# Let $\ V \$ be a vector space and suppose that $\{U_{n}: n \in \mathbb{N} \}$ are subspaces of $\ V \$

Let $\ V \$ be a vector space and suppose that $\{U_{n}: n \in \mathbb{N} \}$ are subspaces of $\ V \$ . Then prove that , if for every $\ k,\ m\ \in \mathbb{N}$ there exists an $\ n \in \mathbb{N} \$ such that $\ U_{k} \cup U_{m} \subseteq U_{n}$ then $\ \bigcup_{n=1}^{\infty}U_{n} \$ is a subspace of $\ V \$.  I know that union of two subspaces is a subspace iff one is contained in another. I want to use this property but I am not sure how to prove the result ? Any help is appreciating .

• what is $k$ in you statement? – fonfonx Jun 3 '17 at 21:11
• Did you ever hear of direct limits? – Bernard Jun 3 '17 at 21:13
• Shouldn't the condition be that for every $m, n \in \mathbb{N}$ there exists a $k \in \mathbb{N}$ such that $U_{m} \cup U_{m} \subseteq U_{m}$? – José Carlos Santos Jun 3 '17 at 21:14

Clearly, $0\in \ \bigcup_{n=1}^{\infty}U_{n} \$
If $x\in \ \bigcup_{n=1}^{\infty}U_{n} \$
Then $x\in U_p$ for some $p\in \mathbb{N}$, so $\alpha x\in U_p$ which implies $\alpha x \in \ \bigcup_{n=1}^{\infty}U_{n} \$.
Also if $x,y\in \ \bigcup_{n=1}^{\infty}U_{n} \$, then $x\in U_k$ and $y\in U_m$ for some natural numbers $k,m$ hence they belong to a subspace $U_n$ for some integer $n$ (By the conditions given in the question,) which implies $x+y \in U_n$, so $x+y \in \ \bigcup_{n=1}^{\infty}U_{n} \$.
Hence $\ \bigcup_{n=1}^{\infty}U_{n} \$ is clearly a subspace of $V$.