vector space over an infinite field is not a finite union of proper subspaces

Theorem 1.

A vector space $V$ over an infinite field $\\mathbbmss{F}$ cannot bea finite union of proper subspaces of itself.

Proof.

Let $V=V_{1}\\cup V_{2}\\cup\\ldots\\cup V_{n}$ whereeach $V_{i}$ is a proper subspace of $V$ and $n>1$ is minimal.Because $n$ is minimal, $V_{n}\ot\\subset V_{1}\\cup V_{2}\\cup\\ldots\\cup V_{n-1}$ .

Let $u\ot\\in V_{n}$ and let $v\\in V_{n}\\setminus\\left(V_{1}\\cup V_{2}\\cup\\ldots\\cup V_{n-1}\\right)$ .

Define $S=\\left\\{v+tu:t\\in\\mathbbmss{F}\\right\\}$ . Since $u\ot\\in V_{n}$ is not the zero vector and the field $\\mathbbmss{F}$ is infinite, $S$ must be infinite.

Since $S\\subset V=V_{1}\\cup V_{2}\\cup\\ldots\\cup V_{n}$ one of the $V_{i}$ must contain infinitely many vectors in $S$ .

However, if $V_{n}$ were to contain a vector, other than $v$ , from $S$ there wouldexist non-zero $t\\in\\mathbbmss{F}$ such that $v+tu\\in V_{n}$ .But then $tu=v+tu-v\\in V_{n}$ and we would have $u\\in V_{n}$ contrary to the choice of $u$ . Thus $V_{n}$ cannot containinfinitely many elements in $S$ .

If some $V_{i},1\\leq i<n$ contained two distinct vectors in $S$ ,then there would exist distinct $t_{1},t_{2}\\in\\mathbbmss{F}$ such that $v+t_{1}u,v+t_{2}u\\in V_{i}$ . But then $\\left(t_{2}-t_{1}\\right)v=t_{2}\\left(v+t_{1}u\\right)-t_{1}\\left(v+t_{2}u\\right%)\\in V_{i}$ and we would have $v\\in V_{i}$ contrary tothe choice of $v$ . Thus for $1\\leq i<n,V_{i}$ cannot containinfinitely many elements in $S$ either.∎

单词	VectorSpaceOverAnInfiniteFieldIsNotAFiniteUnionOfProperSubspaces
释义	vector space over an infinite field is not a finite union of proper subspaces Theorem 1. A vector space $V$ over an infinite field $\\mathbbmss{F}$ cannot bea finite union of proper subspaces of itself. Proof. Let $V=V_{1}\\cup V_{2}\\cup\\ldots\\cup V_{n}$ whereeach $V_{i}$ is a proper subspace of $V$ and $n>1$ is minimal.Because $n$ is minimal, $V_{n}\ot\\subset V_{1}\\cup V_{2}\\cup\\ldots\\cup V_{n-1}$ . Let $u\ot\\in V_{n}$ and let $v\\in V_{n}\\setminus\\left(V_{1}\\cup V_{2}\\cup\\ldots\\cup V_{n-1}\\right)$ . Define $S=\\left\\{v+tu:t\\in\\mathbbmss{F}\\right\\}$ . Since $u\ot\\in V_{n}$ is not the zero vector and the field $\\mathbbmss{F}$ is infinite, $S$ must be infinite. Since $S\\subset V=V_{1}\\cup V_{2}\\cup\\ldots\\cup V_{n}$ one of the $V_{i}$ must contain infinitely many vectors in $S$ . However, if $V_{n}$ were to contain a vector, other than $v$ , from $S$ there wouldexist non-zero $t\\in\\mathbbmss{F}$ such that $v+tu\\in V_{n}$ .But then $tu=v+tu-v\\in V_{n}$ and we would have $u\\in V_{n}$ contrary to the choice of $u$ . Thus $V_{n}$ cannot containinfinitely many elements in $S$ . If some $V_{i},1\\leq i<n$ contained two distinct vectors in $S$ ,then there would exist distinct $t_{1},t_{2}\\in\\mathbbmss{F}$ such that $v+t_{1}u,v+t_{2}u\\in V_{i}$ . But then $\\left(t_{2}-t_{1}\\right)v=t_{2}\\left(v+t_{1}u\\right)-t_{1}\\left(v+t_{2}u\\right%)\\in V_{i}$ and we would have $v\\in V_{i}$ contrary tothe choice of $v$ . Thus for $1\\leq i<n,V_{i}$ cannot containinfinitely many elements in $S$ either.∎
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