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It is well known that the axioms of a ring R with unity 1 imply that the underlying group must be commutative.

For if a and b are elements of R, and writing + for the group operation then applying the distributive property one has $$ \begin{align} a+a+b+b&=a*(1+1)+b*(1+1)\\\\ &=(a+b)*(1+1)\\\\ &=(a+b)*1+(a+b)*1\\\\ &=a+b+a+b, \end{align} $$ whence $a+b=b+a$.

For educational purposes, are there more (not only algebraic) examples of such superfluous definitions?

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    $\begingroup$ Same thing happen with vector spaces. The distributivity of scalar multiplication implies the commutativity of vector addition. $\endgroup$
    – Nick S
    Commented Feb 9, 2011 at 22:00
  • $\begingroup$ Isn't this just a problem of "popular definitions" (like Wikipedia ones)? $\endgroup$
    – Hans-Peter Stricker
    Commented Feb 15, 2011 at 16:06
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    $\begingroup$ It is an epsilon off topic, but, `All mathematics is built up by combinations of a certain number of primitive ideas, and all its propositions can, but for the length of the resulting formulae, be explicitly stated in terms of these primitive ideas; hence all definitions are theoretically superfluous.' -Bertram Russell, Principles of Mathematics, Chapter XILX $\endgroup$
    – ADL
    Commented Feb 16, 2011 at 16:08
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    $\begingroup$ "Bertram" Russell? $\endgroup$
    – Gerald Edgar
    Commented Feb 6, 2022 at 12:59
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    $\begingroup$ I just want to point out that there is a difference between a "superfluous definition" and a "superfluous condition". Example: An algebra homomorphism, in the sense of universal algebra, must respect all of the algebra operators. This definition has no superfluous parts---it is parsimonious and captures the main idea. However, when applied to groups, we get that group homomorphisms must respect not only the group multiplication, but also the identity and inverses. That some of these are superfluous conditions does not make the definition superfluous in any way. $\endgroup$
    – Pace Nielsen
    Commented Feb 6, 2022 at 16:44

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A unique factorization domain is typically defined as:

a domain $D$ such that each non-zero and non-invertible element $d$ can be factored as a product of irreducible elements.

And, for any factorizations $d=u_1 \dots u_m$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has that $m=n$ and there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.

Yet, this can be replaced by:

And, for any factorizations $d=u_1 \dots u_n$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has that there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.

In other words, it is sufficient that all factorizations of an element with the same number of factors are essentially equal. The fact that there, then, cannot be any factorization with a different number of factors can be proved and thus not have to be included in the definition.

Side note: If one replaces 'domain' by 'commutative cancellative semigroup with identity' the weakened definition actually is different.

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  • $\begingroup$ So what goes wrong with the "obvious counterexample" C[x,y,z,u,v]/xyz=uv? $\endgroup$
    – Noah Snyder
    Commented Feb 15, 2011 at 19:53
  • $\begingroup$ One can construct an element with different factorizations with the same number of factors. An explicit example is, except I made an error, (x+u) (xyz+v) = x ((x+u+1)yz + v) $\endgroup$
    – user9072
    Commented Feb 15, 2011 at 23:43
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A rectangle is defined as a quadrilateral having four right angles, but one could replace "four" in the definition with "three".

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    $\begingroup$ What if we were in a non-Euclidean space? $\endgroup$
    – Zhen Lin
    Commented Feb 14, 2011 at 7:38
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A distributive lattice is usually defined as a lattice for which the operation $\vee$ distributes over $\wedge$, and dually $\wedge$ distributes over $\vee$. See https://mathworld.wolfram.com/DistributiveLattice.html and Link. However, either of these distributivity axioms implies the other, and Wikipedia in fact uses just one of them: https://en.wikipedia.org/wiki/Distributive_lattice.

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A Lie subgroup of a Lie group is usually defined as a subgroup which is also a submanifold. But actually any closed subgroup of a Lie group is automatically a manifold, hence a Lie subgroup. Similarly, any continuous group homomorphism between Lie groups is automatically smooth, i.e. a morphism in the category of Lie groups.

NB Needless to say, the usual definitions (maybe due to Chevalley?) are there for very good reasons and shouldn't be changed. I just find it interesting that formally weakening them doesn't change the concepts studied.

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    $\begingroup$ Related: if a smooth manifold is equipped with a group structure, then smoothness of the multiplication map implies smoothness of the inversion map by the inverse function theorem. $\endgroup$
    – Steven Sam
    Commented Feb 9, 2011 at 22:36
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    $\begingroup$ The first example isn't in quite the same spirit - since a Lie subgroup need not be closed. $\endgroup$
    – ndkrempel
    Commented Feb 13, 2011 at 23:09
  • $\begingroup$ In the same spirit: thanks to the deep results of Montgomery-Zippin, Gleason, and Yamabe resolving Hilbert's fifth problem, the definition of a Lie group can be relaxed from "smooth finite-dimensional group" to "locally Euclidean topological group". $\endgroup$
    – Terry Tao
    Commented Feb 6, 2022 at 17:41
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Common axioms for groups are associativity, existence of two-sided identity and existence of two-sided inverses. (Sometimes even uniqueness is required too.) However, it is enough to require associativity, existence of right identity and existence of right inverses.

If we mix directions and require right identity and left inverses, we get something not too far removed from a group (I'll leave it as an exercise...)

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    $\begingroup$ You have to be careful with how you define "inverse" here, though. You need a right identity $e$ and for each $a$ an $a^{-1}$ such that $aa^{-1}=e$ - note that here when we multiply by the inverse we get a specific right-identity. If we merely require for each $a$ there exists an $a^{-1}$ such that $aa^{-1}$ is some right-identity, it doesn't work. $\endgroup$
    – Harry Altman
    Commented Feb 14, 2011 at 0:16
  • $\begingroup$ @Harry: Thanks for making that clear. $\endgroup$
    – ndkrempel
    Commented Feb 14, 2011 at 3:26
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A $\sigma$ algebra of subsets of a set X is defined as a collection of subsets of X which is invariant by taking complements and denumerable unions. And which contains the empty set.

But this last condition is (almost) superfluous. If there exists an element, say $A$, in the $\sigma$-algebra, then it must contain $(A\cup A^c)^c$, which is the empty set. Hence the sole purpose of requiring the empty set to be in the $\sigma$-algebra, is to deny the empty set the right to be a $\sigma$-algebra itself.

I don't really know why the empty set should not be a $\sigma$-algebra, and I don't see any result that would suddenly fail badly if we give the empty set this promotion.

EDIT: Bourbaki, Topology, ch5, section 6 no 3 (TG IX.60) does not require the empty set to be in the $\sigma$-algebra. So I think this is really superfluous.

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  • $\begingroup$ Isn't there something strange where the empty set has a pretty good supply of localic sigma-algebras? $\endgroup$
    – Harry Gindi
    Commented Feb 13, 2011 at 21:27
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    $\begingroup$ The empty set axiom of ZF (en.wikipedia.org/wiki/Axiom_of_empty_set) is similar, although in that case it is actually superfluous because the axiom of infinity implies the existence of a set. $\endgroup$
    – George Lowther
    Commented Feb 14, 2011 at 0:48
  • $\begingroup$ Also, I think $1\not=0$ in integral domains is similar. $\endgroup$
    – George Lowther
    Commented Feb 14, 2011 at 0:49
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    $\begingroup$ To me it is artificial to require invariance under countable unions without understanding "finite" as a special case of "countable". And the empty set is the union of a finite set of members of any set, namely the empty set of members. $\endgroup$
    – Tom Goodwillie
    Commented Feb 14, 2011 at 3:15
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    $\begingroup$ I agree with Tom. The empty subset of $X$ is the union of zero subsets of $X$, and zero subsets make a very countable collection. $\endgroup$
    – S. Carnahan ♦
    Commented Feb 14, 2011 at 5:59
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A complete lattice is a poset in which every subset has both an infimum and a supremum. Existence of the infimum for every subsets is already enough.

P: Every element is a lower bound for the empty set, so the infimum is the largest element and a largest element exists, so every set has an upper bound. The infimum of all upper bounds is easily seen to be an upper bound- and a smallest one at that.

Nonnegativity in the definitions of metrics and norms is redundant but often included.

P: $0=d(x,x)\leq d(x,y)+d(y,x)=2d(x,y)$ for all x,y. The case of norms is similar.

A regular Hasudorff space is just a regular space that is $T_0$.

P: Take two points, one is not in the closure of the other. Separate that point and the closure of the other by open sets. This also separates the points.

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