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Let $V$ be a vector space and suppose that $\lbrace U_n : n \in \Bbb N\rbrace$ are subspaces of $V$. Show that for every $k,m \in \Bbb N$ there exists an $n \in \Bbb N$ such that $U_k \cup U_m \subseteq U_n$ then $\bigcup^\infty_{j=1}U_j $ is a subspace of $V$.

I understand that I need to prove that the infinite union satisfies the properties of a subspace, But how does the given assumption aid in this proof? As I see it, $\bigcup^\infty_{j=1}U_j = U_1 \cup U_2 \cup U_3 \cup ...$ So applying $U_k \cup U_m \subseteq U_n$ wouldn't give any insight as it is still an infinite union.

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    $\begingroup$ Hint: the assumption comes into play when proving the union is closed under addition. So, suppose $x, y \in \bigcup_{j=1}^\infty U_j$; then $x \in U_k$ and $y \in U_m$ for some $k, m$... $\endgroup$ – Daniel Schepler Jun 5 '17 at 15:16
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If $v,w\in V$, then $v\in U_k$ and $w\in U_n$, for some $k,n\in\mathbb N$. Take $m\in\mathbb N$ such that $U_k\cup U_n\subset U_m$. Then $v,w\in U_m$ and therefore $v+w\in U_m\subset V$.

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To see why some assumption on the collection is needed, consider the case when $V=\mathbb{R}^2$ and $U_n = \operatorname{span}\{(1,n)\}$ (i.e. the line through $\vec{0}$ of slope $n$).

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