Given a finite number of vector or affine subspaces with dimension less than $n$( $n$ being the total space dimension), we need to find a vector that doesn´t belong to any of the subspaces . How can we prove this is possible?
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You need to assume the field $K$ of scalars is infinite, otherwise counterexamples are easily obtained.
Also, the comparison to $n-1$ need not be strict.
Restating the problem so as to include those revisions . . .
Claim:
If $V$ is a vector space of dimension $n > 0$, over an infinite field $K$, then $V$ cannot be expressed as the union of finitely many proper affine subspaces.
Note:
Once the claim is proved, it follows immediately that for any finite collection $S$ of proper affine subspaces of $V$, there is some $v\in V$ such that for all $A\in S$, we have $v\notin A$.
Proof of the claim:
Proceed by induction on $n$.
Since $K$ is infinite, so is $V$, hence $V$ is not the union of finitely many singletons.
It follows that the claim holds for $n=1$.
Suppose the claim holds for some fixed positive integer $n \ge 1$.
Let $V$ be a vector space over $K$ of dimension $n+1$, and suppose $V$ is the union of finitely many proper affine subspaces, $A_1,...,A_m$.
Our goals is to derive a contradiction.
Note that $V$ has infinitely many subspaces of dimension $n$, so we can choose one of them, $W$ say, such that $W$ is not equal to any of $A_1,...,A_m$.
For $1 \le k \le m$, let $B_k = A_k\cap W$.
Since $W$ is not equal to any of $A_1,...,A_m$, it follows that each nonempty $B_k$ is a proper affine subspace of $W$.
Then \begin{align*} W &= W\cap V\\[4pt] &=W\cap \left({\small{\bigcup_{k=1}^m}} A_k\right)\\[4pt] &= {\small{\bigcup_{k=1}^m}} \left(A_k\cap W\right)\\[4pt] &= {\small{\bigcup_{k=1}^m}} \,B_k\\[4pt] \end{align*} contrary to the inductive hypothesis.
This completes the proof.
quasi
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The dimension is $n$ thus exists a basis of n vectors.
For the basis we can choose the $n-1$ vectors basis $v_i$ of the subspaces and another vector $w$ linearly independent from the others, that is
$$\sum a_kv_k +a_nw=0\iff a_i=0$$
therefore $w$ doesn't belong to the subspaces.
user
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