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.


marked as duplicate by user26857, happymath, hardmath, Chris Godsil, Joey Zou Aug 27 '16 at 16:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Well, the linked questions have answers with self-contained proofs for the infinite case. So what makes this question a non-duplicate? $\endgroup$ – Hagen von Eitzen Aug 31 '15 at 20:15
  • $\begingroup$ 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. $\endgroup$ – user152715 Aug 31 '15 at 20:20
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    $\begingroup$ If you are unhappy with the answers in the question you linked to, would the answers to this one be easier to follow? $\endgroup$ – Jyrki Lahtonen Aug 31 '15 at 20:31
  • $\begingroup$ @JyrkiLahtonen That's a really elegant proof for the claim by the way $\endgroup$ – Hagen von Eitzen Aug 31 '15 at 20:40

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.

  • $\begingroup$ Thanks thats why I was confused in mathoverflow.net/questions/26/… 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. $\endgroup$ – user152715 Aug 31 '15 at 20:48

$V$ 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 $$V=W_{n}$$

  • $\begingroup$ 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 $|F|\le n$) $\endgroup$ – Hagen von Eitzen Aug 31 '15 at 20:34
  • $\begingroup$ @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. $\endgroup$ – user118494 Aug 31 '15 at 20:36
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    $\begingroup$ 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) $\endgroup$ – Hagen von Eitzen Aug 31 '15 at 20:39
  • $\begingroup$ @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. $\endgroup$ – user118494 Aug 31 '15 at 20:43
  • $\begingroup$ For my result can you prove your claim that the chain in $W_i$ comes? $\endgroup$ – user152715 Aug 31 '15 at 20:49

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