# Is the lattice generated by finitely many subspaces in a finite-dimensional vector space finite?

Let $$V$$ be a finite-dimensional vector space, let $$U_1,\dots,U_n$$ be subspaces, and let $$L$$ be the lattice they generate; namely, the smallest collection of subspaces containing the $$U_i$$ and closed under intersections and sums. Is $$L$$ finite?

This is well-known to hold if $$n\le3$$: if $$n=3$$ then there are at most $$28$$ elements in $$L$$, indpendently of $$V$$'s dimension.

Note that I suspect the answer to be "no" if $$V$$ is allowed to be infinite-dimensional: there exist infinite modular lattices generated by $$4$$ elements; here $$L$$ is a bit more than modular ("arguesian", see Is the free modular lattice linear?) and finiteness of free arguesian lattices doesn't seem to be known. It would be nice to have an example.

• Maybe mention (it seems implicit) that the answer is "yes" when $n\le 3$, regardless of finite dimension, because the free modular lattice on 3 generators is finite (see golem.ph.utexas.edu/category/2015/09/…), an old result of Dedekind. – YCor Sep 21 '18 at 8:04
• It's great we have a definite answer (grok's answer below, pointing to Todd Trimble's answer). I would not vote to close as duplicate, because the question is asked more clearly here than there (where one should have an intuition of what "being in general position" means). – YCor Sep 21 '18 at 8:26
• Note on the upper bound 28: the bound is rather 30 is one considers the modern definition of lattice, for which $\{0\}$ (as sup of the empty set) and $V$ (as inf of the empty set) and $V$ belongs to it. This is discussed in Baez's blog I pointed above, which also shows that this upper bound (28 or 30) is sharp. – YCor Sep 21 '18 at 8:45

The answer is in fact "no", and appears in another MO post, How many subspaces are generated by three or more subspaces in a Hilbert space? : starting from four points in $$P^2(\mathbb R)$$, infinitely many points may be generated by intersecting lines and joining points.