Simple Notational Issue in Roman: "Advanced Linear Algebra"

[Math Amateur]

I am reading Steven Roman's book, Advanced Linear Algebra and am currently focussed on Chapter 1: Vector Spaces ... ...

I need help/clarification with respect to a notational issue regarding Roman's definition of the direct product and external direct sum of a family of vector spaces ... ...

Roman defines the direct product and the external direct sum of a family of vector spaces on page 41 as follows:View attachment 5088In the above definitions Roman does not define \(\displaystyle K\) ... nor as far as I can see, does he define it earlier in the book ...

... ... so my question is ... ... what is the exact nature of the set \(\displaystyle K\) ... can it be any set? ... looks like it should be something like \(\displaystyle \mathbb{N}\) ... but maybe it can be more general ...

Can someone please clarify this issue for me?

Hope someone can help ... help will be appreciated ...

Peter

 

[Deveno]

Yes, $K$ can be any set, although for many applications, $K$ is the set $\{1,2,\dots,n\}$ or $\Bbb N$. More formally, $\mathcal{F}$ is a $K$-indexed family, with indexing set $K$.

While for finitely-indexed sets, it is unusual to use any OTHER index BUT $\{1,2,\dots,n\}$, with infinite sets, it is often desirable to establish a correspondence with the indexing sets and the indexed sets, and infinite sets can be in one-to-one correspondence with sets that "look quite different" (for example, there exists a bijection between $\Bbb Q[x]$ and $\Bbb N$, via the axiom of choice).

This is more than just a "notational convenience", while countable indices can be satisfactorily be indicated by ellipses (the ''...") this is outright impossible with uncountable indexing sets.

It is not uncommon in functional analysis to consider the vector space $F^K$, for a field $F$, which can be given a vector space structure by using the operations of $F$. If $K = [a,b] \subseteq \Bbb R$, this is an uncountably indexed set (the $x$-th "coordinate", for $x \in [a,b]$ of a vector $f:[a,b] \to F$ is simply the value $f(x)$).

 

[Math Amateur]

Yes, $K$ can be any set, although for many applications, $K$ is the set $\{1,2,\dots,n\}$ or $\Bbb N$. More formally, $\mathcal{F}$ is a $K$-indexed family, with indexing set $K$.

While for finitely-indexed sets, it is unusual to use any OTHER index BUT $\{1,2,\dots,n\}$, with infinite sets, it is often desirable to establish a correspondence with the indexing sets and the indexed sets, and infinite sets can be in one-to-one correspondence with sets that "look quite different" (for example, there exists a bijection between $\Bbb Q[x]$ and $\Bbb N$, via the axiom of choice).

This is more than just a "notational convenience", while countable indices can be satisfactorily be indicated by ellipses (the ''...") this is outright impossible with uncountable indexing sets.

It is not uncommon in functional analysis to consider the vector space $F^K$, for a field $F$, which can be given a vector space structure by using the operations of $F$. If $K = [a,b] \subseteq \Bbb R$, this is an uncountably indexed set (the $x$-th "coordinate", for $x \in [a,b]$ of a vector $f:[a,b] \to F$ is simply the value $f(x)$).

Thanks Deveno ... very much appreciate your assistance ...

Reflecting on what you have said ... and the implications of what you have said ...

Peter