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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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  • $\begingroup$ Would the use of "find" instead of "show" be more appropriate? $\endgroup$
    – Ilya
    Mar 20, 2012 at 17:08
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    $\begingroup$ I 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? $\endgroup$
    – MJD
    Mar 20, 2012 at 17:10
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    $\begingroup$ @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. $\endgroup$ Mar 20, 2012 at 17:12
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    $\begingroup$ @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. $\endgroup$ Mar 20, 2012 at 17:20
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    $\begingroup$ I'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. $\endgroup$ Mar 20, 2012 at 17:34

3 Answers 3

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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:

  1. Kechris, A. Classical Descriptive Set Theory. Springer-Verlag, 1994.
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    $\begingroup$ Unfortunately, $\mathbb R^{\mathbb N}$ 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 $\mathbb R^{\mathbb N}$. $\endgroup$
    – GEdgar
    Mar 21, 2012 at 0:06
  • $\begingroup$ @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? $\endgroup$
    – Asaf Karagila
    Mar 21, 2012 at 0:09
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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.

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    $\begingroup$ Ha! Excellent point. $\endgroup$ Mar 20, 2012 at 18:33
  • $\begingroup$ f(x)=c if x=c(1,1,1,1..), 0 otherwise. $\endgroup$
    – Nick Alger
    Mar 20, 2012 at 18:53
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    $\begingroup$ @Nick: That's not linear; for instance, add $(2,0,1,1,\dotsc)$ and $(0,2,1,1,\dotsc)$. $\endgroup$
    – joriki
    Mar 20, 2012 at 19:02
  • $\begingroup$ ah ha good point joriki, thanks. $\endgroup$
    – Nick Alger
    Mar 20, 2012 at 19:03
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    $\begingroup$ Just out of curiosity, is it still open that the statement 'each real vector space has a basis' implies the Axiom of Choice? $\endgroup$ Apr 5, 2012 at 18:10
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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 $\mathbb{R}^\mathbb{N}$.

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    $\begingroup$ It 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. $\endgroup$
    – Asaf Karagila
    Mar 20, 2012 at 23:10
  • $\begingroup$ Ah, excellent, thanks. Editing that in. $\endgroup$
    – Desiato
    Mar 21, 2012 at 0:50
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    $\begingroup$ One 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). $\endgroup$
    – Asaf Karagila
    Mar 21, 2012 at 1:26
  • $\begingroup$ 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. $\endgroup$
    – Desiato
    Mar 21, 2012 at 1:42
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    $\begingroup$ Well, 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. $\endgroup$
    – Asaf Karagila
    Mar 21, 2012 at 8:36

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