Alonzo Odom

2022-07-20

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 $\bigcup _{i=1}^{r}{W}_{i}\ne V.$
I tried the following:
for all $j\in \left\{1,...,r\right\}$, I take ${w}_{j}\in {W}_{j}$ such that ${w}_{j}\notin {W}_{i}$ whenever $j\ne i$, 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 $l\in \left\{1,...,r\right\}$ 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?

Caylee Davenport

Expert

The assertion seems to be false. Take $k={\mathbb{F}}_{2}$ and $V=k\oplus k=\left\{\left(0,0\right),\left(1,0\right),\left(0,1\right),\left(1,1\right)\right\}$. Now you can take r=3 and the following proper subspaces of ${W}_{1}=\left\{\left(0,0\right),\left(1,0\right)\right\}$, ${W}_{2}=\left\{\left(0,0\right),\left(1,1\right)\right\}$, and ${W}_{3}=\left\{\left(0,0\right),\left(1,1\right)\right\}$