Trying to prove F^infinity is infinite dimensional

[PhillipKP]

Homework Statement

Theorem: Prove that \mathbb{F}^{\infty} infinite dimensional

Homework Equations

Definition of Infinite Dimensional Vector Space: A vector space that
is not finite dimension

Definition of Finite Dimensional Vector Space: \exists list of vectors
in it that spans the space

Definition of List of Length n: An ordered collection of n objects.
It has finite length.

The Attempt at a Solution

1) \mathbb{F}^{n} is the set of all list of length n

2) \mathbb{F}^{\infty} is the set of all list of length \infty

2.1) This contradicts the definition of list \therefore there is
no list of infinite length \therefore it cannot be a finite dimensional
vector space because it doesn't contain a list \therefore it is
an infinite dimensional vector space.

My solution seems weak in that it doesn't use any math. But does
it actually prove the theorem? If not can someone point me in the right direction?

 

[JG89]

That's a good outline for a proof. Suppose that F^{\infty} isn't of infinite dimension. Then it must be of dimension n. So the set (consisting of n vectors) \beta = (1, 0, 0, ... 0), (0, 1, 0, ... 0), ... , (0, 0, ... , 1) must span F^{\infty}.

Obviously (1, 2, 3, ... , n+1) \in F^{\infty}. Is that vector in \beta ?

 

[PhillipKP]

It's not in the list \beta, it's in the span of \beta. Right?

 

[JG89]

Sorry, my question should have been, "is that vector in the span of \beta?"

 

[PhillipKP]

I don't understand where you are trying to go with this. There is no contradiction if you suppose that it is not infinite dimensional.

 

[PhillipKP]

Never mind this was due last week and my proof was correct according to the solutions.

 

[HallsofIvy]

I don't understand where you are trying to go with this. There is no contradiction if you suppose that it is not infinite dimensional.

Since you are trying to prove it is infinite dimensional there had better be!

 

[slider142]

... \therefore it cannot be a finite dimensional
vector space because it doesn't contain a list \therefore...

This reasoning is not clear. A finite dimensional vector space does not need to contain a list; it need only satisfy the axioms defining a vector space and be spanned by a finite amount of elements of that space under scalar multiplication and vector addition. Ie., the set of sequences of the form (n, 0, 0, ...) for all real n forms a 1-dimensional vector space under pointwise addition, but it does not consist of any lists.
J89's outline has a similar missing link: ie., it refers to a list of vectors that are not themselves elements of F^oo as potential spanning vectors for F^oo, which is unfounded.