# Can a vector space over finite field be written as union of finite number of proper subspaces?

Recently, I solved a problem that says-
If $$V$$ is a vector space over an infinite field. Prove that, V cannot be written as set-thoretic union of a finite number of proper subspaces.
But is this result true in case of finite field?
. I can't get such an example where a vector space over finite field can be written as union of finite number of proper subspaces.
Can anybody give such an example? Thanks for assistance in advance.

• See here for the theory on how many subspaces you need at minimum, Nov 4, 2018 at 5:31
• That thread is linked to many others with more discussion and variants. Search them! Nov 4, 2018 at 5:41

The answer is yes; if $$V$$ is a finite vector space (so over a finite field and of finite dimension), then it has only finitely many elements. Let $$\langle v\rangle$$ denote the span of a vector $$v\in V$$. Then clearly $$V=\bigcup_{v\in V}\langle v\rangle,$$ because $$v\in\langle v\rangle$$ for every $$v\in V$$, and the union is finite because $$V$$ is finite. Hence $$V$$ is the union of finitely many proper subspaces (if $$\dim V>1$$, otherwise the subspaces aren't proper).

For a very concrete example, consider $$\Bbb{F}_2^2$$, a $$2$$-dimensional vector space over the finite field $$\Bbb{F}_2$$ of two elements. Then $$\begin{eqnarray*} \Bbb{F}_2^2&=&\{(0,0),(1,0),(0,1),(1,1)\}\\ \bigcup_{v\in\Bbb{F}_2^2}\langle v\rangle&=&\{(0,0)\}\cup\{(0,0),(1,0)\}\cup\{(0,0),(0,1)\}\cup\{(0,0),(1,1)\}. \end{eqnarray*}$$

• But is the result true when the vector space is of infinite dimension? The vector space may not be finite. Example: set of all polynomials over $\Bbb{Z}_2$ Nov 3, 2018 at 19:17
• Servaes, if $V$ is not finite dimensional then your method will not work, because in that case $V$ has infinitely many elements. Nov 5, 2018 at 6:59
• What if the vector space is one-dimensional? Then the span of v is not a proper subspace. Nov 27, 2019 at 11:34

It seems to me that the case of an infinite dimensional space is creating extra headache. Spelling out an example to show what can be done.

Consider the field $$\Bbb{F}_2$$. Let $$V=\Bbb{F}_2[x]$$ be the space of univariate polynomials. Let $$U$$ be the subspace of polynomials with zero constant and linear terms. Consider the following three subspaces of $$V$$: $$V_1=\langle 1\rangle\oplus U=\{a_0+a_1x+a_2x^2+\cdots +a_nx^n\mid n\in\Bbb{N}, a_i\in\Bbb{F}_2, a_1=0\},$$ $$V_2=\langle x\rangle\oplus U=\{a_0+a_1x+a_2x^2+\cdots +a_nx^n\mid n\in\Bbb{N}, a_i\in\Bbb{F}_2, a_0=0\},$$ and $$V_3=\langle 1+x\rangle\oplus U=\{a_0+a_1x+a_2x^2+\cdots +a_nx^n\mid n\in\Bbb{N}, a_i\in\Bbb{F}_2, a_0=a_1\}.$$ It is clear that every polynomial of $$V$$ belongs to at least one of the subspaces $$V_1,V_2,V_3$$ according to what its constant and linear terms look like. Therefore $$V=V_1\cup V_2\cup V_3.$$

What just happened?

Consider the quotient space $$V/U$$. It is 2-dimensional. Its non-zero elements are the cosets $$1+U,x+U$$ and $$1+x+U$$. So we can write $$V/U$$ as a union of three 1-dimensional subspaces following the recipe of Servaes' answer each spanned by one of those cosets. The resulting subspaces are exactly the spaces $$V_i/U, i=1,2,3$$. No wonder that $$V_1,V_2,V_3$$ cover all of $$V$$!