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A subset $A$ of a metric space is nowhere dense when the interior of its closure is empty. One form of the Baire category theorem says that a nonempty complete metric space cannot be a countable union of nowhere dense closed sets; equivalently, a countable intersection of dense open sets is dense.

Fix $\epsilon >0$ and work in the complete interval $[1,2]$. For $N\ge 1$ define the closed setThe sets are closed by the continuity of $f$, and the assumed pointwise convergence gives $[1,2]={\bigcup }_N{E}_N$. The Baire category theorem therefore gives an $N$ and a nonempty open interval $(a,b)\subseteq [1,2]$ contained in $E_N$.

The intervals $[na,nb]$ overlap once $(n+1)a\le nb$, which holds for every sufficiently large natural number $n$. Their union consequently contains a half-line. Every sufficiently large $t$ can therefore be written as $t=nx$ with $n\ge N$ and $x\in (a,b)$, and then $|f(t)|=|f(nx)|\le \epsilon$. Since $\epsilon$ was arbitrary, $f(t)\to 0$ as $t\to \infty$.

A set is of first Baire category, or meagre, when it is a countable union of nowhere dense sets. The statement is true: a countable union of countable unions of nowhere dense sets is again a countable union of nowhere dense sets.

The relevant Baire category theorem says that a nonempty complete metric space is not meagre in itself.

The Baire category theorem states that every countable intersection of open dense subsets of a complete metric space is dense. Equivalently, no nonempty open subset of a complete metric space is a countable union of nowhere dense sets.

To prove the first form, let $G_1,G_2,\dots$ be open dense subsets of a complete metric space $X$, and let $V\subseteq X$ be nonempty and open. Choose a closed ballInductively, density and openness of $G_{n+1}$ allow a closed ballThe balls are nested and $d(x_m,x_n)<2^{-n}$ for $m>n$, so $(x_n)$ is Cauchy. Completeness gives $x_n\to x\in X$. For every $n$, all later centres lie in $\overline{B}(x_n,r_n)$; closedness gives $x\in \overline{B}(x_n,r_n)\subseteq G_n$. The first ball also lies in $V$, henceSince every nonempty open $V$ meets the intersection, that intersection is dense.

A subset $A$ of a metric space $X$ is a nowhere dense set whenEquivalently, every nonempty open set contains a nonempty open subset disjoint from $A$.

The Baire category theorem says that if $X$ is a complete metric space and $G_1,G_2,\dots$ are open dense subsets of $X$, then ${\bigcap }_{n\ge 1}{G}_n$ is dense in $X$. Equivalently, no nonempty open subset of $X$ is a countable union of nowhere-dense sets.

To prove it, take a nonempty open set $V$. Since $G_1$ is open and dense, there is a closed ballwith $0<r_1<2^{-1}$. Inductively, openness and density of $G_{n+1}$ allow us to chooseThe balls are nested, and for $m>n$ their centres satisfyThus $(x_n)$ is Cauchy and has a limit $x\in X$. Every closed ball contains the tail of the sequence, so it contains $x$. HenceAs $V$ was arbitrary, the intersection is dense.