Countably many proper subspaces never cover M_n

Idea. Proper subspaces are "small" in the topological sense (closed nowhere dense); a complete space is "large" (Baire).

Step 1: proper subspaces are closed nowhere dense. By the earlier problem on subspaces, a proper subspace $V\subsetneq M_n(\mathbb{R})$ is closed with empty interior, i.e. nowhere dense.

Step 2: $M_n(\mathbb{R})$ is a complete metric space. It is isometric to $\mathbb{R}^{n^2}$ with the Frobenius norm.

Step 3: Baire. Suppose $M_n(\mathbb{R})=\bigcup_{k\ge1}V_k$ with each $V_k$ a proper subspace. This is a countable union of closed nowhere dense sets covering a non-empty complete metric space, contradicting the Baire category theorem. The finite case is a special case (repeat the subspaces).

Elementary proof for finitely many. Suppose $V_1,\dots,V_m$ are proper subspaces covering $M=M_n(\mathbb{R})$, with $m$ minimal. Pick $x\in V_1$ not in $V_2\cup\dots\cup V_m$ (it exists by minimality) and $y\notin V_1$. For $t\in\mathbb{R}$, the points $y+tx$ are distinct for distinct $t$; none lies in $V_1$ (else $y\in V_1$ as $tx\in V_1$). Each of the infinitely many points $y+tx$ lies in some $V_j$ with $j\ge2$, so one $V_j$ contains two of them, $y+tx$ and $y+t'x$, hence $(t-t')x\in V_j$ and $x\in V_j$: contradiction.

Remark.
- The finite statement is a property of *infinite* fields: over $\mathbb{F}_2$ the space $\mathbb{F}_2^2$ is the union of the three lines through the origin.
- The countable statement is false in a non-complete space: $\mathbb{Q}^{n^2}$ is a countable union of points.
- The example $T=\{\operatorname{tr}\in\mathbb{Q}\}$ from the previous problem shows that a union of *countably many hyperplanes (affine ones)* can be dense without being everything.