Is there a minimal generating set of reals which additively generate all the reals?

Question

Is there a set $S$ of real numbers such that the submagma generated by $S$ under addition is the entire set of real numbers, but such that no proper subset of $S$ generates the entire set of real numbers?

Answer 1

Note: My answer here shows that the additive group of real numbers admits no minimal generating set. As Keith Kearnes notes in the comments, it does not resolve the question as stated, which is whether the additive magma of real numbers admits a minimal generating set.

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No. In fact, no nontrivial divisible abelian group admits a minimal generating set. I found the argument in the paper Abelian Groups with a Minimal Generating Set by Pavel Růžička (Theorem 1.1), which, unsurprisingly, contains lots more information about minimal generating sets in abelian groups.

Here's Růžička's argument, slightly rephrased:

Let $(G,+)$ be a nontrivial divisible abelian group, and suppose for contradiction that $S$ is a minimal generating set. Let $s\in S$, and let $G'$ be the subgroup of $G$ generated by $S\setminus \{s\}$. By minimality, $G'$ is a proper subgroup of $G$, so $G/G'$ is a nontrivial group. But since $G'\cup \{s\}$ generates $G$, $G/G'$ is cyclic, generated by the coset $s+G'$. Now a quotient of a divisible abelian group is divisible, but no nontrivial cyclic group is divisible, so we have a contradiction.

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A bit more concretely, without thinking about quotients: Since $G'$ is a proper subgroup of $G$, $s\notin G'$.

Case 1: $ns\in G'$ for some $n\in \mathbb{N}^+$. Since $G$ is divisible, there is some $g\in G$ such that $ng = s$. Writing $g$ as a sum of elements of $S$ and multiplying by $n$, we can write $s = ng$ as a sum of elements of $G'$, so $s\in G'$, contradiction.

Case 2: $ns\notin G'$ for all $n\in \mathbb{N}^+$. Since $G$ is divisible, there is some $g\in G$ such that $2g = s$. Write $g$ as a sum of elements of $S$, and say $s$ appears in this sum $m$ times. Multiplying by $2$, we can write $s = 2g$ as a sum of elements of $G'$ plus $2ms$. Thus $(2m-1)s\in G'$, contradiction.

Answer 2

I don't know the answer to this question. Here I describe some examples which illustrate some relevant facts.

Example 1. There is an abelian group $(G,+,-,0)$ with a subset $S\subseteq G$ that is NOT an irredundant generating for the group $(G,+,-,0)$, but IS an irredundant generating subset of the magma $(G,+)$. This illustrates the fact that solving the problem negatively for $(\mathbb R, +, -, 0)$ does not resolve the problem for $(\mathbb R,+)$. This is relevant to the answer posted on August 17, 2022.

Example 2. There is a sequence of algebraic structures ${\mathbf A}_0, {\mathbf A}_1, {\mathbf A}_2, \ldots$ such that

1. ${\mathbf A}_{n+1}$ is an expansion of ${\mathbf A}_n$ by one unary operation for each $n$, and
2. ${\mathbf A}_k$ has an irredundant generating subset for even $k$ and does not have an irredundant generating subset for odd $k$.

The purpose of this example is to show that the question of whether an algebra has an irredundant generating subset depends sensitively on the signature of the algebra -- a slight expansion in signature can change in either direction whether the algebra has an irredundant generating subset. This example is also relevant to the answer posted on August 17, 2022.

Example 3. There is an algebra $\mathbf B$ with the following property: If $\mathcal P$ is any poset consisting of some of the subsets of $B$ such that

1. the order on ${\mathcal P}$ is the inclusion order,
2. any subset of an irredundant generating subset of $\mathbf B$ is a member of $\mathcal P$, and
3. $\mathcal P$ contains only subsets $S\subseteq B$ with the property that for any $s\in S$ one has $\langle S\rangle \neq \langle S\setminus \{s\}\rangle$,

then ${\mathcal P}$ is NOT an inductively ordered poset.

The purpose of this example is to show that even if an algebra has irredundant generating subsets, Zorn's Lemma might not be able to produce one. This is relevant to the answer posted and then deleted on July 23, 2022.

Construction of Example 1. Take $G$ to be the additive group of integers. The subset $S=\{1,-1\}$ is not an irredundant generating subset of the group $(\mathbb Z, +, -, 0)$, but it is an irredundant generating subset of the magma $(\mathbb Z, +)$.

Construction of Example 2. First some background: the structure $(\mathbb Z,\emptyset)$ with underlying set $\mathbb Z$ and no operations has irredundant generating subset $S=\mathbb Z$. If we expand the algebra to include the operation $f(x)=x+1$ we get a new structure $(\mathbb Z, f)$. This structure has no irredundant generating subset for the following reasons: any generating subset $S$ requires infinitely many negative integers but if $s\in S$ is negative, then $S\setminus \{s\}$ will still generate.

Now suppose that we add the inverse operation $f^{-1}(x)=x-1$ to this structure to get $(\mathbb Z, f, f^{-1})$. Any singleton $S=\{n\}$ is an independent generating subset.

The previous two paragraphs yield a chain of structures of length $3$, $(\mathbb Z,\emptyset)$, $(\mathbb Z, f)$, $(\mathbb Z, f, f^{-1})$, where each is an expansion of the previous one by one unary operation and where this chain alternates with respect to the property of having an independent generating subset. One can extend the idea to get an infinite alternating sequence as follows. All structures in the sequence will have underlying set $\mathbb Z^{\omega}$. For each $k\in \omega$, let $f_k\colon \mathbb Z^{\omega}\to \mathbb Z^{\omega}$ be the unary operation that increments the value in coordinate $k$ by $1$.
$(f_k(a_0,\ldots,a_{k-1},a_k,a_{k+1},\ldots) = (a_0,\ldots,a_{k-1},a_k+1,a_{k+1},\ldots)$.)
Let $f_k^{-1}$ be the inverse of $f_k$. (It is the operation that decrements the value in coordinate $k$ by $1$.) I claim that the sequence of structures

$$ (\mathbb Z^{\omega},\emptyset), (\mathbb Z^{\omega},f_0), (\mathbb Z^{\omega},f_0,f_0^{-1}), (\mathbb Z^{\omega},f_0,f_0^{-1},f_1), (\mathbb Z^{\omega},f_0,f_0^{-1},f_1,f_1^{-1}), \ldots $$ has the desired properties. (I.e., each structure is an expansion of the preceding one by one unary operation, and the structures have irredundant generating subsets iff they have an even number of operations in the signature.) The reason the even-indexed structures have irredundant generating subsets is that they are definitionally equivalent to $G$-sets, so by taking one element from each orbit we get an irredundant generating set. For odd-indexed structures one can show that any generating subset of $(\mathbb Z^{\omega},f_0,f_0^{-1},\ldots,f_k)$ must contain tuples $(a_0,a_1,\ldots,a_k,0,0,0,\ldots)$ with arbitrarily large negative $a_k$-values, and that any one of these is an inessential generator.

Construction of Example 3. This algebra $\mathbf B = (\mathbb Z, \{f_{b,n}\;|\;(b,n)\in \mathbb Z\times \mathbb Z^+\})$ has $\mathbb Z$ as its underlying set and it has an operation $f_{b,n}(x_0,\ldots,x_n)$ of arity $(n+1)$ for each choice of an element $b\in \mathbb Z$ of the algebra and a positive integer $n$. The operation $f_{b,n}$ is defined so that $f_{b,n}(-n,1,2,3,\ldots,n)=b$ and $f_{b,n}(x_0,\ldots,x_n)=x_0$ if $(x_0,\ldots,x_n)\neq (-n,1,2,3,\ldots,n)$.

Any of the operations $f_{b,n}$ has little generating power. If you apply $f_{b,n}$ to elements of any subset $U\subseteq \mathbb Z$ such that $\{-n,1,2,\ldots,n\}\not\subseteq U$, then you will generate no new elements. If you apply $f_{b,n}$ to elements of any subset $U\subseteq \mathbb Z$ such that $\{-n,1,2,\ldots,n\}\subseteq U$, then the only new element you can generate is $b$ and you can generate $b$ in only one way.

From this, one can deduce that

1. The irredundant generating subsets of $\mathbf B$ are exactly the subsets $S_n:=\{-n,1,2,\ldots,n\}, n>0$.
2. Each of the subsets $\{1,2,\ldots,n\}$ are subsets of irredundant generating sets of $\mathbf B$, but there is no irredundant generating subset of $\mathbf B$ that is a superset of all $\{1,2,\ldots,n\}$'s.

If you wanted to apply Zorn's Lemma to some poset $\mathcal P$ of 'partial irredundant generating subsets' of $\mathbf B$, you would likely want to include as elements of $\mathcal P$ all subsets of $\mathbf B$ that are subsets of irredundant generating subsets. Thus you would include each subset $\{1,2,\ldots,n\}$. To employ Zorn's Lemma, you would like $\mathcal P$ to be inductively ordered, so $\mathcal P$ would have to contain an element that dominates $\bigcup_n \{1,\ldots,n\}=\{1,2,\ldots\}=\mathbb Z^+$. In fact, you would like some maximal element of $\mathcal P$ to dominate $\bigcup_n \{1,\ldots,n\}=\mathbb Z^+$ and you also want this maximal element to be a partial or total irredundant generating subset. But there is no 'partial irredundant generating subset' of $\mathbf B$ that contains $\mathbb Z^+$. If $\mathcal P$ truly contained all partial irredundant generating subsets of $\mathbf B$ and was inductively ordered, then its maximal elements could not all be irredundant generating subsets.

Answer 3

I'm familiar with this question for a long time (the immediate argument below is contained in the proof of Proposition 4.11 in my 2014 J. Inst. Math. Jussieu paper with Bieri, Guyot, Strebel — arXiv link).

Fact: if $G$ is a divisible abelian group, $S$ a generating subset, $s\in S$, then $S-\{s\}$ generates $G$.

Proof: the quotient of $G$ by the subgroup generated by $S-\{s\}$ is then cyclic and divisible, hence zero. $\Box$.

Corollary: Such $G$ has no minimal generating subset unless $G=\{0\}$. $\Box$