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Let $k$ be an infinite field.

Let $V$ be a vector space over $k$ and $W_1,...,W_r$ proper subspaces of $V$.

Show that $\bigcup_{i=1}^r W_i \not = V.$

I tried the following:

for all $j \in \{1,...,r\}$, I take $w_j \in W_j$ such that $w_j \not\in W_i$ whenever $j \not=i$, so I know that $w_1+\cdots+w_r \in V$. If $w_1+\cdots+w_r \in \bigcup_{i=1}^r W_i$, then there is $l \in \{1,...,r\}$ such that $w_1+\cdots+w_r \in W_l$. I don't find because $w_1+\cdots+w_r \in W_l$ is absurd.

Is this correct reasoning, or is there other way for me to prove this?

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    $\begingroup$ Your idea isn't entirely the right one: Consider what would happen if, say, $W_r$ contains all the other $W_i$. Then $\bigcup W_i=W_r$ is actually a vector space. Also, are you absolutely certain that $k$ is supposed to be finite? Because for finite $k$, and finite dimensional $V$, this just isn't true. $\endgroup$ – Arthur Feb 4 at 17:10
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The assertion seems to be false. Take $k=\mathbb{F}_2$ and $V=k\oplus k=\{(0,0),(1,0),(0,1),(1,1)\}$. Now you can take $r=3$ and the following proper subspaces of $V$: $W_1=\{(0,0),(1,0)\}$, $W_2=\{(0,0),(1,1)\}$, and $W_3=\{(0,0),(1,1)\}$.

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Here is a counterexample due to Jean-Pierre Merx.

Hypothesis: Given a vector space $V$ over a finite field $k$. Define the proper subspaces $W_1,...,W_r \subset V$. Then $\bigcup_{i=1}^{r} W_i \neq V$.

Suppose that our hypothesis is true, and consider a vector space $V$ over the finite field $\mathbb Z_2$, with a canonical basis $(e_1, e_2)$, and define the following: $$W_1 = \mathbb Z_2 \cdot e_1, \space W_2 = \mathbb Z_2 \cdot e_2, \space and \space W_3 = \mathbb Z_2 \cdot (e_1 +e_2) \\$$

Clearly, $V = \{(0,0), (1,0), (0,1), (1,1)\}$, since these are the vectors generated by our canonical basis. Thus, $V = W_1 \cup W_2 \cup W_3$, which is a contradiction.

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