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I have a question about one example in Linear Algebra.

Let $\mathbb R^∞$ be the vector space of infinite sequences $(\alpha_1, \alpha_2, \alpha_3, \ldots )$ of real numbers.

Scalar multiplication are defined in the natural way: the sum of $(\alpha_1 , \alpha_2 , \alpha_3 , \ldots )$ and $(\beta_1,\beta_2,\beta_3,\ldots)$ is $(\alpha_1 +\beta_1, \alpha_2 + \beta_2, \alpha_3 + \beta_3,\ldots)$ the product of $(\alpha_1,\alpha_2,\alpha_3,\ldots)$ by a scalar $\lambda$ is the sequence $(\lambda \alpha_1, \lambda\alpha_2, \lambda\alpha_3, \ldots )$.

There exists infinite linear independent set of vectors $(e_1, e_2, e_3, \ldots)$ \begin{align} e_1 &= (1, 0, 0, \ldots)\\ e_2 &= (0, 1, 0, \ldots)\\ & \,\,\,\vdots \end{align}

The problem is that this set (lets call it $X$) is not a basis of this vector space. Because for example $v = (1, 1, 1, \ldots)$ cannot be written as a linear combination of set $X$ (Linear combination must be a finite sum).

My task is to add "some vectors" to the set $X$ to create a basis of that vector space. If I add $v$ its not basis ($\langle X, v\rangle \ne\mathbb R^∞$)

Is there any proof that the process of adding vectors to set $X$ is not finite? Or is it possible to create a basis with adding vectors to $X$?

Thanks for answers

Michael Hardy
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CptSwaggo
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  • I once had the same question and my teacher told me you could construct a basis but I guess this basis should be uncountable. For example, surely the set of all sequences generates $\mathbb{R}^{\infty}$. From there on, in principle you can throw away all vectors that are linear combinations of other vectors in this set, e.g. you don't need $(1,1,1,\ldots)$ and $(2,2,2,\ldots)$ – Václav Mordvinov Mar 22 '18 at 11:21
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    Example 2 of this pdf consider your example. The final comment is that `` it can be proved that no countable set of vectors in $\mathbb R^\infty$ spans $\mathbb R^\infty$, so even if $\mathbb R^\infty$ has a basis, we will not be able to construct it by simply adding in elements one at a time to this set.'' Moreover, such as @ziggurism notice in his answer, if you want $\mathbb R^\infty$ to have a basis you will need to consider the Zorn's lemma, which implies the Axiom of Choice. – Dog_69 Mar 22 '18 at 11:26
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    You cannot explicitly write down such a basis. Indeed, you could prove that it exists by extending this idea of adding vectors to it, but it requires Zorn's lemma to actually finish this procedure. – Mathematician 42 Mar 22 '18 at 11:26
  • @Dog_69 I have read that document and i know that the basis has uncountable number of vectors (basic knowledge of linear algebra). What I need is exact proof that its not possible with adding some countable number of vectors to set X. – CptSwaggo Mar 22 '18 at 12:36
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    Zorn's lemma is what you would use to show the existence of a basis. If all you need to do is prove that any basis is uncountable, you can try modifying Cantor's diagonal argument. – Jonny Lomond Mar 22 '18 at 13:43
  • Related: https://math.stackexchange.com/questions/122571/is-there-a-constructive-way-to-exhibit-a-basis-for-mathbbr-mathbbn?noredirect=1&lq=1 – Watson Mar 22 '18 at 17:01
  • This appears not to be a duplicate because the other question explicitly asked about constructiveness and this one does not. – Michael Hardy Oct 18 '19 at 16:52

2 Answers2

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You have noticed the difference between $\prod^\infty R$ (the vector space of all sequences of elements in $R$) and $\bigoplus^\infty R$ (the vector space of sequences which are only finitely often nonzero). The latter has a basis of cardinality $\aleph_0$ (the basis you write down in your question), the former has a basis of cardinality $2^{\aleph_0}$, and some choice principle must be invoked to prove the existence of this basis (and there even exist models of set theory where this space provably does not have a basis). In particular, yes, you cannot get a basis for $\prod^\infty R$ by adding finitely many basis elements to the standard basis for $\bigoplus^\infty R$, since it has strictly larger cardinality.

How can we describe a Hamel basis for $\prod^\infty R$? We can't, not with standard set theoretic constructions. Since it is consistent with ZF set theory that no such basis exist, no ZF constructions (like natural numbers, tuples, powersets, set builder) will allow you to write down this basis.

How can we nonconstructively prove that such a basis exists then, using the axiom of choice or Zorn's lemma? Well linearly independent sets are ordered by inclusion. Each chain of such sets has a maximum (take the union). Therefore by Zorn's lemma there is a maximal linearly independent set, also known as a basis. Or see this answer by Michael Hardy.

ziggurism
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Take all non-zero sequences, and impose a well-order on them (which is possible assuming the Axiom of Choice). If you want to have the partial basis you identified as part of the final basis, arrange that the well-ordering starts with that partial basis.

Now take the set of those vectors that cannot be written as a linear combination of vectors preceding it in that well-order. Those vectors will form a basis of the vector space.

Linear independence of that set is quite obvious from the construction, and that it spans the complete vector space is seen from the fact that any non-zero vector that is not a linear combination of the vectors in that set would in particular not be a linear combination of those that precede it in the total order; but then by construction it should be in that set.

celtschk
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  • Im not sure if I understand it right. The problem is I dont know how to describe the set of vectors that cannot be written as a linear combination. My basis would look like (X,Y), where X is my partial basis I described and how to describe Y? – CptSwaggo Mar 22 '18 at 13:41
  • @CptSwaggo: What I gave is a description of the basis, in terms of the imposed well-order. You cannot explicitly specify a well-order, as if you could, you'd be able to use that specification as proof that the well-order exists (you would just have specified it!); however without the Axiom of Choice, that may not be true. – celtschk Mar 22 '18 at 13:54