# There exists a smallest linear subspace that contains every $M\in\mathcal{M}$

This is Exercise 7 from Berberian (1992) Linear Algebra, p.23.

Let $$V$$ be a vector space and let $$\mathcal{M}$$ be any set of linear subspaces of $$V$$. Let A be the union of the subspaces in $$\mathcal{M}$$. Apply Exercise 6 to show that there exists a smallest linear subspace of $$V$$ that contains every $$M\in\mathcal{M}$$.

P.S. Here is Exercise 6. Let $$V$$ be a vector space and let $$A$$ be any subset of $$V$$. There exist linear subspaces of $$V$$ that contain $$A$$. Let $$\mathcal{N}=\{N:N \text{is a linear subspace of}\,V \text{containing}\,A \}.$$ Show that $$\mathcal{N}$$ contains a smallest element. Thus there exists a smallest linear subspace of $$V$$ containing $$A$$

Proof.

As far as I understood, no Span arguments can be used at this point since it is not yet introduced and the sets $$M\in\mathcal{M}$$ may not necessarily be finite.

Now, since the linear subspace $$M_1+M_2$$, for $$M_1,M_2\in\mathcal{M}$$ is the smallest linear subspace in $$V$$ that includes both $$M_1$$ and $$M_2$$, my conjecture is that the answer to this problem should be something like $$M_1+M_2+\ldots$$, but here is where I'm stuck. I'm not sure this sum is over a finite number of elements since we do not know if $$\mathcal{M}$$ is finite.

What am I missing?

• What you want to show is that the intersection of subspaces is again a subspace. Commented Aug 24, 2019 at 9:10
• Ok, now I see what you mean. Commented Aug 24, 2019 at 10:27

There is at least one subspace of $$V$$ that contains $$A$$: namely $$V$$ itself. Therefore we can take the intersection of all subspaces that contain $$A$$ and note that:
• It contains $$A$$, because every subspace being intersected contains $$A$$.
• It's contained in every subspace that contains $$A$$.