Show that a vector space over an infinite field cannot be the union of finitely many proper linear subspaces.
Solution
Let \(V\) be the given vector space and let \(F\) be its scalar field.
1. Suppose, for contradiction, that\[ V=W_1\cup\cdots\cup W_m \]for proper linear subspaces \(W_1,\ldots,W_m\), and choose such a cover with\(m\) minimal.
If any finite cover by proper subspaces existed, the positive integers indexing the possible numbers of members would have a least element. Since a single proper subspace cannot equal \(V\), one has \(m\ge2\).
2. For each \(i\), there is a vector\[ v_i\in W_i\setminus\bigcup_{j\ne i}W_j. \]
If no such \(v_i\) existed for some \(i\), then \[ W_i\subseteq\bigcup_{j\ne i}W_j. \] Removing \(W_i\) would still leave a cover of \(V\) by \(m-1\) proper subspaces, contradicting the minimality in step 1.
3. Put \(x\coloneqq v_1\) and \(y\coloneqq v_2\). The affine line\[ L\coloneqq\{x+ty:t\in F\} \]has infinitely many points.
By step 2, \(y\notin W_1\). Since \(0\in W_1\), it follows that \(y\ne0\). Hence the map \[ F\longrightarrow L, \qquad t\longmapsto x+ty \] is injective. The field \(F\) is infinite, so \(L\) is infinite.
4. Every subspace \(W_i\) meets \(L\) in at most one point.
4.1. The intersection \(W_1\cap L\) has at most one point.
If \(x+sy\) and \(x+ty\) were distinct points of \(W_1\), then \[ (s-t)y\in W_1. \] Since \(s-t\ne0\), this would imply \(y\in W_1\), contradicting step 2.
4.2. The intersection \(W_2\cap L\) has at most one point.
In fact \(W_2\cap L\) is empty. If \(x+ty\in W_2\) for some \(t\in F\), then \(y\in W_2\) gives \[ x=(x+ty)-ty\in W_2, \] contradicting step 2.
4.3. For every \(i\ge3\), the intersection \(W_i\cap L\) has at most onepoint.
If \(x+sy\) and \(x+ty\) were two distinct points of \(W_i\), their difference would imply \(y\in W_i\), again contradicting step 2.
4.4. Q.E.D.
Steps <2>1–<2>3 cover all indices \(1\le i\le m\).
5. The assumed finite cover cannot contain all of \(L\).
By step 4, the union \[ W_1\cup\cdots\cup W_m \] contains at most \(m\) points of \(L\). Step 3 shows that \(L\) has infinitely many points. Thus \(L\) is not contained in the union, contradicting the equality assumed in step 1.
6. Q.E.D.
The contradiction in step 5 proves that no finite union of proper linear subspaces can equal \(V\).