0

Let $U,W$ two subspaces in $V$. Assuming that $U\cup W = V$, show that: $U=V$ or $W=V$.

So, by definition of union:
$$U \cup W = \left\{ {u + w|u \in U,w \in W} \right\}$$

  • How to proceed?
  • And a technical issue: Is $U \cup W$ equivalent to $U + W$?

EDIT:

Consider the following scenario (Following @Learner approach):
$U = V \backslash \{x\}$ and $W = V \backslash \{y\}$.

$U\cup W = V$, but none of the subspaces is equal to $V$.

Where is the error?

SuperStamp
  • 993
  • 1
    The definition of union should read $U \cup W = {v| v \in U \text{ or } v \in W}$. One important point of the exercise is precisely that $U \cup W$ is not equivalent to $U + W$: notice that if $V$ is a space with inner product, $U + U^\perp = V$ – Pedro M. Feb 18 '14 at 11:22
  • 2
    Your definition of $U \cup W$ is wrong; in fact $U \cup W := {v \in V | v \in U \text{ or } v \in W}$. What you define is $U + W$. – Wolfgang Spindeler Feb 18 '14 at 11:24
  • Maybe, But I have a specific question which stayed open. @nik – SuperStamp Feb 18 '14 at 12:44
  • 1
    @SuperStamp: $U$ and $V$ aren't vector subspaces in your example. – Najib Idrissi Feb 18 '14 at 12:49

2 Answers2

3

You can show more generally that $U\cup W$ can only be a vector space if one of $U,W$ is contained in the other, in which case the union is of course the latter (larger) of the two subspaces. To prove this, suppose to the contrary that neither of $U,W$ is contained in the other; choose $u\in U\setminus W$ and $w\in W\setminus U$, and prove that $u+w\notin U\cup W$ for a contradiction.

Marc van Leeuwen
  • 115,048
1

Suppose $U \subsetneq V$ and $W \subsetneq V$, then $U \cup W \subsetneq V$.

Pedro M.
  • 3,061
  • 18
  • 19