# Return to Revisions

2 of 2 improved formatting

Here's a reduction to the finite dimensional case. Let $$F$$ be a finite set of subspaces of $$X$$. For each finite dimensional subspace $$Y$$ of $$X$$, let $$u(Y)$$ be the set of elements $$Z$$ of $$F$$ such that $$Y$$ is contained in $$Z$$. By assumption, $$u(Y)$$ is non-empty for every $$Y$$. Since any two finite dimensional subspaces are contained in a third, the intersection of the sets $$u(Y)$$, as $$Y$$ runs among all finite dimensional subspaces of $$X$$, is non-empty. Hence there is at least one set in $$F$$ that contains every finite dimensional subspace of $$X$$, hence contains $$X$$.

For the finite dimensional case, let $$F$$ be a finite set of subspaces of $$X$$. By induction, every codimension 1 subspace of $$X$$ is contained in some $$Y$$ from $$F$$. But there are infinitely many codimension $$1$$ subspaces, so some $$Y$$ in $$F$$ contains more than one such subspace. Any two distinct codimension 1 subspaces $$\operatorname{span} X$$ (if $$\dim X > 1$$) so $$Y = X$$.

 answered Oct 1 '09 at 4:03 Jonathan Wise 5,65411 gold badge3232 silver badges4848 bronze badges