= Solution
The <Baire category theorem> states that in a complete <metric space>, a <countable> intersection of open <dense> sets is <dense>. Equivalently, a nonempty <complete metric space> cannot be a <countable union> of <nowhere dense sets>, where a set is <nowhere dense> if its closure has empty interior.
Here is the proof. Let $G_1,G_2,\ldots$ be open <dense> sets and let $U$ be any nonempty <open set>. Choose a point $a_1\in U\cap G_1$ and a radius $0<r_1<1/2$ such that the <closed ball> $B_1=\overline B(a_1,r_1)$ lies inside $U\cap G_1$. Inductively choose $a_{m+1}$ in the nonempty <open set> $B(a_m,r_m)\cap G_{m+1}$, and then a radius $0<r_{m+1}<2^{-m-1}$ such that
$$
B_{m+1}=\overline B(a_{m+1},r_{m+1})\subseteq B(a_m,r_m)\cap G_{m+1}.
$$
Thus the <closed> balls are nested and their radii tend to zero. For $j,k\geq m$, both centers lie in $B_m$, so $d(a_j,a_k)\leq2r_m$. They form a <Cauchy sequence>. <Completeness> gives a limit $a$, and closedness puts it in every $B_m$. Consequently $a\in U\cap\bigcap_mG_m$. As $U$ was arbitrary, \b[the intersection is <dense>].
To obtain the equivalent formulation, take complements of the closures of the <nowhere dense> sets; those complements are open <dense>. Conversely, complements of open <dense> sets are <closed> with empty interior. In particular, a <countable> <closed> cover of a nonempty <complete metric space> must contain a set with nonempty interior.
Back to article page
