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Let be $ V $ vector space :

a) If we have an infinite chain of subspaces $ {U}_{1}\mathrm{\subseteq}{U}_{2}\mathrm{\subseteq} $

Show that $ {U}\mathrm{{=}}\mathop{\mathrm{\cup}}\limits_{{k}\mathrm{{=}}{1}}\limits^{\mathrm{\infty}}{U}_{K} $ is a subspace .

b) if $ V $ is finite dimensional show that there is a some subspace ,call it $ {U}_{P} $ ,such that $ {U}\mathrm{{=}}{U}_{p} $ and hence $ {U}_{P}\mathrm{{=}}{U}_{{P}\mathrm{{+}}{1}}\mathrm{{=}}{\mathrm{....}} $

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    Any ideas how to solve it ? – 2017-01-06
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    Really I am just thinking about the addition and the scalar multiplication – 2017-01-06
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    We proved before that the union of M and N is a subspace if and only if : M $ \mathrm{\subset} $ N or N $ \mathrm{\subset} $ M. – 2017-01-06
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    Can we proved that or depend on that ? – 2017-01-06

2 Answers 2

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a)

Take any $x\in U$ and any scalar $\alpha$. Then, prove that $\alpha\cdot x$ is also in $U$.

Take $x,y\in U$ and prove that $x+y\in U$.

Hint:

  1. $x\in U\iff \exists i\in\mathbb N: x\in U_i$
  2. If $x\in U_j$, then $x\in U_k$ for all $k\geq j$.

b)

Hint:

If $A\subseteq B$ is a subspace and $\dim(A)=\dim(B)\in\mathbb N$, then $A=B$.

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    Thanks a lot really i just want to ask more about the second part of the question that the basis of the subspace cant be bigger than the space – 2017-01-06
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    @user401187 I don't understand what you are trying to say. – 2017-01-06
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    Sorry I think that I didn't get it . – 2017-01-06
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    I didn't understand the second part of the answer – 2017-01-06
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    @user401187 It's a hint, not an answer. The second hint, maybe, is to say try to prove the statement by contradiction. – 2017-01-06
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    Can you help .me more – 2017-01-06
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    I just think the second part of the question is clear but I can't express that .can you tell me how is the hint meaning ? – 2017-01-10
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Your title seems a bit different than the question you ask. In general a union of subspaces is not a subspace. For instance, take the x axis and then the y axis. Each is a 1 dimensional subspace, but the union is not closed under addition for instance.

a) These are chains of subspaces, so the union is easy to show closure. Let $x,y \in U$, then $x \in U_i, y \in U_j$ for some $j \leq i$. As $U_j \subseteq U_i,\ y \in U_i$. Then, $U_i$ is a subspace, so $x+y \in U_i$ and $\lambda x \in U_i$, so they are both in U.

b) Track the basis of the vector space. Eventually you will get all of them in that union, so it will terminate.

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    Can you help me more for the second part – 2017-01-06