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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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    @Theorem What goes wrong roughly is that there are too many hyperplanes. (Write down the plane for the field with three elements to see it.) I would prove this first for the case that the $V_i$ are hyperplanes and then expand to the general case. (Don't know if this is the most effective way to do it.)2012-11-19