Can one turn finite-dimensional vector subspaces into a cancellative semigroup? Let $V$ be a vector space over some field and let ${\rm Fin}\,V$ be the family of all finite-dimensional subspaces of $V$. Is it possible to turn ${\rm Fin}\,V$ into an commutative cancellative semigroup $(S,+)$ such that $F,G\subseteq F+G$ for all $F,G\in S$.
Note that here $+$ is not the familiar Minkowski sum.
 A: If $V$ is finite-dimensional, Wojowu explained that the answer is no. If $V$ is infinite-dimensional, we'll show that the answer is yes, and we can even have it be a monoid with the zero-dimensional space $0$ as the identity.

Let $S$ be the set of finite-dimensional subspaces of $V$, excluding the zero-dimensional space $0$. Let $\prec$ be a well ordering on $S$ whose order type is the initial ordinal of the cardinality of $S$.
Let $T_n$ be the set of finite multisets of at least $n$ elements whose elements lie in $S$. We well-order $T_2$ by 'colexicographical ordering':


*

*If $A \subsetneq B$, then $A < B$;

*If neither $A$ nor $B$ are subsets of each other, then let $a := \max(A \setminus B)$ and $b := \max(B \setminus A)$. Then $A < B$ if and only if $a \prec b$.


Then $T_2$ is also well-ordered by the initial ordinal of the cardinality of $S$.
Now, we construct a map $f : T_2 \rightarrow S$ by transfinite induction:
Given a partial function $g : T_2 \rightarrow S$ (initially the empty partial function), let $A$ be the first element of $T_2$ (in its well-ordering) such that:


*

*$A$ is not already in the domain of $g$;

*no element of $A$ is in the image of $g$.


Then we let $g'$ be the partial function which extends $g$ by setting $g'(A)$ to be an arbitrary element $W \in S$ such that:


*

*For all $B \subsetneq A$ with $|B| \geq 2$, $g(B)$ is a subspace of $W$ (note that the colexicographical ordering on $T_2$ ensures that we've already defined $g$ on all such subsets $B$);

*For all $U \in A$, $U$ is a subspace of $W$;

*$W$ occurs later in the lexicographical ordering on $S$ than any element in $A$ or in the image of $g$.


Such an element $W$ necessarily exists because the first two conditions only narrow the set of candidate spaces to a set of equal cardinality to $S$ (this is where we require that $V$ is infinite-dimensional), and the third condition removes strictly fewer elements from the set of candidates (this is where we require the well-ordering to be with the initial ordinal [of that cardinality], rather than an arbitrary ordinal) so the set of candidates is non-empty.
Proceed inductively, and let $f$ be the limit of this process.
$f$ is defined on all finite multisets of $\geq 2$ elements from $S \setminus \textrm{Image}(f)$. It's injective, and when $A \subseteq B$, $f(A)$ is a subspace of $f(B)$.

We now define a function $h : S \rightarrow T_1$ as follows:


*

*If $U$ is in the image of $f$, then $h(U)$ is its preimage;

*otherwise, $h(U)$ is the singleton set containing $V$.


We can think of $h$ as 'decomposing' a vector space into its 'prime factor' vector spaces, and $f$ as 'multiplying' a multiset of 'prime factors' back into a vector space. Given that, we can now define:
$U + W := f(h(U) \sqcup h(W))$
where $\sqcup : T_1 \times T_1 \rightarrow T_2$ is the disjoint union operation on multisets. Then, it's easy to see that:
$U + W + X = f(h(U) \sqcup h(W) \sqcup h(X))$
for all spaces $U,W,X$, so we indeed have a commutative semigroup. It's also cancellative, because $f$ is injective, $h$ is injective, and $\sqcup$ is cancellative.

We can then throw in the zero-dimensional space $0$ as an identity element, obtaining a monoid.
