Assuming the Axiom of Choice, every vector space has a basis, though it can be troublesome to show one explicitly. Is there any constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$, the vector space of real sequences?
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Would the use of "find" instead of "show" be more appropriate? – SBF Mar 20 '12 at 17:08
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1I hope this isn't a super-dumb question, but wouldn't $b_i = { \delta^i_n }$, where $\delta^i_n$ is the delta function, be such a basis? Or are you looking for a basis where every element is a combination of a finite number of basis elements? – MJD Mar 20 '12 at 17:10
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13@Mark: No, because the span of those vectors consists only of vectors that have at most finitely many nonzero entries. The vector $(1,1,1,\ldots)$ is not in their span. – Arturo Magidin Mar 20 '12 at 17:12
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@Ilya: Maybe, English isn't my first language so I'll let the rest of you decide. – Fernando Martin Mar 20 '12 at 17:14
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@Arturo: I thought that restriction was what distinguishes a Hamel basis from a generic basis; am I misunderstanding? – MJD Mar 20 '12 at 17:14
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7@Mark: Hamel basis means "basis of a vector space" in the usual sense: linearly independent, and spanning (and the span is the collection of all linear combinations of elements of the set, and by definition linear combinations have only finitely many nonzero coefficients). Hamel bases are distinguished from Hilbert Bases (not from "generic basis"; I'm not sure what you mean by a "generic basis"). A Hilbert Basis (of a complete inner product space) is a maximal orthonormal subset; its span is dense in the space, but not necessarily equal to it. – Arturo Magidin Mar 20 '12 at 17:20
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@arturo: Thanks. – MJD Mar 20 '12 at 17:27
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@ilya: I think "exhibit" is often used in this context. – MJD Mar 20 '12 at 17:29
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1I'd be extremely surprised if the answer to this question is “yes”, but I don't know how to prove a “no” answer, other than in the most general sense. – Harald Hanche-Olsen Mar 20 '12 at 17:34
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Relevant: http://en.wikipedia.org/wiki/Basis_(linear_algebra)#Related_notions – Mar 20 '12 at 17:44
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@ArturoMagidin: +1. I like your words about "Hamel basis" and "Hilbert basis". – Mar 20 '12 at 17:45
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@Harald: What's the most general sense? – joriki Mar 20 '12 at 18:01
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@joriki: The idea is to exhibit a set of consistent axioms of set theory, including ZF but excluding the axiom of choice, in which $\mathbb{R}^{\mathbb{N}}$ has no basis. – Harald Hanche-Olsen Mar 20 '12 at 18:28
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@Harald: I see -- "in the most general sense" referred to "know" -- I thought it referred to "prove" :-) – joriki Mar 20 '12 at 18:29
3 Answers
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Assume that every set of real numbers has the Baire property, as well the axiom of dependent choice (it is true if we assume AD, or live in Solovay's model but we can get away with less than large cardinals too):
Since every set of $\mathbb R^\mathbb N$ has the Baire property, and $\mathbb R^\mathbb N$ is a Polish group, every homomorphism of it into itself is continuous [1, Th. 9.10].
Given a Hamel basis has to have cardinality $\frak c$, it defines $2^\frak c$ many endomorphisms of $\mathbb R^\mathbb N$.
Now, given that $\mathbb R^\mathbb N$ is a separable space (by rational sequences which are eventually zero) this means that a continuous function is defined uniquely on the countable dense set, in particular this implies that we can only have $\frak c$ many continuous functions from $\mathbb R^\mathbb N$ to itself.
Contradiction.
Bibliography:
- Kechris, A. Classical Descriptive Set Theory. Springer-Verlag, 1994.
Asaf Karagila
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1Unfortunately, $\mathbb R^{\mathbb N}$ is not a Banach space. Solovay has another model where every set (in a Polish space) has the property of Baire. Then your method works for $\mathbb R^{\mathbb N}$. – GEdgar Mar 21 '12 at 00:06
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@GEdgar: Thanks for the comment. I think that in Solovay's model every set has BP to begin with. I always recalled it extends to all Polish spaces, no? – Asaf Karagila Mar 21 '12 at 00:09
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Can you add some explanation of how this relates to the question? It seems you've argued that (dependent choice) + (every set of reals has the Baire property) is inconsistent. I see references to $\mathbb R^{\mathbb N}$ and a Hamel basis thereof, but I'm unclear on what's being shown. That no such basis exists? Ok, but then why assume every set has the Baire property? – WillG Feb 28 '24 at 19:23
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@Will: Where do you see references to such Hamel basis? Is the context that if ZFC? Because ZFC proves that "every set of reals has the Baire property" is in fact false. – Asaf Karagila Feb 28 '24 at 23:19
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I just meant: I see that you have referred to a Hamel basis of $\mathbb R^{\mathbb N}$ in your answer (at least, that's how I'm interpreting "Given a Hamel basis..."). But I don't understand how your answer relates to the question, namely, whether one can constructively exhibit a basis for $\mathbb R^{\mathbb N}$. Does your answer show that this is impossible? – WillG Feb 29 '24 at 04:45
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@WillG: As is very standard in mathematics "If there is such and such, then there is such and such, therefore such and such is false". So, if the last such and such happened to be true, the first such and such must be false. In particular, since the existence of a Hamel basis implies the existence of discontinuous functionals and endomorphisms, something which implies that all functionals are continuous will imply there is no Hamel basis. Something like "all sets of reals have the Baire Property", for example. – Asaf Karagila Feb 29 '24 at 10:28
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Sorry, I'm still confused. The questions asks whether one can exhibit a basis of $\mathbb R^{\mathbb N}$. I think you're saying "If $P$, then no Hamel basis of $\mathbb R^{\mathbb N}$ exists," where $P$ is the claim "all sets of reals have the Baire property." Why not assume $\neg P$ then? Why are we assuming $P$ to begin with? – WillG Feb 29 '24 at 16:09
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We are not interested in the existence per se, we are interested in an explicit construction of a basis. In the standard understanding, this means that this can be done without appealing to the axiom of choice, or without appealing to the entirety of the set theoretic universe, but perhaps to more internal structure of the vector space. Since $P$ is consistent with $\sf ZF+DC$, the answer is "no, we cannot explicitly exhibit a Hamel basis". – Asaf Karagila Feb 29 '24 at 17:43
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Ah, ok, so is the following correct? We already know (for other reasons) that $P$ is consistent with ZF + DC, but in ZF + DC + ($\mathbb R^{\mathbb N}$ has an explicit basis), $P$ is false. Hence $\mathbb R^{\mathbb N}$ doesn't have an explicit basis. – WillG Feb 29 '24 at 17:58
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"Constructively" "exhibiting" a basis for $\mathbb R^{\mathbb N}$ means "constructively" "exhibiting" a lot of linear functionals on $\mathbb R^{\mathbb N}$; one coordinate functional for each element of the basis. So, in particular, it would mean "constructively" "exhibiting" a linear functional on $\mathbb R^{\mathbb N}$ that is linearly independent of the point-evaluations. Can you "constructively" "exhibit" even one such functional? I think not.
Martin Sleziak
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GEdgar
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2@Nick: That's not linear; for instance, add $(2,0,1,1,\dotsc)$ and $(0,2,1,1,\dotsc)$. – joriki Mar 20 '12 at 19:02
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1Just out of curiosity, is it still open that the statement 'each real vector space has a basis' implies the Axiom of Choice? – Jan Veselý Apr 05 '12 at 18:10
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1What does "linearly independent of the point evaluations" mean? – The_Sympathizer Jan 23 '17 at 01:22
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2@mike4ty4... A "point-evaluation" at the point $k$ is the linear functional that maps $f \in \mathbb R^{\mathbb N}$ to the scalar $f(k)$. There are thus countably many "point-evaluations". They are linear functionals on $\mathbb R^{\mathbb N}$. But (given a basis for $\mathbb R^{\mathbb N}$) there are linear functionals not in the linear span of the set of point-evaluations. These would be linear functionals linearly independent of the point-evaluations. – GEdgar Jan 23 '17 at 13:09
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@GEdgar: So the point evaluation is a functional returning the $k$th element of the sequence? – The_Sympathizer Jan 24 '17 at 07:02
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1@WillG ... the domain of $f \mapsto \sum 2^{-k} f(k)$ is not all of $\mathbb R^{\mathbb N}$. – GEdgar Feb 28 '24 at 17:49
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This seems extremely unlikely, because if you could do this "constructively" enough to (obviously not countably) enumerate each member of the basis, you could use that enumeration to construct a well-ordering of $\mathbb{R}$ from this because we could then identify each $x \in \mathbb{R}$ with its unique final rational representation in terms of this basis.
While being able to well-order $\mathbb{R}$ is obviously a consequence of AoC, it is widely considered impossible to "constructively" give such a well-ordering.
All this is a consequence of the finiteness of our language and thus the inherent countability of everything we can describe precisely enough to "separate" each element of a set.
So the best you can get really is describing the basis as some structured collection of uncountable sets.
Edit to add that the second argument still holds if you're looking for an $\mathbb{R}$-basis, because this basis would need to be uncountable as well.
An easy way to see this is considering $\{(n^x)_{n \in \mathbb{N}}:x \in \mathbb{R}^+\}$ - this is an uncountable and $\mathbb{R}$-linearly independent set in $\mathbb{R}^\mathbb{N}$.
Martin Sleziak
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Desiato
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1It is easy to see that a basis cannot be countable since $\mathbb R^\mathbb N$ is the dual of all eventually-zero sequences (or if you will $\mathbb R[x]$), which has a countable basis. In particular this means that if there was a basis it could not be countable as it would have to contain continuum many elements. – Asaf Karagila Mar 20 '12 at 23:10
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2One minor (read: major) nitpick: Uncountable need not imply of size continuum. It is perfectly reasonable that there are uncountable sets whose cardinality is unspeakable and can be embedded into $\mathbb R$ (from the Dedekind-finite set introduced by Cohen to the set which disproves the continuum hypothesis in the Feferman-Levy model). – Asaf Karagila Mar 21 '12 at 01:26
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I'm aware of this, but I don't really see where I implied this. The argument that whatever we can define is countable means that we cannot describe any uncountable set in a way that allows us to distinguish all its elements - not just the continuum, but also, should we chose to accept them, sets with cardinality in between, as long as they're not countable. – Desiato Mar 21 '12 at 01:42
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1Well, my point is that assuming that the continuum is $\aleph_2$ (for example) saying just that a basis would have to be uncountable might imply that it can be of size $\aleph_1$. Without choice it can get worse. Assume that the continuum is singular (in the sense that it is a union of countably many countable sets; or $\aleph_1$ many sets of size $\aleph_1$ but still greater than $\aleph_1$ in size...) these models have even stranger sets which are uncountable. However it will not help, a basis for $\mathbb R^\mathbb N$ has to have size continuum, not just "uncountable" size. – Asaf Karagila Mar 21 '12 at 08:36