Matrices with no zero entry: open and dense

Idea. The two sets in the statement are complements of each other. It is enough to show that $Z=\{A:\exists(i,j),\,a_{ij}=0\}$ is closed with empty interior; then its complement is open and dense.

Step 1: $Z$ is closed. $Z=\bigcup_{i,j}H_{ij}$ with $H_{ij}=\{a_{ij}=0\}$, a finite union of closed sets (each is the kernel of a continuous linear functional).

Step 2: each $H_{ij}$ has empty interior. $H_{ij}$ is a proper linear subspace, so it has empty interior (earlier problem on subspaces). Concretely, $A+\varepsilon E_{ij}\notin H_{ij}$ if $a_{ij}=0$ and $\varepsilon\neq0$.

Step 3: $Z$ has empty interior. We prove: *the union of two closed sets with empty interior has empty interior* (then induct on the number of sets). Let $F,G$ be closed with empty interior and suppose an open ball $B\subset F\cup G$. Since $F$ has empty interior, $B\not\subset F$, so $B\setminus F$ is a non-empty open set contained in $G$, so $G$ has non-empty interior: contradiction.

Step 4: conclusion. $Z$ is closed with empty interior, so $M_n(\mathbb{R})\setminus Z$ is open, and it is dense (a set is dense iff its complement has empty interior).

Remark.
- This is a model for "generic" statements: a property that holds outside a finite union of proper algebraic subsets (zero sets of non-zero polynomials) holds on an open dense set. Examples: invertible, distinct eigenvalues, all entries non-zero.
- For *countable* unions the argument must be replaced by the Baire theorem, and one only gets a dense $G_\delta$ complement, not an open one (see the Baire problem below).