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$F^{I}$ is the set of functions with range $F$ and domain $I$. I already know that this is a vectorspace but I don't know how a basis of this vectorspace would look like if $I$ is infinite. One can also say that $F^{I}$ is the vectorspace of sequences.

If I don't restrain the elements of my set to finitely nonzero entries then how can I say that a basis exists?

A basis must be a generating system that means it bust be a combination of finitely many basisvectors.

Is there a set of vectors such that to each vector I propose I can provide a finite linearcombination and the set is not $F^{I}$ itself ? Otherwise $F^{I}$ must be a basis because it is a minimal generating system.

Suppose $Z$ is such a set then ... this vector cannot be described as a linearcombination of the vectors in $Z$ which is a contradiction. That would be the plan of my proof, can someone help me to fill the gap or give a hint how I can proceed?

Edit: F^{I} cannot be a basis because it is not maximally linear independent if $F$ is not $\{0,1\}$, so if we suppose $F\neq \{0,1\}$ the proof wouldn't work

md2perpe
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New2Math
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  • What you write seems to indicate the notion of basis isn't very clear in your head (e.g. "A basis must be a generating system that means it bust be a combination of finitely many basisvectors"). Perhaps you should clarify that first. Next, bases for $F^I$ are very complicated objects and it's not clear that we can even describe them : their existence is proved through a very non constructive axiom, the axiom of choice. – Maxime Ramzi Jul 13 '19 at 17:15
  • Tell at least what are $;F,,I\ldots;$ – DonAntonio Jul 13 '19 at 17:15
  • It sounds to me like $F$ is a scalar field, $I$ is some arbitrary set, and the bases you're referring to are Hamel bases, i.e. a set of vectors so that every vector in the space can be written uniquely as a finite linear combination of basis vectors. Am I correct in thinking this? – Theo Bendit Jul 13 '19 at 17:18
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    @TheoBendit yes that's what I meant. And I am looking at the case now where I is infinite – New2Math Jul 13 '19 at 17:20
  • @Max I don't know the proof that the axiom of choice is equivalent to zorn's lemma but I have read the proof that the zorn lemma implies that every vecorspace has a basis and with basis I mean Hamelbases which TheoBendit has described in his comment. I am trying to construct a set of vectors which is a basis of $F^{I}$ – New2Math Jul 13 '19 at 17:23
  • Possibly relevant is the 23 July 2000 sci.math post (corrections and additions here) in which the existence of continuum many almost disjoint subsets of a given countably infinite set is employed. – Dave L. Renfro Jul 13 '19 at 17:24

1 Answers1

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There is no algorithm for defining a basis.

Even if $F=\Bbb R$ and $I=\Bbb N$, it is consistent that the axiom of choice fails, and $F^I=\Bbb{R^N}$ does not have a basis at all. In particular, there is no explicit definition (using $F$ and $I$, or some other fixed sets perhaps) of a basis.

The axiom of choice lets us prove that there is a basis, but it is not telling us what this basis might be. And the above tells us that this is indeed the best we can hope for.

Asaf Karagila
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  • +1 I was just about to point the asker to your old answer here. – Theo Bendit Jul 13 '19 at 17:25
  • I don't understand first you say if we pick $\mathbb{R}=F $ and $I=\mathbb{N}$ then one can prove that $\mathbb{R}^{N}$ has no basis. But in the next sentence you say a basis must exist because of the axiom of choice. What is true now? – New2Math Jul 13 '19 at 17:32
  • No. It is consistent, it is possible that the axiom of choice fails and there is no basis of $\Bbb{R^N}$. – Asaf Karagila Jul 13 '19 at 17:33
  • Okey without assuming AC it is not possible I understood but how do you know that there exists no algorithm? – New2Math Jul 13 '19 at 17:37
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    @New2Math: Because an algorithm, or rather an explicit definition, would work even in the absence of choice. – Asaf Karagila Jul 13 '19 at 17:38
  • I don't understand how you can be so sure that if you don't assume AC there cannot exist another proof why not? Is it somehow proved that AC is necessary for the argument for $F^{I}$ in particular? Can you not just take some disjoint unions? For ex. you first take $1000...,01000...,0010...$ then you take $0010,1001,0100...$ and then $1000,010001....,0010001...$ i.e you just take the disjoint union of the sets where the only difference is the pattern of $0$ because it is disjoint it linear independent and one can also probably show generating sys. but you say it is not possible, why so sure? – New2Math Jul 13 '19 at 17:53
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    There cannot be a proof, because under some assumptions (which are consistent with the failure of AC) we can prove there is no basis. – Asaf Karagila Jul 13 '19 at 17:54
  • Last question why does giving an explicit definition means that AC is unnecessary? – New2Math Jul 13 '19 at 18:02
  • @New2Math: Because the AC is not involved in the definition. Since the axiom of choice is not constructive, it is not usually involved in what we consider as "explicit definitions", and the proofs that these are indeed valid definitions. – Asaf Karagila Jul 13 '19 at 18:06
  • usualy? but what if? – New2Math Jul 13 '19 at 18:34
  • The axiom of choice is not constructive what does that mean? An explicit definition of a basis does mean that there is a rule to describe the elements of the basis, the existence of ths rule means it is constructive right? You also said AC is not usualy involved in "a rule to describe the elements of a set" but what if? – New2Math Jul 13 '19 at 18:40
  • @New2Math: When we say that AC is not constructive we meant that it tells us that there exists a set with certain properties, but it does not identify an explicit set with said properties (as opposed to most axioms in mathematics that tell you "this object exists" rather than "a set exists"). Yes, there are axioms in set theory that do provide you with an explicit basis, always, and they are in some sense, very strong versions of the axiom of choice. But since AC does not guarantee a unique object to exist (indeed most of the time if there is one choice function, there are many) [...] – Asaf Karagila Jul 13 '19 at 19:06
  • [...] it is not something that we use in definitions. Since definitions define explicit objects. And if you have to resort to AC, that means that we had to rely on some object which we cannot control, and if there is one of that object, then there are many of that object, and each one might give a different end-result. In that sense, at least, this means that this is not an explicit definition. – Asaf Karagila Jul 13 '19 at 19:07
  • @New2Math: The point is that if I tell you that I have an explicit definition of a real number, and I just tell you "There is some $x$ such that $x^2$ is that real number", you will protest that I haven't given you a very explicit definition, unless $x$ was predetermined by some fixed "oracle". Similarly, if you fix a choice function, or a well-ordering, of some large enough set, you can indeed give an explicit definition of a basis. But (1) there might not be one; and (2) even if there is one, which one is the one you're using? – Asaf Karagila Jul 13 '19 at 22:58