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If $V$ is a $K$-vector space with $|K|\ge n$ and $V_1, \ldots, V_n$ are subspaces of $V$ such that $V=V_1 \cup \cdots \cup V_n$, then $V=V_i$ for some $i$. Is there a counterexample that for $|K|< n$ it doesn't hold ?

If $K$ has characteristic zero, it follows directly from induction that one of the subspaces has to be equal to $V$. Some hints are appreciated. Thanks

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Hint: Draw a picture of $K^2$ where $K$ is the field of $2$ elements. Then try to write it as a union of three 1-dimensional subspaces.

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Might be hard to find 4 distinct 1-d subspaces... :) –  Erick Wong Nov 18 '12 at 21:02
    
@ Thats just the combination of $(0,0)$ with three points respectively . $|K|=2$ but $K^2$ can be written as union of these four 1-d vector spaces , is that right ? but when $|K|\ge n$ what actually happens we don't get this property ? –  Theorem Nov 18 '12 at 21:07
    
@Theorem To be fair, what you're asking about was not really posed as a question above. –  Erick Wong Nov 18 '12 at 21:10
    
@ErickWong : I know but i realized that i just thought of getting more information on the problem . I apologize if that breaks the norms of MSE. –  Theorem Nov 18 '12 at 21:13
    
@ErickWong: Thanks adjusted to 3, did not think long enough before posting. –  Julian Kuelshammer Nov 18 '12 at 21:16

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