And I'd guess that a normed space with a Schauder basis must be complete, although I haven't thought that through in detail.) – Rob Arthan Feb 13 '16 at 22:49
@NoahSchweber: the way the OP has phrased the question means that your answer doesn't apply to a normed space with an uncountable Hamel basis. – Rob Arthan Feb 13 '16 at 22:53
So to summarize: a countable Hamel basis is also a Schauder basis; an uncountable Hamel basis might not be a Schauder basis? (or possibly, I should substitute might not be with is not?) – Fequish Feb 13 '16 at 23:09
No. It doesn't make sense for a Schauder basis to be uncountable. – Rob Arthan Feb 13 '16 at 23:36

Jay Zha · Answer 2 · 2022-05-16T21:52:46.780

In terms of one comment from OP question - "just extened the finite linear combination into an infinite one by adding zero coeeficients":

Hamel basis could be defined on any linear space X. While for Schauder basis, we need some good topology defined on space X so that "limit" or "dense" makes sense. Thus in lots of spaces, you could not even define infinite sums.

When Banach space is infinite-dimensional, the cardinality of Hamel basis is actually uncountable. By contrast, Schauder basis is always a sequence, which means that it is, by definition, at most countable. I feel this could also illustrate a bit why sometimes it does not make sense to consider Schauder basis as an extension from Hamel basis - Hamel basis could be much "larger" compared with Schauder basis.

Sequence order matters for Schauder basis that is not unconditional, a permutation on the order of a Schauder basis might make it no longer a Shauder basis; however since Hamel basis only deals with finite linear combination, different ordering does not cause any issue.

Hamel basis always exists, assuming Axiom of Choice. But Schauder basis does not always exist. Since by definition, Schauder basis is at most countable, it requires the space to be separable in order for the closure of its span to be the whole space. But separability itself is not a guarantee for the existence of Schauder basis: Per Enflo showed in 1973 that even for some separable Banach space, there is no Schauder basis on it - yet it definitely has a Hamel basis. This is another example why it's not a good idea to say "extend a Hamel basis to Schauder basis"

score 1 · Answer 3 · answered Mar 25 '21 at 10:59

Interestingly, once once knows that not every Schauder basis is a Hamel basis, this implies almost immediately that not every Hamel basis is a Schauder basis:

A Schauder basis is clearly linearly independent, so it can be extended to a Hamel basis. If it wasn't a Hamel basis to begin with, then this new Hamel basis contains an element not contained in the original Schauder basis. But this element can be written as a countable linear combination of the Hamel basis in two different ways: Once just one times itself, and once as countable linear combination of elements from the original Schauder basis. So the new Hamel basis cannot be a Schauder basis.

Why isn't every Hamel basis a Schauder basis?

3 Answers3