Infinite Basis and Supernatural Numbers

[FallenApple]

I've read that it is unsatisfactory to consider infinitely many basis vectors to span an infinite dimensional space. For example, for the infinite dimensional Hilbert space, {e1,e2,e3...} we could use this to make an arbitrary infinite tuple (a,b,c,...). If this is looked down upon, then why are Supernatural Numbers, which are the product of infinitely many primes, considered ontologically valid ?

Surely if we do a transformation, we could express those supernatural numbers as vectors, since its just converting multiplication to vector addition.

 

[TeethWhitener]

I've read that it is unsatisfactory to consider infinitely many basis vectors to span an infinite dimensional space.

Citation needed. For example, the fact that the infinite set of basis vectors $e^{ikx}, k\in \mathbb{Z}$ spans the space of periodic functions is the cornerstone of Fourier theory. Also, the axiom of choice is equivalent to the statement that every vector space--even if it has uncountably infinite dimensions--has a basis.

 

[FallenApple]

Citation needed. For example, the fact that the infinite set of basis vectors $e^{ikx}, k\in \mathbb{Z}$ spans the space of periodic functions is the cornerstone of Fourier theory. Also, the axiom of choice is equivalent to the statement that every vector space--even if it has uncountably infinite dimensions--has a basis.

http://www.math.lsa.umich.edu/~kesmith/infinite.pdf

 

[Mark44]

For the question in your quoted text, I don't see the problem. It seems to me that (1, 1, 1, ..., ) + (2, 2, 2, ... , ) + (3, 3, 3, ... , ) = (6, 6, 6, ... , ).

IOW, component-wise vector addition, even though the vectors belong to an infinite-dimension vector space.

I wouldn't consider the sum of 3 vectors to be an infinite sum, being as there are only three of them being added. I will take a look at the link you gave to see if there is some context that is missing from what you quoted.

 

[Mark44]

IMO, the author of the PDF from which you quoted is quite sloppy.
In Example 2 she says

The vector addition and scalar multiplication are defined in the natural way: the sum of $(\alpha_1, \alpha_2, \alpha_3, \dots, \dots)$ and
$(\beta_1, \beta_2, \beta_3, \dots, \dots)$ is $(\alpha_1 + \beta_1, \alpha_2 + \beta_2, \alpha_3 + \beta_3, \dots, \dots)$

Later in the same example, she says

Since the vector v = (1, 1, 1, 1, ..., ) can not be written as a linear combination of
the vectors e_i = (0, ..., 0, 1; 0; 0, ...), it must be that together the e_i's and v form
a linearly independent set.

In the sentence above she meant that v cannot be written as a finite linear combination, but neglected to write it.

Finally, from the text you quoted, she calls (1, 1, 1, ..., ) + (2, 2, 2, ..., ) + (3, 3, 3, ..., ) an infinite sum even though there are only three vectors. She asks what would be the sum of these vectors, a question she answers more generally in the first part of Example 2.

 

[FallenApple]

IMO, the author of the PDF from which you quoted is quite sloppy.
In Example 2 she saysLater in the same example, she says
In the sentence above she meant that v cannot be written as a finite linear combination, but neglected to write it.

Finally, from the text you quoted, she calls (1, 1, 1, ..., ) + (2, 2, 2, ..., ) + (3, 3, 3, ..., ) an infinite sum even though there are only three vectors. She asks what would be the sum of these vectors, a question she answers more generally in the first part of Example 2.

Ok that makes sense. I was thinking somewhere alone those lines of totaling up the coefficients per coordinate, which is example 2. That makes sense.

In an infinite vector space, it is a bit self explanatory that infinite combinations are needed unless the resultant vector is non zero in finitely many places; of which we still get around this by placing 0 in front of the remaining basis vectors. So it's strange for her to attach significance to the fact that some vectors wouldn't be able to be written as a finite linear combo, given the context of infinite dimensions.

 

[Mark44]

In an infinite vector space, it is a bit self explanatory that infinite combinations are needed unless the resultant vector is non zero in finitely many places; of which we still get around this by placing 0 in front of the remaining basis vectors. So it's strange for her to attach significance to the fact that some vectors wouldn't be able to be written as a finite linear combo, given the context of infinite dimensions.

Yes. Although it's obvious that you can't generate a particular vector with a finite linear combination of basis vectors, it seemed very sloppy to me to omit "finite".
And her example didn't make any sense, namely the one with (1, 1, ...) + (2, 2, ...) + (3, 3, 3, ...) and calling that an infinite sum.

 

[FallenApple]

Yes. Although it's obvious that you can't generate a particular vector with a finite linear combination of basis vectors, it seemed very sloppy to me to omit "finite".
And her example didn't make any sense, namely the one with (1, 1, ...) + (2, 2, ...) + (3, 3, 3, ...) and calling that an infinite sum.

Thats true. It's a tri-nary sum producing a single answer infinitely many separate times , which is not an infinite sum of a sequence.

 

[Erland]

For the question in your quoted text, I don't see the problem. It seems to me that (1, 1, 1, ..., ) + (2, 2, 2, ... , ) + (3, 3, 3, ... , ) = (6, 6, 6, ... , ).

I think there is a typo in the text quoted by OP, so that it should be "(1, 1, 1, ...) + (2, 2, 2, ...) + (3, 3, 3, ...) + ... "

 

[Mark44]

I think there is a typo in the text quoted by OP, so that it should be "(1, 1, 1, ...) + (2, 2, 2, ...) + (3, 3, 3, ...) + ... "

Maybe, but I would be embarrassed to put out something with such a glaring error.