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Let $V=\bigcup_{i=1}^n W_i$ where $W_i$'s are subspaces of a vector space $V$ over an infinite field $F$. Show that $V=W_r$ for some $1 \leq r \leq n$.

I know the result "Let $W_1 \cup W_2$ is a subspace of a vector space $V$ iff $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$."

Now can I extend this to some $n$ subspaces.

I have some answers here & here.

So before someone put it as a duplicate I want to mention that I want a proof of this problem using basic facts which we use in proving the mentioned result.

Community
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Ri-Li
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  • Well, the linked questions have answers with self-contained proofs for the infinite case. So what makes this question a non-duplicate? – Hagen von Eitzen Aug 31 '15 at 20:15
  • In the first case the proof uses covering things which I don't know clearly. And in math overflow if you see "Steve D"s answer it uses basic techniques. But I have a question that how he gets the conclusion "so there is some $V_j, j≠1$, with infinitely many of these vectors, so it contains $y$, and thus contains $x$." I can't ask him in a comment because I am not a member in mathoverflow and can't wait for the answer untill $50$ reputations. – Ri-Li Aug 31 '15 at 20:20
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    If you are unhappy with the answers in the question you linked to, would the answers to this one be easier to follow? – Jyrki Lahtonen Aug 31 '15 at 20:31
  • @JyrkiLahtonen That's a really elegant proof for the claim by the way – Hagen von Eitzen Aug 31 '15 at 20:40

2 Answers2

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The general result is that if $\lvert K\rvert\ge n$, $V$ cannot be the union of $n$ proper subspaces (Avoidance lemma for vector spaces).

We'll prove that if $V=\displaystyle\bigcup_{i=1}^n W_i$ and the $W_i$s are proper subspaces of $V$, then $\;\lvert K\rvert\le n-1$.

We can suppose no subspace is contained in the union of the others.

Pick $u\in W_1\smallsetminus\displaystyle\bigcup_{i\neq1} W_i$, and $v\notin W_1$. The set $v+Ku$ is disjoint from $W_1$, and it intersects each $W_i\enspace(i>1)$ in at most $1$ point (otherwise $u$ would belong to $W_i$). As this set is in bijection with $K$, there results that $K$ has at most $n-1$ elements.

Bernard
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  • Thanks thats why I was confused in http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces that why in Steve D"s answer "there is some $V_j,j≠1$, with infinitely many of these vectors, so it contains $y$, and thus contains $x$." Is true because every $V_j$ should contain at most $1$ element. – Ri-Li Aug 31 '15 at 20:48
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$V$ is a vector space over an infinite field. If $\ \ $ $\cup_{i=1}^{n} W_{i}=V$ is a vector space itself then from what you already know we can write $$W_{1}\subseteq W_{2}\subseteq W_{3}........\subseteq W_{n}$$ possibly with a re-ordering of the subspaces $W_{i}$'s .

Then what do we see here? $W_{n}$ containing the other subspaces of $V$. So intuitively we could suspect that $V=W_{n}$ might be the thing .

Let's prove it. $$W_{n}\subseteq V$$ is known. For the reverse inclusion ; if possible let $$V\neq W_{n}$$ let $w$ be a vector such that $$w\in V \ \ but\ \ w\notin W_{n}$$ . Now $$\cup_{i=1}^{n} W_{i}=V$$ implies $w$ is in $W_{i}$ for some $i=1,2...,n$. But $$W_{i}\subseteq W_{n}$$ for all $i=1,2,....n$ . Hence $$w\in W_{n}$$ . A contradiction. Thus $$V\subseteq W_{n}$$ must be true. Hence proved $$V=W_{n}$$

user118494
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  • Where do you use that the ground field is infinite? If you don't your proof must be wrong (because the result is wrong if $|F|\le n$) – Hagen von Eitzen Aug 31 '15 at 20:34
  • @HagenvonEitzen : Apologies, Sir. Actually read the previous comment of the OP and had infinite field in mind. Should have wrote it, I agree. I have added one line to my answer . If there is still some mistake , please point out. – user118494 Aug 31 '15 at 20:36
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    Yes, now you mention that $F$ should be infinite. But where do you use it? More to the point, how do you justify the "from what you already know" (it can't be justified) – Hagen von Eitzen Aug 31 '15 at 20:39
  • @HagenvonEitzen :I meant this "Let $W_1 \cup W_2$ is a subspace of a vector space $V$ iff $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$." by " from what you already know" . And yes , this also requires an infinite field to be true Actually OP's question was wholly $\mathcal over $ an infinite field, I thought. – user118494 Aug 31 '15 at 20:43
  • For my result can you prove your claim that the chain in $W_i$ comes? – Ri-Li Aug 31 '15 at 20:49
  • @user152715 : Sorry , I don't understand , which part exactly are you referring to? – user118494 Aug 31 '15 at 20:51
  • Just proof how do you get $$W_{1}\subseteq W_{2}\subseteq W_{3}........\subseteq W_{n}$$? – Ri-Li Aug 31 '15 at 20:54
  • @user118494 "Let $W_1\cup W_2$ is a subspace of a vector space $V$ iff $W_1\subseteq W_2$ or $W_2\subseteq W_1$." happens to be true alo over finite fields and at any rate cannot readily be extended to $n>2$. Henc $W_1\subseteq \ldots \subseteq W_n$ is not something we already know – Hagen von Eitzen Sep 01 '15 at 06:24