0

Let us say we have $\textsf W$ as a vector space over some field $F$. Let $U_α$ be a collection of subspaces of $\textsf W$. Now is $\bigcup_αU_α$ a subspace of $\textsf W$?

I know that union of two subspaces is a subspace if and only if one is included in another and the proof. For a collection of subspaces, I believe the condition have to be the same with that the collection of subspaces have to form a chain of subsets. Is there a more rigorous proof?

I cannot seem to find one anywhere that proves union a finite collection of subspaces.

azif00
  • 20,792
james black
  • 1,893
  • For a finite collection of subspaces over an infinite field the claim is true. For a finite collection of spaces over a finite field, the claim is NOT true. For infinite collections it is not true either. – N. S. Sep 13 '19 at 04:30
  • The union of an infinite chain of subspaces is not a subspace. – Kavi Rama Murthy Sep 13 '19 at 05:38
  • If you're looking to define supremum for subspaces, try using summing instead of taking unions. – AlvinL Sep 13 '19 at 05:53
  • In the finite case, you can find the top element of your chain. You can't always do the same with an infinite family. – AlvinL Sep 13 '19 at 05:58
  • You don't need a chain; it's enough for one of the $U_{\alpha}$ to contain the others. – Qiaochu Yuan Sep 13 '19 at 06:27
  • Related: https://mathoverflow.net/questions/43538/wonderful-applications-of-the-vandermonde-determinant and https://math.stackexchange.com/questions/2083837/when-is-the-union-of-a-family-of-subspaces-of-a-vector-space-also-a-subspace and https://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces and many others. – Gerry Myerson Sep 13 '19 at 06:48
  • Just to point out that an infinite family of subspaces doesn't necessarily need to be a chain for the union to be a subspace of W. For instance, $\bigcup_{w\in W} Fw = W$. –  Sep 13 '19 at 10:05

1 Answers1

1

If the subspaces form a chain, then the union is a subspace as well.

Let $(U_i)_{i \in I}$ be a chain of subspaces of $W$. That is for $i,j\in I$ we have $U_i\subset U_j$ or $U_j\subset U_i$.

The claim is that $U:=\cup_{i\in I}U_i$ is a subspace as well.

Let $u,v\in U$. Then there are $i,j$ such that $u\in U_i$, $v\in U_j$. Due to the chain property, $U_i \subset U_j$ or $U_j\subset U_i$. In any case, $u+v$ is in $U_i$ or in $U_j$. Hence $u+v\in U$.

Take $u\in U$, $t\in F$. Then $u\in U_i$ for some $i\in I$, $tu\in U_i$, and $tu\in U$ follows.

The converse result is not true: Let $F$ be a finite field. Then $F^n$ is a finite set, $F^n = \{0, v_1, \dots v_m\}$. Then we can write $F=\cup_i \{0,v_i\}$, but the subspaces $\{0,v_i\}$ do not form a chain. Also, none of the subspaces is equal to the union.

daw
  • 49,113
  • 2
  • 38
  • 76