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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?

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    $\begingroup$ "Vector space" should read "Banach space", shouldn't it? $\endgroup$
    – Rob Arthan
    Feb 13, 2016 at 21:49
  • $\begingroup$ It's "vector space" in Kreyszig's functional analysis text. $\endgroup$
    – Fequish
    Feb 13, 2016 at 22:13
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    $\begingroup$ @Batominovski: Please stop flooding the front page with Hamel basis questions. $\endgroup$
    – user856
    Aug 12, 2016 at 21:54
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    $\begingroup$ @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 . . . $\endgroup$ Aug 12, 2016 at 22:13
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    $\begingroup$ @Batominovski: Relevant meta discussions: Retagging old questions that get bumped. How much bumping is too much? Tag editing etiquette on old questions $\endgroup$
    – user856
    Aug 12, 2016 at 22:17

4 Answers 4

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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.

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  • $\begingroup$ In more detail, a Schauder basis of an infinite dimensional Banach space will not be countable. $\endgroup$
    – PhoemueX
    Feb 13, 2016 at 21:26
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    $\begingroup$ @PhoemueX I don't really see what that has to do with my answer? $\endgroup$ Feb 13, 2016 at 21:27
  • $\begingroup$ 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. $\endgroup$
    – PhoemueX
    Feb 13, 2016 at 21:34
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    $\begingroup$ @RobArthan I wasn't saying it wasn't relevant to the question, just that it wasn't relevant to my answer. $\endgroup$ Feb 13, 2016 at 22:17
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    $\begingroup$ 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. $\endgroup$
    – Rob Arthan
    Feb 13, 2016 at 22:28
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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"

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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.

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Vectors in a Hamel basis is only "linear independent", but vectors in a Schauder basis is by definition, required to be "countably linear independent". It may happen that a set $\{x_i\}_{i=1}^\infty$ is linear independent, but there's a sequence of nubmers $(\alpha_i)$ such that $\sum_{i=1}^\infty \alpha_i x_i=0$.

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