# Question about vector space and union of proper subspaces

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?

• 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. – Arthur Feb 4 at 17:10

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 $$W_3=\{(0,0),(1,1)\}$$.
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.