Can a vector space over an infinite field be a finite union of proper subspaces ?

  • 2
    $\begingroup$ See here. The result is stated there for characteristic zero, but it can be adapted so only infinitude of the field is used. $\endgroup$ Feb 27, 2014 at 7:14

2 Answers 2


Short answer: No.

Long answer: No. Assume that our vectorspace $V$ is a finite union of proper subspaces, hence $$V=\bigcup_{i=1}^n U_i.$$

Now, pick a non-zero vector $x\in U_1$, and pick another vector $y\in V\,\backslash U_1$.

There are infinitely many vectors $x+ky$, where $k\in K^*$ ($K$ is our infinite field). Note that $x+ky$ is not in $U_1$, hence must be contained in some $U_j$ where $j\neq 1$.

Then since $k\in K^*$, we can have $x+k_1y,x+k_2y\in U_j$, which implies that it also contains $y$ and hence also $x$, hence $U_1\subset U_j$. Hence $$V=\bigcup_{i=2}^n U_i.$$ Evidently, this can be continued, hence a contradiction arises.

  • $\begingroup$ What is $K^*$ and how is it different from $K$? $\endgroup$
    – ubadub
    Mar 9, 2019 at 17:47
  • 1
    $\begingroup$ @ubadub $K^* = K \setminus \{0\}$ $\endgroup$ Mar 16, 2019 at 21:39
  • $\begingroup$ But what if union of all $U_j$, where $j\neq 1$, are all subsets of $U_1$. Is the answer still No? $\endgroup$
    – Roy Huang
    Aug 19, 2022 at 5:25
  • $\begingroup$ Why do we know, that all $x+ky$ are in one $U_j$? And not spread over more subspaces. $\endgroup$
    – Incompl33t
    Mar 18, 2023 at 19:34

Denote by $K$ the field over which $V$ is a vector space. Suppose the answer is true and consider write $V=\bigcup\limits_{i=1}^{n} V_i$, where $V_i$ are proper subspaces of $V$ and $n \in \mathbb{N}$ is minimal.

Because $n$ is minimal, there exists an element $v_n \in V_n \setminus \bigcup\limits_{i=1}^{n-1}V_i $. There exists also an element $v \in \bigcup\limits_{i=1}^{n-1}V_i \setminus V_n$.

Now if we look at the infinite set $M=\{ v_n+k \cdot v : k \in K \}$, using Pigeonhole Principle we can see that there exists $i \in \{ 1, \dots, n\}$ such that $|M \cap V_i|\geq 2$ (we can actually say that the cardinal is infinite, but we don't need that for the rest of the proof).

If $V_n \cap M$ contains some element other than $v_n$, then there exists $k \in K \setminus \{ 0 \}$ such that $k \cdot v \in V_n$, which contradicts the choice of $v$.

If $V_i \cap M$ has more than two elements for some $i \in \{1,\dots, n-1\}$ then, looking at their difference, we obtain that $v_n \in V_i$ which contradicts the choice of $v_n$.

Therefore, $V$ is never a finite union of proper subspaces.

Edit: Repetition was not intended, I was posting while BlackAdder posted.