Alonzo Odom

Answered

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 $j\in \{1,...,r\}$, 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 $l\in \{1,...,r\}$ 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?

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 $j\in \{1,...,r\}$, 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 $l\in \{1,...,r\}$ 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?

Answer & Explanation

Caylee Davenport

Expert

2022-07-21Added 14 answers

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 ${W}_{1}=\{(0,0),(1,0)\}$, ${W}_{2}=\{(0,0),(1,1)\}$, and ${W}_{3}=\{(0,0),(1,1)\}$

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