Language to describe $\forall$

Question

In a document that I am working on, I have written the following definition for a group:

Definition: (Group) $\hspace{2pt}$ A group (G) is a set together with a binary operation, usually called multiplication, such that the following hold: $$ \begin{array}{l} -\text{For all $g_{1}, g_{2}, g_{3} \in G$ we have $(g_{1}g_{2})g_{3} = g_{1}(g_{2}g_{3})$};\\ -\text{There is an element $1_{G} \in G$ such that $g1_{G} = 1_{G}g = g$ for every $g \in G$};\\ -\text{For each \(g \in G\) there exists \(g^{-1} \in G\) such that \(gg^{-1} = g^{-1}g = 1_{G}\)}. \end{array} $$

I currently have "for all", "for every", and "for each" to describe the universal quantifier $\forall$ since I typically try to avoid repeating myself over and over.

Each of them are correct (supported by this answer), though I wonder whether it would be more consistent to stick to just one of "for all", "for every", and "for each" in cases like this, or is the use of all three better as it sounds less repetitive? Is it up to personal preference, or is there a general convention?

Answer 1

This is a very much a matter of style, and different authors will have different preferences. If this is for a publication, I would ask your reviewers / copy editor / publisher what they prefer. Otherwise, do what you like. That being said, I personally prefer a parallel phrasing for parallel structures. Indeed, I think that repetitive language in mathematics can be useful, and that it can be potentially confusing to use too many synonyms (unless, of course, you are looking for synonyms for "hence," "thus," and "therefore"—there will never be enough of these $\ddot\smile$). Ergo (there's another one!), I would think that something like the following would be good:

Definition: (Group) A group $(G,\cdot)$ is a set $G$, along with a binary operation $\cdot$ (usually called multiplication), such that the following hold:

• for any $g_1,g_2,g_3 \in G$ we have $g_1 \cdot (g_2 \cdot g_3) = (g_1 \cdot g_2) \cdot g_3$;
• there is an element $1_G \in G$ such that for any $g\in G$ we have $g \cdot 1_G = 1_G \cdot g = g$; and
• for any $g\in G$ there exists an element $g^{-1} \in G$ such that $g\cdot g^{-1} = g^{-1} \cdot g = 1_G$.

Here, I've used the phrase "for any" in each case, which (to me) emphasizes the fact that each of the three axioms has basically the same structure, i.e. we pick out arbitrary elements of $G$ and check that they do stuff. Having read this out loud a couple of times to make sure, I think it reads rather well. Your milage may vary. $\ddot\smile$