# The union of linear subspaces is not necessarily a subspace

$W$ and $U$ are subspaces of vector space $V$. Show that $U\cup W$ is not necessarily a subspace of $V$.

I know that one counter-example is enough to show that. What example could fit here?

• Of course it is a subspace. You cannot get out of a vector space by addition – Mister Benjamin Dover Jan 4 '15 at 15:18
• Read the problem statement again. Could it be about $U\cup W$? – Hagen von Eitzen Jan 4 '15 at 15:19
• Yes I meant U∪W. – David Jan 4 '15 at 15:19
• Then take any distinct one-dimensional subspaces of a vector space of dimension $>1$ – Hagen von Eitzen Jan 4 '15 at 15:21
• If $U \cap W = 0$ (i.e the zero vector), and $u \in U$ and $w \in W$ is $u+w \in U \cup W$ ? – Tom Collinge Jan 4 '15 at 15:24

Let $V=\mathbb{R}^2$, then it is clear that $W:=\mathbb{R}\times\{0\}$ and $U:=\{0\}\otimes\mathbb{R}$ are subspaces. But, their union is not closed under addition (for example $(1,0)+(0,1) = (1,1)\notin U\cup W$)). Therefore, their union is not a vector space, so cannot be a subspace (although it is a subset).

• The only thing I didn't get is why (1,1)∉U∪W ? – David Jan 4 '15 at 15:37
• If $(1,1)\in U \cup W$, then EITHER $(1,1) \in U$ OR $(1,1) \in W$ (this is just the definition of union). Points in $U$ are all of the form $(0,x)$ with $x\in \mathbb{R}$ and points in $W$ are all of the form $(x,0)$ with $x\in \mathbb{R}$ but $(1,1)$ is not of this form and hence cannot lie in the union. – user180850 Jan 4 '15 at 15:43
• I guess I just forgot the definition of union... Thanks a lot for clearing this up, it's such a small thing but without it it's impossible to understand. – David Jan 4 '15 at 15:48
• I guess $U$ should be defined by $U:=\{0\}\times\mathbb{R}$ instead of $\{0\}\otimes\mathbb{R}$ – Adren Jan 12 '17 at 15:12

Hint: look at the coordinate axes in $\mathbb{R}^2$.

Hint: Show that if $U,W$ are subspaces of $V$ and $U\cup W$ is also a subspace, then $U\subseteq W$ or $W\subseteq U$.

Suppose neither $U \subset W$ nor $W \subset U$. Take $u\in U\setminus W$, $w\in W\setminus U$. If $\,U\cup W$ were a subspace of $V$, it would contain $u+w$. The ask yourself: ‘Where does $u+w$ live?’.

Note 1. This can generalised: If a subspace $V$ of some vector space over an infinite field is contained in the union of a finite number of subspaces, it is contained in one of them. (Avoidance lemma for vector spaces).

Note 2. There is an an analog result for groups: a group cannot be the union of two (strict) subgroups However, examples are known of groups that are the union of 3 subgroups.