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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?

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

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You're better off using the fact that the intersection of linear subspaces is a subspace. (Make sure you know how to prove this)

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 is a subspace, as the intersection of a collection of subspaces.
  • It contains $A$, because every subspace being intersected contains $A$.
  • It's contained in every subspace that contains $A$.

This pretty much works as a definition for the sum of subspaces.

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