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Show that if $V$ is a nonzero vector space over $\mathbb{R}$, and $V_{1}, V_{2},\ldots V_{k}$ are proper subspaces of $V$ then there exists $v\in V$ such that $v\not\in V_{i}$ for any $1\leq i\leq k$

I can prove the case for $k=2$ but I cannot produce an induction argument.

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    It can certainly happen that $V_i \neq V$ for all $i$, but $\bigcup V_i = V$, right? For example, if you take $V = \mathbb R^2$, and $V_1 = \overline{\{(0,1)\}}$, and $V_2 = \overline{\{(1,0)\}}$, then clearly no vector can belong to neither $V_i$. This hints that the question could be wrong. – 2017-02-20
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    Yes i know that but it doesn't help proving the original statement does it? – 2017-02-20
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    I think it shows that the question is incorrect. – 2017-02-20
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    $(1,1)$ does not belong to $span\{(0,1)\}$, nor does it belong to $span\{(1,0)\}$ – 2017-02-20
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    @астонвіллаолофмэллбэрг: No, not right. It can happen that $\bigoplus V_i=V$, but not $\bigcup V_i=V$. Say, in your example, $v=(1,1)$ works - it's neither in $V_1$ not in $V_2$. – 2017-02-20
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    Ah, i See. thank you. – 2017-02-20
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    @астонвіллаолофмэллбэрг;here is the answer – 2017-02-20
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    @learnmore I see the difference, the proof is also very nice. Thank you for informing me. – 2017-02-20
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    My pleasure@астонвіллаолофмэллбэрг – 2017-02-20
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    @learnmore I will point out that subspaces tag has been discussed on meta previously and it has been removed (even blacklisting was discussed): See http://meta.math.stackexchange.com/questions/21345/what-to-do-with-the-subspaces-tag/21374#21374 and http://chat.stackexchange.com/rooms/20352/conversation/removal-of-subspaces-tag – 2017-02-21
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    Okay ;thank you very much @MartinSleziak; – 2017-02-22

2 Answers 2

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Let $V=V_1\cup V_2\cup \cdots \cup V_n$ where $V_1\not\subset V_2\cup V_3\cup \cdots \cup V_n$.

Let $v\in V_1\setminus( V_2\cup V_3\cup \cdots \cup V_n)$.

Also since $V_1$ is proper so $\exists w\in V\setminus V_1$.

Consider $P=\{w+rv:r\in \Bbb R\}$ which is definitely an infinite subspace of $V$.

Claim: $V_i$ contains at-most one member of $P$ for each $1\le i\le n$.

If $w+rv\in V_1$ for some $r\in \Bbb R\implies w\in V_1$ which is false.

If $|V_i|>1$ and for $r_1\neq r_2$$;w+r_1v,w+r_2v\in V_i;i\ge 2\implies (r_1-r_2)v\in V_i\implies v\in V_i$ which is false and hence $|V_i|<1$.

Hence $V_i$ contains at-most one member of $P$ for each $1\le i\le n\implies V$ contains finite number of members of $P$ which is false since $P$ is an infinite subspace of $V$.

Hence $V\neq V_1\cup V_2\cup \cdots \cup V_n$ ,hence there exists $v\in V$ such that $v\notin V_i$ for any $i$.

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    I'm lost already at "$V=V_1\cup V_2\cup \cdots \cup V_n$" -- the whole question is to prove that this isn't the case. – 2017-02-20
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    You did not go through the proof either..and you are busy downvoting@zipirovich – 2017-02-20
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    It's a proof by contradiction. The first line is to assume the contrary for the sake of contradiction – 2017-02-20
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    Oh, so the first word must be "assume", not "let"?Then it makes more sense. But what about "$V_1\not\subset V_2\cup V_3\cup \cdots \cup V_n$"? Is it part of the assumption to be contradicted? Or is it a claim that's supposed to be true? If the latter, then on what grounds? Say, what if $V=\mathbb{R}^2$, $V_1=V_2=\operatorname{span}\{(1,1)\}$? Actually, I think your proof probably works, but the exposition is way too sloppy. – 2017-02-20
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    1. Huge difference. "Let" defines an entity, and you can't contradict it later on. 2. That's why I said that your proof is probably correct -- it's nice in content, but very sloppy in exposition. You can't just assume arbitrary things that are not given in the original statement. What you said in the comment should've been in the proof -- in fact, it had to be there, and better with some justification. – 2017-02-20
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A proof using some topology in the finite dimensional case (in the general case and using only linear algebra, see learnmore's answer) :

If each of the $V_i$ is proper then they're all closed sets with empty interior, which, according to the Baire Category theorem implies that their union also has empty interior : therefore it isn't $V$.