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

1. Aug 26, 2014

### V0ODO0CH1LD

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?

2. Aug 26, 2014

### mathman

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

3. Aug 26, 2014

### V0ODO0CH1LD

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!

4. Aug 27, 2014

### pwsnafu

You really want the notion of net not sequence. It has the same properties but only needs a directed set.

5. Aug 27, 2014

### V0ODO0CH1LD

I though nets generalized sequences not series.

6. Aug 27, 2014

### mathman

An Abelian group (or Z module) has an addition operation (by definition).

7. Aug 27, 2014

### WWGD

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.

8. Aug 27, 2014

### WWGD

A series is a sequence of partial sums.

9. Aug 27, 2014

### V0ODO0CH1LD

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.

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.

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).

10. Aug 28, 2014

### pwsnafu

I'm pretty sure you don't need abelian. A semigroup seems strong enough.

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.

11. Aug 28, 2014

### economicsnerd

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.

12. Aug 28, 2014

### economicsnerd

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.

13. Aug 28, 2014

### 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.

Last edited: Aug 28, 2014
14. Aug 28, 2014

### disregardthat

You COULD, as you mention in your other posts, define the free semigroup on X, call it S, composed of all finite strings of elements $x_1x_2...x_n$ where $x_i \in X$, and define a series defined by a sequence in X as a sequence in S to be on the form

$x_1, x_1x_2, x_1x_2x_3,...$

However, this would leave the set X, and the series would be defined by a sequence of terms not in X, but in S, the free semigroup generated by X. To speak of convergence in such a construction, we depend on what topology we give S. Now, as a set, S is on the form $S=X \sqcup X \times X \sqcup X \times X \times X \sqcup ... = \bigsqcup^{\infty}_{n=0} X^n$. We could give it its natural topology: $X^n$ has the product topology for each n, and the infinite disjoint union the naturally induced topology generated by the topologies on $X^n$ for each n.

Now.. the big question is what does convergence mean here? The problem is that the terms of the sequence (which is on the form $x_1, x_1x_2, x_1x_2x_3,...$ are in separate (connected or not) components of S. That means it can never converge. The reason being that if it supposedly converged to x, then x must be a point in one of these components (the m'th component $X^m$, say). But the terms of the partial sums will eventually leave the m'th component indefinitely.