MHB A fundamental fact about Linear Algebra

AI Thread Summary
In the discussion, a proof is presented that a vector space \( V \) over an infinite field \( F \) cannot be expressed as a set-theoretic union of a finite number of proper subspaces. The proof assumes the contrary and uses the concept of linear combinations to show that the resulting set must be infinite, leading to a contradiction. The author emphasizes the importance of allowing sufficient time for responses, suggesting a week for participation. The conversation highlights the challenge of engaging others in mathematical discussions. Overall, the proof effectively demonstrates the fundamental property of vector spaces in linear algebra.
caffeinemachine
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Hello MHB,
This is probably my first challenge problem which falls in the 'University Math' category.

$V$ is a vector space over an infinite field $F$, prove that $V$ cannot be written as a set theoretic union of a finite number of proper subspaces.
 
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Nobody participated :(

Here's my solution:

Assume contradictory to the problem. Let $n$ be the minimum integer such that $V$ can be written as $V=V_1\cup\cdots\cup V_n$ where each $V_i$ is a proper subspace of $V$. Thus, \begin{equation*}\forall i,\exists x_i\in V \text{ such that } x_i\in V_j\iff j=i\tag{1}\end{equation*}Now consider $S=\{f_1x_1+\cdots+f_nx_n:f_i\in F\}$. Clearly this set is infinite, thus, by PHP, there is a $k$ such that $a,b\in V_k$ for distinct $a$ and $b$ in $S$. This contradicts $(1)$. Hence we achieve the required contradiction and the proof is complete.
 
caffeinemachine said:
Nobody participated :(

I would give everyone at least a week to participate, since not everyone checks in on a daily basis. Some may only have time once a week.
 
MarkFL said:
I would give everyone at least a week to participate, since not everyone checks in on a daily basis. Some may only have time once a week.
My bad then. Next time I'll wait for a week.
 
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