Generators and relations for $\mathbb{R}_{\geq 0} \cup \{\infty\}$ involving infinite sums

Question

A countably-complete semiring is basically a semiring with some additional structure making infinite sums possible. Formally, it is a tuple $R=(|R|,\Sigma,\cdot,1)$, where $|R|$ is a set, $(R,\cdot,1)$ is a monoid, and $\Sigma$ is an operation which assigns to every countable family $(a_i)_{i \in I}$ of elements in $|R|$ an element $\sum_{i \in I} a_i$ in $|R|$, such that the following equations hold:

• $\sum_{i \in \{j\}} a_i = a_j$
• If a countable set $I$ is the disjoint union of countably many sets $I_j$, $j \in J$, then $\sum_{i \in I} a_i = \sum_{j \in J} \sum_{i \in I_j} a_i$.
• $(\sum_{i \in I} a_i ) \cdot b = \sum_{i \in I} (a_i \cdot b)$
• $b \cdot \sum_{i \in I} a_i = \sum_{i \in I} (b \cdot a_i)$

It is clear how to define homomorphisms of countably-complete semirings. Every countably-complete semiring has an underlying semiring: We define $0 := \sum_{i \in \emptyset} a_i$ for the unique $\emptyset$-indexed family $a$ and $a_1 + a_2 := \sum_{i \in \{1,2\}} a_i$. (The usual definition of a complete semiring in the literature starts with a semiring, but this is redundant.)

My interest for countably-complete semirings comes from the observation that the Lebesgue-integral for non-negative measurable functions has a purely algebraic characterization within countably-complete commutative monoids: We need monotone convergence, which just means that the integral is countably-additive, and the extension property $\int \chi_A \, d\mu = \mu(A)$. This also motivates the following:

Question. What is a description of the countably-complete semiring $(\mathbb{R}_{\geq 0} \cup \{\infty\},\Sigma,\cdot,1)$ by generators and relations? Here, we define $\sum_{i \in I} a_i := \sup_{F \subseteq I \text{ finite}} \sum_{i \in F} a_i$, where $\sup$ refers to the usual order on $\mathbb{R}_{\geq 0} \cup \{\infty\}$.

It is easy to check that the initial countably-complete semiring is $(\mathbb{N} \cup \{\infty\},\dotsc)$ and that the initial (usual) semiring in which all $n \in \mathbb{N}_{>0}$ are invertible is $(\mathbb{Q}_{\geq 0},\dotsc)$. Since $(\mathbb{R},\leq)$ is the order-completion of $(\mathbb{Q},\leq)$, I suspect that $(\mathbb{R}_{\geq 0} \cup \{\infty\},\dotsc)$ is the initial countably-complete semiring in which all $n \in \mathbb{N}_{>0}$ are invertible. But I don't think that this is trivial at all.

So here is explicitly what has to be proven for my claim: Let $R$ be a countably-complete semiring and assume that all $n \in \mathbb{N}_{>0}$ become invertible in $R$. We obtain a map $f : \mathbb{Q}_{\geq 0} \cup \{\infty\} \to |R|$. We want to define an extension $\overline{f} : \mathbb{R}_{\geq 0} \cup \{\infty\} \to |R|$ as follows: If $r$ is a non-negative real number, or $\infty$, find some countable sum expression $r = \sum_{i \in I} q_i$ with rational numbers $q_i$, and define $\overline{f}(r) := \sum_{i \in I} f(q_i)$.

Question. Why is $\overline{f}$ well-defined?

If this was the case, the rest of the proof would be clear. Well-definedness seems to have a quite strong interpretation, namely that every equation of infinite series of rational numbers is "purely algebraic". For example, the equation $\sum_{n=1}^{\infty} \frac{1}{2^n} = 1$ holds, in fact, in every countably-complete semiring containing the non-negative rationals.

Edit: I'm not so sure anymore about the equation $\sum_{n=1}^{\infty} \frac{1}{2^n} = 1$. Let $s$ denote the sum. Then one observes $2 \cdot s = s + 1$. But since we are in a semiring, this does not imply $s=1$! Also, I've tried to show the equation $\sum_{n=1}^{\infty} \frac{1}{n} = \infty$ algebraically, i.e. in a countably-complete semiring containing $\mathbb{Q}_{\geq 0}$. Using prime factor decompositions, one reduces this to geometric series. But $s := \sum_{n=1}^{\infty} \frac{1}{q^n}$ (for $q \in \mathbb{Q}_{>1}$) only satisfies $q s = s + 1$, and I don't know how to deduce $s = \frac{1}{q-1}$.

Edit: Ok, my claim is wrong, but my question remains, namely how to describe $(\mathbb{R}_{\geq 0} \cup \{\infty\},\Sigma,\cdot,1)$ by generators and relations.

Answer 1

Part 1

Assume $f:\mathbb{Q}_{\geq0}\cup\{\infty\}\to|R|$ is a morphism commutes with infinite sums of rationals that add up to a rational. Let us show that the extension to $\mathbb{R}_{\geq0}\cup\{\infty\}$ you defined above is well defined.

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Assume that $r=\sum_{i\in I} q_i=\sum_{i\in I'} q_i'$.

Then you can do the following:

If $q_1=q_1'$ we look at the next index.

If $q_1>q_1'$ (let us assume that it is), then we write $q_1=q_1'+a$, we pass to look at $a$ from the first series and $q_2'$ from the second, and repeat:

If $a_1>q_2'$ we decompose $a_1=q_2'+a_2$. If $a_1<q_2'$ then it is $q_2'$ the one we decompose as $q_2'=a_1+a_2$.

In this manner we produce decompositions $q_i=\sum_{k\in I_i}a_k$ and $q_i'=\sum_{k\in I_i'}a_i$ and moreover $\{a_i|\ i\in\bigcup I_i\}=\{a_i|\ i\in\bigcup I_i'\}$ and the terms appear in the same order.

Notice that each $I_i,I_i'$ is finite except for possibly the case in which one of $I$ or $I'$ (say $I$) is finite and the decomposition of $q_i$ has infinitely many terms only for the last element of $I$.

We use now the second property of the countably complete semi-ring, and that $f(q_i)=f(\sum_{i\in I_i}a_i)=\sum_{i\in I_i}f(a_i)$ (which we have because $q_i$ is rational) and the same thing for the numbers with primes.

Each sum $\sum_{i\in I} f(q_i)$ and $\sum_{i\in I'} f(q_i')$ is equal to the sum $\sum_{i\in\bigcup I_i=\bigcup I_i'}f(a_i)$.

Answer 2

Part 2

I just noticed that there is no axiom linking sums of finitely many elements to sums of infinitely many elements.

This means that we can have the following countably-complete semi-ring $(\mathbb{R}_{\geq0}\cup\{\infty\},\Sigma',\cdot,1)$ such that $\Sigma'$ is the usual sum when the set of indexes is finite and $\infty$ when the set of indexes is infinite (more precisely, when infinitely many of the terms of the sequence are non-zero.), or when $\infty$ is one of the summands.

This $\Sigma'$ satisfies the axioms because whenever a set of indexes is infinite in one side of the axioms there is also an infinite set of indexes on the other side of the axiom, turning it into $\infty=\infty$. When all set of indexes are finite the axioms are the usual properties of finite sums.

We see that each $n\in\mathbb{N}$ is invertible in $(\mathbb{R}_{\geq0}\cup\{\infty\},\Sigma',\cdot,1)$ with inverse $1/n$.

We can't have a morphism from $(\mathbb{R}_{\geq0}\cup\{\infty\},\Sigma,\cdot,1)$ to $(\mathbb{R}_{\geq0}\cup\{\infty\},\Sigma',\cdot,1)$.

Such a morphism $f$ would have to send $f(1)=1$, and would have to satisfy $$1=f(1)=f\left(\sum_{n>0}2^{-n}\right)=\sum_{n>0}f(2^{-n})=\infty.$$

Therefore $(\mathbb{R}_{\geq0}\cup\{\infty\},\Sigma,\cdot,1)$ can't be the initial object.

Answer 3

Part 3 (a form of answer to your first question)

Lemma: Assume $q_i$, $q$ are in $\mathbb{Q}_{\geq0}$ such that $\sum_{i\geq0}q_i=q$ and we know that $\sum_{i\geq0}f(q_i)=f(q)$, and that $\sum_{i>n}f(q_i)=f(q-\sum_{i=0}^{n}q_i)$. Then, for every sequence $q_i'$ in $\mathbb{Q}_{\geq0}$ such that $\sum_{i\geq0}q_i'=q\in\mathbb{Q}_{i\geq0}$ we must have $\sum_{i\geq0}f(q_i')=f(q)$.

Proof: We first apply the technique in the other answer to split the terms in both series $\sum q_i$ and $\sum q_i'$ into a common series $\sum_{i\in I}a_i$ with $q_i=\sum_{k\in I_i}a_k$ and $q_i'=\sum_{k\in I_i'}a_k$.

By construction either all $I_i,I_i'$ are finite or say $q_i'$ were almost all zero and the last term $q_n'=\sum_{n>N}q_i$.

In each case, since we know that $f$ commutes with finite sums of rationals and we are given also the relations on the tails we get that

$$\begin{align}f(q)&=f\left(\sum_{i}q_i\right)\\&=\sum_if(q_i)\\&=\sum_if\left(\sum_{k\in I_i}a_i\right)\\&=\sum_i\left(\sum_{k\in I_i}f(a_k)\right)\\&=\sum_{i}\sum_{k\in I_i'}f(a_k)\\&=\sum_{i}f\left(\sum_{k\in I_i'}a_k\right)\\&=\sum_if(q_i')\end{align}$$

Lemma: If $\sum_i q_i=q$, with $q_i,q$ non-negative rationals and we know that $\sum_if(q_i)=f(q)$ (and its tails), then we also know a relation (and its tails) for every other rational $q'>q$.

Proof: In fact, $q'=(q'-q)+\sum_i q_i$ and we can apply $f$ to both sides to get

$$f(q')=f((q'-q)+\sum_iq_i)=f(q'-q)+f(\sum_iq_i)=f(q'-q)+\sum_if(q_i).$$

And the tails of this series are just tails for the series of $q=\sum_i q_i$.

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Due to lack of cancellation for the sum I can't see right now if it is possible to get from the relation for the series for $q$, relations for series giving rational number smaller that $q$.

In any case, what the results we have so far imply is that it is enough to give the following set of relations:

Partition the non-negative reals according the orbits of translations by rational numbers. In for each equivalence class consider one sequence $r_1,r_2,...$ that tends to zero. For each element $r_i$ on each sequence give a relation $f(r_i)=\sum_jf(q_{i,j})$ corresponding to one series of non-negative rationals $\sum_jq_{i,j}=r_i$.

Given such a set of relations in $|R|$, the two lemmas above imply that $(\mathbb{R}_{\geq0},\Sigma,\cdot,1)$ embeds in $|R|$.