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Let $V$ be a vector space with infinite dimensions. A Hamel basis for $V$ is an ordered set of linearly independent vectors $\{ v_i \ | \ i \in I\}$ such that any $v \in V$ can be expressed as a finite linear combination of the $v_i$'s; so $\{ v_i \ | \ i \in I\}$ spans $V$ algebraically: this is the obvious extension of the finite-dimensional notion. Moreover, by Zorn Lemma, such a basis always exists.

If we endow $V$ with a topology, then we say that an ordered set of linearly independent vectors $\{ v_i \ | \ i \in I\}$ is a Schauder basis if its span is dense in $V$ with respect to the chosen topology. This amounts to say that any $v \in V$ can be expressed as an infinite linear combination of the $v_i$'s, i.e. as a series.

As far as I understand, if a $v$ can be expressed as finite linear combination of some set $\{ v_i \ | \ i \in I\}$, then it lies in its span; in other words, if $\{ v_i \ | \ i \in I\}$ is a Hamel basis, then it spans the whole $V$, and so it is a Schauder basis with respect to any topology on $V$.

However Per Enflo has constructed a Banach space without Schauder basis (ref. wiki). So I guess I should conclude that my reasoning is wrong, but I can't see what's the problem.

Any help appreciated, thanks in advance!

UPDATE: (coming from the huge amount of answers and comments) Forgetting for a moment the concerns about cardinality and sticking to span-properties, it has turned out that we have two different notions of linear independence: one involving finite linear combinations (Hamel-span, Hamel-independence, in the terminology introduced by rschwieb below), and one allowing infinite linear combinations (Schauder-stuff). So the point is that the vectors in a Hamel basis are Hamel independent (by def) but need not be Schauder-independent in general. As far as I understand, this is the fundamental reason why a Hamel basis is not automatically a Schauder basis.

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    $\begingroup$ @DanielFischer Sorry, I meant something besides the indexing by $\Bbb N$ axiom (which is begging the question about how big the basis is). We're bound to countable sums, I believe that. Hamel bases are bound to finite sums, but they are not restricted to be finite themselves. Is there something wrong with taking an uncountable set and talking about their countable linear combinations? $\endgroup$
    – rschwieb
    Jan 7, 2014 at 21:14
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    $\begingroup$ @rschwieb If the family is summable, no problem. But for a Schauder basis, summability is not required. Note that for a Schauder basis you have not only countability, but a specific ordering. If $\{ e_j : j \in \mathbb{N}\}$ is a Schauder basis that is not unconditional, there are permutations $\pi$ of $\mathbb{N}$ such that $\{ e_{\pi(j)} : j \in \mathbb{N}\}$ is not a Schauder basis. $\endgroup$
    – Daniel Fischer
    Jan 7, 2014 at 21:19
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    $\begingroup$ @DanielFischer Sorry, I guess I'm still overlooking something :( The last comment still looks like you're saying "A Schauder basis does not satisfy this Hilbert axiom." I agree with that, but I'm asking the other direction. (The last sentence of that comment is super interesting though, thanks for telling me about it. I did not expect order to come into play.) $\endgroup$
    – rschwieb
    Jan 7, 2014 at 21:28
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    $\begingroup$ @rschwieb A Schauder basis can lack a nice property a Hilbert basis automatically has. Therefore, we must impose a restriction on Schauder bases that we need not impose on Hilbert bases. Let us call a Friendly basis of a(n infinite dimensional Hausdorff locally convex) topological vector space $E$ a sequence $\{ f_k : k\in \mathbb{N}\}$ such that for every $x\in E$ there is a unique sequence $\{\varphi_k(x)\}$ of coefficients such that the family $\{\varphi_k(x)\cdot f_k : k\in\mathbb{N}\}$ is summable with sum $x$. A Friendly basis is a Schauder basis, and if $E$ is a Hilbert space and the $\endgroup$
    – Daniel Fischer
    Jan 7, 2014 at 21:45
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    $\begingroup$ family is orthonormal, also a Hilbert basis. Now, the concept of "Friendly bases" can be generalised in different directions. We can drop the countability, and we can drop the summability requirement. Dropping both leads to problems. A Schauder basis drops the summability requirement and replaces it with the convergence of the sequence of partial sums in the specified order. $\endgroup$
    – Daniel Fischer
    Jan 7, 2014 at 21:47

5 Answers 5

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People keep mentioning the restriction on the size of a Schauder basis, but I think it's more important to emphasize that these bases are bases with respect to different spans.

For an ordinary vector space, only finite linear combinations are defined, and you can't hope for anything more. (Let's call these Hamel combinations.) In this context, you can talk about minimal sets whose Hamel combinations generate a vector space.

When your vector space has a good enough topology, you can define countable linear combinations (which we'll call Schauder combinations) and talk about sets whose Schauder combinations generate the vector space.

If you take a Schauder basis, you can still use it as a Hamel basis and look at its collection of Hamel combinations, and you should see its Schauder-span will normally be strictly larger than its Hamel-span.

This also raises the question of linear independence: when there are two types of span, you now have two types of linear independence conditions. In principle, Schauder-independence is stronger because it implies Hamel-independence of a set of basis elements.

Finally, let me swing back around to the question of the cardinality of the basis. I don't actually think (/know) that it's absolutely necessary to have infinitely many elements in a Schauder basis. In the case where you allow finite Schauder bases, you don't actually need infinite linear combinations, and the Schauder and Hamel bases coincide. But definitely there is a difference in the infinite dimensional cases. In that sense, using the modifier "Schauder" actually becomes useful, so maybe that is why some people are convinced Schauder bases might be infinite.

And now about the limit on Schauder bases only being countable. Certainly given any space where countable sums converge, you can take a set of whatever cardinality and still consider its Schauder span (just like you could also consider its Hamel span). I know that the case of a separable space is especially useful and popular, and necessitates a countable basis, so that is probably why people tend to think of Schauder bases as countable. But I had thought uncountable Schauder bases were also used for inseparable Hilbert spaces.

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    $\begingroup$ Your answer is probably the best here, but I think it would be even better if you cleared up the whole misconception about the Schauder basis being countable. $\endgroup$
    – tomasz
    Jan 7, 2014 at 20:03
  • $\begingroup$ This is the kind of things I'm looking for. Please correct me if I'm wrong: if I take a Schauder basis then its Schauder-span will be the whole V (by def). Of course its Hamel-span will normally be smaller, that's ok. Let's consider on the other hand a Hamel basis for V. Then again by def its Hamel-span is the whole V. I think we can say that its Schauder-span (with respect to any nice enough topology on V) is again the whole V...this is what I meant in the question with "so it is a Schauder basis with respect to any topology on V". What am I missing? $\endgroup$
    – Lor
    Jan 7, 2014 at 20:09
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    $\begingroup$ @Lor Let's consider on the other hand a Hamel basis for V. Then again by def its Hamel-span is the whole V. I think we can say that its Schauder-span (with respect to any nice enough topology on V) is again the whole V I think the last thing is that you would have to be certain the Schauder linear independence condition was satisfied, but otherwise yes, I agree with that. The Schauder independence condition is, in principle, stronger, although I don't have any informative examples :S $\endgroup$
    – rschwieb
    Jan 7, 2014 at 20:16
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    $\begingroup$ Informative examples have already come up: In $\ell^2,$ compare your favorite not-eventually-zero sequence to the standard basis vectors, say, the harmonic sequence $(1/n)$. This is Hamel-independent of the set $\{e_i\}$ because a Hamel combination of the basis vectors only has finitely many nonzero coordinates. But it's not Schauder independent, because of course $(1/n)=\sum 1/n e_n$. $\endgroup$
    – Kevin Arlin
    Jan 7, 2014 at 21:08
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    $\begingroup$ @rschwieb: I think I'm rather confused myself. I've only had a brief experience with Schauder bases, and I've been told to think that they're like Hilbert bases only for non-Hilbert spaces. Perhaps I was mistaken, so I'll leave it to others for now. $\endgroup$
    – tomasz
    Jan 7, 2014 at 22:23
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The problem is that an element of a Hamel basis might be an infinite linear combination of the other basis elements. Essentially, linear dependence changes definition.

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    $\begingroup$ Exactly; this gives two different representations of the element, so representations aren't unique. $\endgroup$
    – Brian Rushton
    Jan 7, 2014 at 13:33
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    $\begingroup$ Dear: @BrianRushton : I don't think I get the comment about nonunique representations. Doesn't the analogous definition of "linearly independent" guarantee representations are unique? (I'm not sure what detnvvp is getting at in the first comment, either.) $\endgroup$
    – rschwieb
    Jan 7, 2014 at 19:04
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    $\begingroup$ @rschwieb: the point is that a Hamel basis can't always serve as a Schauder basis. Certainly every element of a vector space can be written as an infinite linear combination of elements of a Hamel basis, since it can be written even as a finite such combination. But there's no reason for that infinite representation to be unique. $\endgroup$
    – Kevin Arlin
    Jan 7, 2014 at 21:01
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    $\begingroup$ I take it you $b_i$ are elements of a Hamel basis? Then the point is simply that your independence axiom does not have to hold for infinite sums just because it holds for finite ones. $\endgroup$
    – Kevin Arlin
    Jan 7, 2014 at 21:10
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    $\begingroup$ @KevinCarlson In the meantime, I've just gone and looked up the wiki. It contains the phrase A Schauder basis is a sequence $\{b_n\}$ of elements of V such that for every element v ∈ V there exists a unique sequence $\{α_n\}$ of scalars in F so that... $\endgroup$
    – rschwieb
    Jan 7, 2014 at 21:18
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Maybe a good point to start is this useful corollary of Baire Cathegory Theorem

the cardinality of an Hamel base of a Banach Space can be finite or uncountable. It can't be countable

The proof is a delightful application of Baire theorem.

Now to give an explicit example, we can consider the space $\ell^2 $ which has the standard base $ M:=$ $\lbrace e_n \rbrace $ which is not an Hamel base, but an Hilbert base (or Schauder, in this case the two coincide). To see the differences consider the linear span of $ M $. It's trivial to see that it is $ c_{00} $ but (using orthonormality property of $ M $) each vector $ v \in \ell^2 $ can be expressed as $ v=\sum_{k=1}^{\infty}(v, e_k) e_k $

In fact the restriction to FINITE linear combinations is a strong restriction. Let me show you another similar example. Consider $ c_0 $ the Banach space of the sequences convergent to 0. $ M$ is a Schauder base of it (verify it) but given for example $ u= \lbrace \frac {1}{n}_n \rbrace $ you can't express u as a finite linear combination of elements of M . So changing the meaning of the base in fact change "how big is its span"

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In the case of an infinite dimensional complete space, if you have a Banach space, then any Hamel basis is not countable. On the other hand, any Schauder basis has to be countable.

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There are two extra requirements on a Schauder basis:

(a) it must be countable;

(b) it needs not only to be linearly independent, but to satisfy the infinitary analogue of this property: given any infinite linear dependency $\sum_{i \in \mathcal{I}} a_i \mathbf{v}_i = 0$, we must have $a_i = 0$ for each $i$.

Both of these will fail for any Hamel basis of an infinite-dimensional Banach space.

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  • $\begingroup$ There is a reason for a Schauder basis to be countable, @tomasz. Let $\xi_j$ be the coordinate functionals for the Schauder basis $\{e_j\}$. The family $\{ \xi_j(x)\cdot e_j \}$ is not required to be summable. $\endgroup$
    – Daniel Fischer
    Jan 7, 2014 at 20:34
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    $\begingroup$ @noname1014: consider the Banach space $\ell^\infty$ — i.e. the space of all absolutely convergent infinite sequences of reals. This has a Schauder basis given by the sequences $e_0 = (1,0,0,0…), e_1 = (0,1,0,0,…), e_2 = (0,0,1,0,…)$ and so on. Then this basis, together with the extra sequence (1,1/2,1/4,1/8,…), is linearly independent in the sense of finite linear combinations, but is not linearly independent in the Schauder sense, where infinite linear combinations are allowed, since (1,1/2,1/4,…) is expressible as $\sum_i 2^{-i} e_i$. $\endgroup$
    – Peter LeFanu Lumsdaine
    Aug 25, 2016 at 16:56
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    $\begingroup$ @Peter in the above comments you mentioned that $(e_{n}) $ is basis (schauder and Hamel) for $l^ \infty $. Is that true really? As much as I know the space $l^ \infty $ is not having any schauder basis. I am confused please answer $\endgroup$
    – Prince Khan
    Dec 12, 2016 at 15:42
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    $\begingroup$ @PrinceKhan: yes, you’re right, I meant $\ell^1$ not $\ell^\infty$. In $\ell^1$, the family $(e_n)$ form a Schauder basis (but not a Hamel basis). Whereas $\ell^\infty$ is not-separable and so has no Schauder basis, as you say. Thanks for catching that! $\endgroup$
    – Peter LeFanu Lumsdaine
    Dec 12, 2016 at 15:49
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    $\begingroup$ Oh thats fine as $l^{p} $ space has schauder basis and thatis what you have presented above. Sir! As every vector space has Hamel basis so the space $l^ \infty $ too. Can we have a presentation of elements of Hamel basis for $l^ \infty $ ? $\endgroup$
    – Prince Khan
    Dec 12, 2016 at 16:15

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