# Partition $Z_2^n \setminus \{0\}$ into disjoint sets. Does one contain a maximal subspace?

Let $n \geq 1$. $V=Z_2^n \setminus \{0\}$. $A \cup B = V$. $A \cap B = \Phi$.

Is it true that either A or B contains a sub-space of dimension $n-1$ (without the zero element)?

(I reduced a homework question to this one).

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Are you using $\Phi$ for $\varnothing$? Please don't ever do that again :-) – Myself Jan 16 '11 at 14:57

This is not true. For $n=4$ let $$A=\{(0,1,0,0), (0,0,1,0), (0,0,0,1), (1,1,0,0), (1,1,1,0), (1,1,0,1), (1,0,1,1), (0,1,1,1)\}$$ $$B=\{(1,0,0,0), (1,0,1,0), (1,0,0,1), (0,1,1,0), (0,1,0,1), (0,0,1,1), (1,1,1,1) \}$$ $B \cup \{0\}$ is not a subspace, and its size (with the zero vector) is $2^3$, so it doesn't contain a subspace of dimension 4-1=3. Notice that if you remove one vector from A, then it will not be a subspace (with the zero vector) so again $A\cup \{0\}$ doesn't contain a 3 dimensional subspace.