Linked Questions

6
votes
2answers
936 views

Basis of infinite Vector Space $\mathbb R^{\infty}$ [duplicate]

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 ...
2
votes
1answer
116 views

Basis for $\mathbb{R}^{\infty}$ [duplicate]

It follows from Zorn's Lemma that every vector space $V$ has a basis (this means, a subset $B$ of $V$ that generate any $v \in V$ by means a finite linear combination, and such that $B$ is LI) . But, ...
0
votes
0answers
39 views

Can we reproduce a basis of a vector space of all sequences? [duplicate]

Can we reproduce a basis of a vector space of all sequences?(if exists since by every vector space has a basis, but can we reproduce it)
1
vote
0answers
15 views

Is a basis for the vector space of all series in $\mathbb{R}$ constructible [duplicate]

Given the $\mathbb{R}$-vectorspace $V =\mathbb{R}^\mathbb{N}$ of real valued series I was wondering if we can construct a Hamel-basis of $V$. First of all I think that the dimension of $V$ is $ \mid\...
38
votes
3answers
15k views

What is a basis for the vector space of continuous functions?

A natural vector space is the set of continuous functions on $\mathbb{R}$. Is there a nice basis for this vector space? Or is this one of those situations where we're guaranteed a basis by invoking ...
23
votes
2answers
16k views

Finding a basis of an infinite-dimensional vector space?

The other day, my teacher was talking infinite-dimensional vector spaces and complications that arise when trying to find a basis for those. He mentioned that it's been proven that some (or all, do ...
17
votes
1answer
3k views

A Hamel basis for $\ell^p$?

I am looking for an explicit example for a Hamel basis for $\ell^{p}$?. As we know that for a Banach space a Hamel basis has either finite or uncountably infinite cardinality and for such a basis one ...
9
votes
2answers
769 views

What is the basis of the vector space $l^\infty$?

We know that every vector space has a Hamel basis and also every normed space need not have a Schauder basis. As the normed space $l^\infty$ is not Separable so can't have the Schauder basis, but on ...
6
votes
1answer
776 views

Does the existence of a $\mathbb{Q}$-basis for $\mathbb{R}$ imply that choice holds up to $\frak c$?

The axiom of choice is, for ZF, equivalent to the statement that every vector space has a basis. The implication of AoC by the existence of a basis for any vector space is shown in this paper. The ...
1
vote
2answers
336 views

Is this a basis for the dual space?

There is an example on Wikipedia that I don't understand and I'd appreciate some help. They define $\mathbb R^\infty$ to be the space of all sequences that are zero except for finitely many indexes. ...
6
votes
1answer
105 views

Basis for $\mathbb{R}^\mathbb{N}$ implies axiom of choice?

Let $\mathbb{R}^\mathbb{N}$ denote the vector space over $\mathbb{R}$ of sequences of real numbers, with multiplication and addition defined by component. It's well-known that though the subspace $\...
4
votes
0answers
175 views

Basis of $\mathbb{F}[[x]]$ over $\mathbb{F}$ without AC

Does the ring of formal power series $\mathbb{F}[[x]]$ as a vector space over $\mathbb{F}$ admit a basis without assuming the Axiom of choice, at least in some special cases of $\mathbb{F}$? I'm ...
3
votes
0answers
61 views

basis for $\mathbb{R}^{\mathbb{N}}:=\left\{f:\mathbb{N}\to\mathbb{R}\right\}$, and its cardinality.

I know that all vector space has a basis. My question is "concrete" example for basis for $\mathbb{R}$-vector space $\mathbb{R}^{\mathbb{N}}:=\left\{f:\mathbb{N}\to\mathbb{R}\right\}$. If I refer ...