Prove that F^infinity is infinite dimensional

[nox]

Hello i am currently struggling with a problem, F denotes either the set of all Real or Complex numbers.
The aim is to prove that $F^\infty$ is infinite dimensional.
First of i am not sure i understand the books definition of this collection. It says it consists of all sequences of elements of F.
My understanding is that it is simply $\{x: x \in F^i, \lim_{i \to \infty}\}$ (Feels like there is a better way to write this) Maybe this way would be better. $\{x: x \in (x_1, x_2,...), x_i \in F\}$
if F where the natural numbers for simplicity would that be like $\{(1,0,...), (1,1,0,...)...(2,0,0...), (2,2,0...),...(n1, n2, ...)\}$
Then to prove it, i am pretty sure doing it by proof contradiction would be good, by assuming it where finite dimensional, and hence it would have exist a basis for it. And then i am thinking maybe showing there is an element i can add to the basis, and it still would be a basis, hence contradicting the fact that all basis has the same size, or that any spanning set that are linear independent, becomes dependent if any element is added to that set.
I could be totally on the wrong track, feel i need some pointers on how to proceed.
Thank you

 

[WWGD]

You could clarify if you mean countable or uncountable infinity ( Not likely to go beyond that) and if by dimension you mean as a vector space over itself.

 

[nox]

You could clarify if you mean countable or uncountable infinity ( Not likely to go beyond that) and if by dimension you mean as a vector space over itself.

This is from the book linear algebra done right from chapter 2. He has not introduced the notion of countable infinity yet. I am not sure i understand his definition of $R^\infty$ and the only definition in the book is the one i wrote in words above. I would love to find out how to write it correct in set notation.
Dimension is defined in the book as the degree or size of the basis.

 

[fresh_42]

The most likely case is, that $\mathbb{F}^\infty$ means the $\mathbb{F}-$vector space of sequences in $\mathbb{F}$, likewise convergent or null sequences, but without specification any. So
$$
x\in \mathbb{F}^\infty \Longleftrightarrow x=(x_1,x_2,\ldots)_{n\in \mathbb{N}} \text{ with } x_i\in \mathbb{F}
$$
Note that a basis means, that all vectors, sequences in this case, are represented by a linear combination with only finitely many non-trivial terms.

 

[WWGD]

The most likely case is, that $\mathbb{F}^\infty$ means the $\mathbb{F}-$vector space of sequences in $\mathbb{F}$, likewise convergent or null sequences, but without specification any. So
$$
x\in \mathbb{F}^\infty \Longleftrightarrow x=(x_1,x_2,\ldots)_{n\in \mathbb{N}} \text{ with } x_i\in \mathbb{F}
$$
Note that a basis means, that all vectors, sequences in this case, are represented by a linear combination with only finitely many non-trivial terms.

Yes,makes sense. Then try to find a basis and show it is infinite. Do you know of the $\delta_i^j$ function? Think of the sequence with first term 1 and every other term zero as an n-ple. Can you use that to construct linearly independent-- infinitely many -- sequences?

 

[mathwonk]

I do not believe you will be able to find a basis for that space of all sequences, described in post #5. Rather it suffices just to produce, for every positive integer n, a set of n independent vectors in the space. That shows that no finite set can span the space, since you are probably allowed to assume the basic result that every spanning set has at least as many elements as every independent set. That is essentially what WWGD's hint suggests doing.

 

[WWGD]

I do not believe you will be able to find a basis for that space of all sequences, described in post #5. Rather it suffices just to produce, for every positive integer n, a set of n independent vectors in the space. That shows that no finite set can span the space, since you are probably allowed to assume the basic result that every spanning set has at least as many elements as every independent set. That is essentially what WWGD's hint suggests doing.

Wjy not? Wouldnt the $\delta_i^j; i=1,2,...$ help create a basis?

 

[fresh_42]

Wjy not? Wouldnt the $\delta_i^j; i=1,2,...$ help create a basis?

Not with finite linear combinations.

 

[WWGD]

Not with finite linear combinations.

Yes, that would be part of the argument fir the basis being infinite.

 

[fresh_42]

Yes, that would be part of the argument fir the basis being infinite.

But it also doesn't span the space, so it cannot be a basis.

 

[PeroK]

Hello i am currently struggling with a problem, F denotes either the set of all Real or Complex numbers.
The aim is to prove that $F^\infty$ is infinite dimensional.
First of i am not sure i understand the books definition of this collection. It says it consists of all sequences of elements of F.
My understanding is that it is simply $\{x: x \in F^i, \lim_{i \to \infty}\}$ (Feels like there is a better way to write this) Maybe this way would be better. $\{x: x \in (x_1, x_2,...), x_i \in F\}$
if F where the natural numbers for simplicity would that be like $\{(1,0,...), (1,1,0,...)...(2,0,0...), (2,2,0...),...(n1, n2, ...)\}$
Then to prove it, i am pretty sure doing it by proof contradiction would be good, by assuming it where finite dimensional, and hence it would have exist a basis for it. And then i am thinking maybe showing there is an element i can add to the basis, and it still would be a basis, hence contradicting the fact that all basis has the same size, or that any spanning set that are linear independent, becomes dependent if any element is added to that set.
I could be totally on the wrong track, feel i need some pointers on how to proceed.
Thank you

Alternatively, consider the obvious subspaces.

 

[nox]

I am reading through your great responses and trying to wrap my head around the definition.
This is exactly what the book say about this space. "The vector space F∞, consisting of all sequences of elements of F"
So going by the definition you gave me above, does this mean that every element is basically an infinite tuple, if such things exist, and he used the language as sequence cause he had not allowed an infinite list or tuple at this point.
So if i understand it
(2,1) would not belong to F∞
but (2,1,0,0,0...) would belong to F∞.

And yes the chapter has many thms about linear independence and basis. Like every linear independent set of vectors that span the space, becomes linear dependent if any vector is added to the set.
And facts like the size of any spaning set is always bigger than or equal any set of linear independent vectors in the space.

 

[PeroK]

Yes, that would be part of the argument fir the basis being infinite.

Assuming the space has a basis may need justification.

It is cleaner to show it cannot have a finite basis than to show it has an infinite basis.

 

[WWGD]

Ok, my approach was confusing. Consider my first post on the $\delta_i^j$ for the independent set and ignore the other , for clarity.

 

[PeroK]

I am reading through your great responses and trying to wrap my head around the definition.
This is exactly what the book say about this space. "The vector space F∞, consisting of all sequences of elements of F"
So going by the definition you gave me above, does this mean that every element is basically an infinite tuple, if such things exist, and he used the language as sequence cause he had not allowed an infinite list or tuple at this point.
So if i understand it
(2,1) would not belong to F∞
but (2,1,0,0,0...) would belong to F∞.

And yes the chapter has many thms about linear independence and basis. Like every linear independent set of vectors that span the space, becomes linear dependent if any vector is added to the set.
And facts like the size of any spaning set is always bigger than or equal any set of linear independent vectors in the space.

Strictly speaking, a sequence is a mapping from $\mathbb{N}$ to $F$. You may choose to represent such a mapping as an infinite tuple.

Following the debate above, I suggest it's simplest to show, perhaps by contradiction, that $F^{\infty}$ cannot be finite dimensional.

See, for example, the hint in post #11.

 

[fresh_42]

Strictly speaking, a sequence is a mapping from $\mathbb{N}$ to $F$.

Just to clarify this statement for the OP:

A mapping $f\, : \,X \longrightarrow Y$ can also be written as the set of all pairs $(x,f(x))$.
Now if the $x$ are all the natural numbers, then we have all pairs $(n,f(n))$.
However, in this case there is a natural numbering, too: $1,2,3,,\ldots$ and we can simply only list all $f(n)$ instead of the pairs. Since $f(n)\in Y = \mathbb{F}$ in our case, we have a list $x_1=f(1),x_2=f(2),x_3=f(3),\ldots$ This list is the infinite tupel $(x_1,x_2,\ldots)$ which is called a sequence to stress the ordering we have given. It is also more precise than "infinite tuple", because a sequence is based of the infinity of natural numbers, and not on other infinities as e.g. the real numbers.

Now those sequences can be added elementwise $(x_n)+(y_n)=(x_n+y_n)$ and multiplied by scalars $c\cdot (x_n)=(c\cdot x_n)$, i.e. they form a vector space over the field our $c$ is taken from. If we take the same field, then it will always work since there is already a multiplication in $\mathbb{F}$. But we could also restrict the values of $c$ to the rationals, in which case the sequences in $\mathbb{F}^\infty$ would be a $\mathbb{Q}-$vector space. This becomes important in cases in which complex vector spaces are considered a real vector space (then of twice the dimension), e.g. $x,y\in \mathbb{C}\, , \,c\in \mathbb{R}$.

 

[nox]

So here is my proof, i am curious what you think.
Assume $F^\infty$ is a finite dimensional vector space, then there exist some basis with finite vectors.
$B = \{v_1,...,v_n\}$
We know that these vectors are linear independent, and that they span the space.
So every element in $x \in F^\infty$ can be written as.
$x= a_1v_1 + ... + a_nv_n, a_1,...; a_n \in F$
But since the vector (0,0...,0,1,0...), (where the one is on the n+1 spot) is in $F^\infty$, but can not be written on the form as x above. This contradict the fact it spans the space, and thus B can not be a basis.
Thus we have shown that there is no basis for the space and thus it is infinite dimensional.
QED

I'm wondering do i need to motivate that the vector can not be written in form of x, if so what would be the best approach for that?

 

[fresh_42]

There is a flaw. What does $(0,0,\ldots,1,\ldots)$ mean? It cannot be an expression in coordinates relative to the basis $B$, since those coordinates only have $n$ places. If it is a sequence in $\mathbb{F}^\infty$, then it could easily be one of the basis vectors, say $v_1=(0,0,\ldots,1,\ldots)$.

So what is it?

 

[nox]

There is a flaw. What does $(0,0,\ldots,1,\ldots)$ mean? It cannot be an expression in coordinates relative to the basis $B$, since those coordinates only have $n$ places. If it is a sequence in $\mathbb{F}^\infty$, then it could easily be one of the basis vectors, say $v_1=(0,0,\ldots,1,\ldots)$.

So what is it?

Ok now i have no idea how to proceed.
Was i on the right track, or do i need a different approach entirely?

 

[fresh_42]

Ok now i have no idea how to proceed.
Was i on the right track, or do i need a different approach entirely?

You can use this approach (as one of the other possibilities mentioned before).
Just forget about a basis. Simply take $n$ linear independent sequences, e.g. those with the gliding $1$. Then show that there is always another linear independent one left outside.

 

[nox]

You can use this approach (as one of the other possibilities mentioned before).
Just forget about a basis. Simply take $n$ linear independent sequences, e.g. those with the gliding $1$. Then show that there is always another linear independent one left outside.

So how about this.
Let $v_i = (0,...1,0...)$ be the sequence where 1 is at the ith position
$v_1 = (1, 0,...)$
$v_2 = (0,1,0,...)$
$v_1, v_2$ are linear independent cause, $a_1v_1 + a_2v_2 = 0$ only when $a_1 = 0, a_2= 0$.
same reasoning shows that $a_1v_1 +...+a_nv_n = 0$ only when $a_1=...a_n=0.$
Lets check this, assume this is true for all n, n-1, n-2...1. We now show it is also true for n+1.
$a_1v_1 + ... + a_nv_n + a_{n+1}v_{n+1}=0, a_1=...a_n=0$
$0 + a_{n+1}v_{n+1}=0, a_1=...a_n=0$
Thus since $v_{n+1}$ is not 0, we know $a_{n+1}$ must be 0.
The principle of induction shows that there are infinitely many independent vectors in this space.
And since the size of any spaning set is always bigger than or equal number of independent vectors, we know there are no finite set of vectors that span the space. Thus We have proven that the space is an infinite dimensional vector space.

Is my reasoning with induction correct? I am having second thoughts on if it was legal.

 

[fresh_42]

Is my reasoning with induction correct? I am having second thoughts on if it was legal.

The proof is correct, especially the finish! I thought you would forget it.
The induction is a bit over the top. You can simply write
$$
a_1v_1+\ldots+a_nv_n=0 \Longleftrightarrow (a_1,\ldots,a_n,0,0,\ldots)=(0,\ldots,0,\ldots) \Longleftrightarrow a_1=a_2=\ldots=a_n=0
$$
for any given $n$. Yes, formally this can be wrapped in an induction, but it is obvious enough without.

 

[nox]

The proof is correct, especially the finish! I thought you would forget it.
The induction is a bit over the top. You can simply write
$$
a_1v_1+\ldots+a_nv_n=0 \Longleftrightarrow (a_1,\ldots,a_n,0,0,\ldots)=(0,\ldots,0,\ldots) \Longleftrightarrow a_1=a_2=\ldots=a_n=0
$$
for any given $n$. Yes, formally this can be wrapped in an induction, but it is obvious enough without.

Thank you so much for your help. I am trying to make sure everything i write is correct in my proofs, to get into the right habit. Good to know induction is not necessary in this case, and can be summarized with what you wrote above. The reason i was second guessing my induction, was that i was not sure if i was allowed to use the induction hypothesis like i did. I am glad i used the right reasoning in the proof at least.

 

[fresh_42]

Well, one can conclude by the arbitrariness of the variable $n$. If a statement $A(n)$ is true for any $n$, then it is true for all $n$. The induction would be $A(1) \wedge A(n-1) \Longrightarrow A(n)$. But here we do not need $A(n-1)$ for the conclusion, i.e. $A(n)$ can be shown to be true without using $A(n-1)$.

 

[mathwonk]

basic rule: whenever you want to prove some statement, you should read the definition of the words in that statement. In your case, definition 2.15 says that "V is infinite dimensional" means that V is not finite dimensional. In particular it has nothing to do, as defined in your book, with V having a basis. So go back and read the definition of "V is finite dimensional", and you find that means V is spanned by a finite "list", which means a finite set of vectors essentially. So you have to prove that no finite set of vectors can span your space.

Now result 2.23 says that every spanning list is at least as long as any independent list. So to prove there is no finite spanning list, it suffices to produce an infinite independent list of vectors. This you have done above. congratulations. but remember, before trying to prove something, read the definition. I.e. a "proof" is merely a verification of a definition. This is what mathematicians mean anyway.

At risk of being banned from the site, I recall the famous remark of Harvard mathematician David Kazhdan, "In physics they have many wonderful theorems, but unfortunately they have no definitions." To a mathematician the joke is that these wonderful "theorems" do not have proofs. I assume this joke opens a worm can, which will I hope be filled at worst with physicists' jokes about mathematicians.

 

[nox]

Now result 2.23 says that every spanning list is at least as long as any independent list. So to prove there is no finite spanning list, it suffices to produce an infinite independent list of vectors. This you have done above. congratulations. but remember, before trying to prove something, read the definition. I.e. a "proof" is merely a verification of a definition. This is what mathematicians mean anyway.

Thank you, so is my last proof correct, by i proved there was infinite independent vectors and then used that fact to prove the thm. The induction did not feel right, but i learned it was not needed in that case. I am curious was my induction correct, or did i break some rules? I want to really learn to do solid mathematical proofs, so all critique is highly welcomed.

 

[PeroK]

Thank you, so is my last proof correct, by i proved there was infinite independent vectors and then used that fact to prove the thm. The induction did not feel right, but i learned it was not needed in that case. I am curious was my induction correct, or did i break some rules? I want to really learn to do solid mathematical proofs, so all critique is highly welcomed.

There's nothing wrong with your proof. An alternative, which might be worth thinking about is:

If you consider the sequences that are zero after the nth term, then these form a copy of $F^n$.

In any case, that is an n-dimensional subspace.

As you have this for any $n$, you can conclude (or show by contradiction if you want) that the space itself cannot have any finite dimension.