# Is there a term for two subsets whose intersection is {0}?

In linear algebra I've just recently learned about the "direct sum", which can be defined as the sum of two vector spaces whose intersection is the null vector.

Basically, I'm wondering if there's a formal term in abstract algebra to define algebraic substructures whose intersection is the identity element, as it seems like it'd be a, in the case of vector spaces, useful way to differentiate between subspaces that can be directly summed and subspaces that can't be.

I'm specifically asking this because algebraic substructures whose intersection is the identity element seem like they could have some interesting interacting properties that differ from the interacting properties of any two substructures (and that the term disjoint wouldn't work in the case of any algebraic substructures).

• There's no term from set theory, but there might be in vector spaces. – Thomas Andrews Oct 7 '15 at 1:11
• Orthogonal subspaces? – Eric Towers Oct 7 '15 at 3:33
• @EricTowers No, they do not need to be orthogonal to satisfy this property. Take for instance $\{(x,y) : x=0\}$ and $\{(x,y): x+y=0\}$ in $\mathbb{R}^2$. – Federico Poloni Oct 7 '15 at 4:37

In the context of subspaces of a vector space (or more generally, subgroups of a group), it is common to abuse terminology and call such subspaces disjoint. This abuse is generally harmless since it is obviously impossible for them to actually be literally disjoint. You could also call them (linearly) independent, though personally I think "disjoint" sounds much more natural. I would not, however, use either term to describe arbitrary subsets (as opposed to subspaces) of a vector space whose intersection is $\{0\}$.

• "This abuse is generally harmless since it is obviously impossible for them to actually be literally disjoint." I like this; in set theory one could take an unusual definition of disjoint: "two sets are disjoint exactly when their only common subset is the trivial subset". I.e., the empty subset is the trivial subset, since it's a a subset of every set. Then the analogy would be two groups are disjoint exactly when their only common subgroups is the trivial subgroup (i.e., the group of just the identity element)." – Joshua Taylor Oct 7 '15 at 4:21

I don't think there is a common term in set theory for that.

But in the context of linear algebra, it is frequent to say that two subspaces are independent when their intersection is the zero space.

• Do you mean linearly independent? – Colin Michael Flaherty Oct 7 '15 at 1:19
• @Colin, linearly is left implicit. Just like we say "complementary subspaces" where "linearly complementary subspaces" might be the pedantic way. Or even "subspace" when "linear subspace" is meant. – lhf Oct 7 '15 at 1:22

Sets that have empty intersection are said to be disjoint. Subgroups of the same group, subspaces of the same vector space, and so on will never be disjoint, as they'll have at least one common element (the identity element of the group, the 0 vector, etc.) Sets with just one element are singletons, but it would be unusual to use that term about algebraic substructures. You can say that their intersection is trivial, or that it's the zero subgroup (in Abelian contexts), the zero subspace, etc. By a slight abuse of notation, people write this as, for example, $G \cap H = 0$, and $V \cap W = 0$.

• Yeah, exactly @ThomasAndrews. I'm specifically asking this because sets whose intersection is {0} seem like they could have some interesting properties that differ from disjoint sets and sets whose intersection has more than 1 element, or has one element that's nonzero. – Colin Michael Flaherty Oct 7 '15 at 1:16
• Exactly, that's the problem though. We have the neat word 'disjoint' when referring to sets, but when we are speaking about algebraic substructures we have no equivalently neat term to describe what is essentially the same thing. It seems ideally we'd use the term disjoint to refer to algebraic substructures whose intersection is the identity element, as a. substructures are essentially as disjoint as they could be in this case. But that would introduce confusion between the standard notion and the notion in the context of algebraic structures. – Colin Michael Flaherty Oct 7 '15 at 1:25
• "Trivial" and "zero" will work, and in the case of modules/vector spaces, "independent" works too. – BrianO Oct 7 '15 at 1:31