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This is a broad question, so I will give a specific example from linear algebra, but I vaguely recall the general idea working in other contexts as well. The general idea is that given $x, y$ in a set we sometimes just have to check that $x - y$ is also in the set (or some other similar condition involving differences), then we get a bunch of nice properties or structure or satisfied conditions.

For example, in linear algebra, if $U$ is a subspace of $V$, and we define an affine subset to be a set of the form $v + U$ for some vector $v \in V$, then we have the following result: for any $v, w \in V$, the following are equivalent: (a) $v - w \in U$, (b) $v + U = w + U$, (c) $(v + U) \cap (w + U) \neq \emptyset$. My intuition for it is as follows:

If $v - w \in U$, this means whenever we have a $v + u$ we can “replace” $v$ by $w$, “up to a change of the subspace element”. We do this by writing $v + u = (v + u) + (w - w) = w + ((v-w) + u)$, which is an element of $w + U$. So to replace $v$ by $w$, we just change the subspace element, which is doable since $v - w \in U$. And vice versa, we can “replace” the $w$ in $w + u’$ by $v$ by replacing $u’$ with $-(v-w) + u’ \in U$. Hence $v + U = w + U$. We might also think of $v - w \in U$ as meaning that the “diagonal” between $v$ and $w$, which “connects” the two vectors, is in $U$. So, we can manipulate any expression of the form $v + u$ to get it to be $w + u’$ (and vice versa).

On the other hand, if $v + U = w + U$, then they have nonempty intersection, so $v + u_1 = w + u_2$ for some $u_1, u_2 \in U$. Hence $v - w = u_2 - u_1 \in U$. The way I might think about this is: because $v + U = w + U$, this forces upon us an equation/relation in terms of $v, w$ and some elements of the subspace (specifically, the equation $v + u_1 = w + u_2$). So we have no freedom/choice left for $v - w$, because rearranging this equation gives $v - w = u_2 - u_1 \in U$. $v - w$ has no choice but to live in the subspace.

Why is this idea of closure under differences important or desirable? Am I thinking about it “correctly”? I would appreciate any general comments, e.g. about how you think of it. Also, I know my example from linear algebra isn’t as general as possible but I don’t really know other examples.

Edit: I know my example wasn't necessarily about "closure under differences" but more so about just "differences", but I mean we still see how it can be important to consider differences. Sorry about that.

twosigma
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1 Answers1

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Saying that the nonempty set is closed under differences gives you two things. I am assuming that we are in a set up whose underlying structure is abelian group.

So, suppose $G$ is an abelian group. $H$ is a nonempty subset of $G$. Then $H$ is a Subgroup of $G$ if and only if $H$ is closed under subtraction.

Because, nonemptiness gives you the existence of additive identity $0$ [$a-a =0 $ where $a\in H$] , it gives The existence of inverse of any element in $H$ inside $H$ [$0-a=-a$ for any $a\in H$], it gives closed under addition [ $a-(-b)=a+b$ for $a,b\in H$]. So, nonempty together with closed under subtraction says that it is a Subgroup.

In general for any group a nonempty subset $H$ is a Subgroup if and only if $ab^{-1} \in H \ \forall a,b\in H$.

Similarly for vector spaces, modules you can modify the statement keeping in mind the closure under the external multiplication.

Now, for an ideal of a ring closure under subtraction is quite interesting. If the ring has identity, a nonempty subset is an ideal (by definition a subgroup of additive group of the ring $R$ which is closed under multiplication by ring elements) if and only if It is closed under addition and multiplication by ring elements.

But if the ring doesn’t contain unity then this condition do not suffice as that doesn’t guarantee closure under additive inverse. So either you put an extra condition saying that it is closed under additive inverses or you just say it is closed under subtraction as subgroup part follows from this condition solely.

Naba Kumar Bhattacharya
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