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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 .

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  • $\begingroup$ what is $k$ in you statement? $\endgroup$ – fonfonx Jun 3 '17 at 21:11
  • $\begingroup$ Did you ever hear of direct limits? $\endgroup$ – Bernard Jun 3 '17 at 21:13
  • $\begingroup$ 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}$? $\endgroup$ – José Carlos Santos Jun 3 '17 at 21:14
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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 $V$.

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You meant to say for every k and m there is an n . Look if x is in U_k and if y is in U_m you are assuming there is n so that both x and y are in U_n (a subspace ) so ... This should do it for you ,

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