# Least amount of structure on a set to define a series on it

I know I can define a sequence on a set ##X## as a function ##a:T\rightarrow{}X##, where ##T## is a countable totally ordered set. But what about series? Can I define a series as a function ##\omega{}:a\rightarrow{}A##, where ##A\in{}X##? Or is this too general to be a series? Do I need to define an operation on ##X## so that I can "add" the terms of the sequence?

mathman
For a sequence to become a series you need addition to be defined.

For a sequence to become a series you need addition to be defined.
So I've been looking into it and I think that might not be the case. As far as I understood it series (over a set ##X##?) behave like elements of a free abelian group or free Z-module with basis ##X##. So that even if there's no operation defined on ##X## I can define series over it. I don't get why does the set of all series over a set ##X## needs to have all properties of a free abelian group. So I am still confused.. Slightly less then before though!

pwsnafu
I know I can define a sequence on a set ##X## as a function ##a:T\rightarrow{}X##, where ##T## is a countable totally ordered set.
You really want the notion of net not sequence. It has the same properties but only needs a directed set.

You really want the notion of net not sequence. It has the same properties but only needs a directed set.
I though nets generalized sequences not series.

mathman
So I've been looking into it and I think that might not be the case. As far as I understood it series (over a set ##X##?) behave like elements of a free abelian group or free Z-module with basis ##X##. So that even if there's no operation defined on ##X## I can define series over it. I don't get why does the set of all series over a set ##X## needs to have all properties of a free abelian group. So I am still confused.. Slightly less then before though!
An Abelian group (or Z module) has an addition operation (by definition).

WWGD
Gold Member
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If you want an infinite series, you need to define a topology so you can talk about convergence, together with the notion of sum you referred to.

WWGD
Gold Member
2019 Award
I though nets generalized sequences not series.
A series is a sequence of partial sums.

If you want an infinite series, you need to define a topology so you can talk about convergence, together with the notion of sum you referred to.
If you want to talk about convergence of a sequence or series you need the sequence or series to be defined on a topological space, but I don't need the notion of convergence to define sequence and series.

An Abelian group (or Z module) has an addition operation (by definition).
Sure, but the Z-module is the set of series over a set not the set itself. That means I can sum series to each other and get a third series, but addition for the terms of the series is still not required.

A series is a sequence of partial sums.
But I am fine with the generalization of sequences on a set as functions from countable totally ordered sets to that set.. My question is regarding generalizations of series given that definition of sequences not about generalization sequences into nets.

My question is: If I have a sequence ##a:\mathbb{N}\rightarrow{}X## and a function ##\sigma:X^\mathbb{N}\rightarrow{}Z## defined as
$$\sigma:a\mapsto\sum_{n=0}^\infty{}a(n)$$
that maps a sequence in ##X## to an infinite series in ##X##, why does the set ##Z## have to be a free Z-module? (note: notice that the set ##X## does not have to be a semigroup with addition defined on it or a topological space).

pwsnafu
Sure, but the Z-module is the set of series over a set not the set itself. That means I can sum series to each other and get a third series, but addition for the terms of the series is still not required.
I'm pretty sure you don't need abelian. A semigroup seems strong enough.

But I am fine with the generalization of sequences on a set as functions from countable totally ordered sets to that set.. My question is regarding generalizations of series given that definition of sequences not about generalization sequences into nets.
I brought up nets, not for the sequence→net generalisation, but to get you to abandon sequence→series and replace it with net→some-net-equivalent-of-series.

Also why do you insist on "countable totally ordered set". Series is more general than that to begin with.

You can refer to a sequence on any set ##X##, and people will know what you mean. You can refer to a series on any pair ##(X,+)## consisting of a set ##X## and an associative binary operation ##+:X\times X\to X##, and people will know what you mean. If you want to ask whether a given series on ##(X,+)## is summable, then you'll want ##X## to also have a topology on it, in which case people will assume you mean the sequence of partial sums converges.

Also, a notational nitpick: It's not standard to call a function ##T\to X## a sequence in ##X## for arbitrary countable totally ordered ##T##. The way most people use the term "sequence", your ##T## has to be (order-isomorphic to) the natural numbers.

disregardthat
The wikipedia article on series generalize series to apply to semigroups. You need only a set to define a sequence. But you need a topological space to talk of convergence, and the space must be hausdorff in order for the limit (if it exists) to be unique. So I would say that what we need is at least a topological semigroup which is hausdorff to talk of convergent series. A topological semigroup is a topological space X with a continuous multiplication function ##m: X \times X \to X## defining an operation * on elements on X. This way we may form partial "sums" ##x_1, x_1*x_2, x_1*x_2*x_3,...## for any sequence ##x_1,x_2,...##

The multiplication function must be continuous because (for example) we want to say that if ##x_n## converges to x, then ##y*x_n## converges to y*x.

If you want to operation to commute with elements, you will require the semigroup to be abelian. If you want an identity for the multiplication, we need a topological monoid. If you want inverses, a topological group. If you want multiplication and addition, you would need a topological ring, etc... finally, ##\mathbb{R}## is a topological field.

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disregardthat