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The Baire category theorem says that in a nonempty complete metric space, every countable intersection of open dense sets is dense. To prove it, start in an arbitrary nonempty open set $O$ and let $G_n$ be open dense. Inductively choose closed balls $\overline{B}(x_n,r_n)$ with $0<r_n\le 2^{-n}$, contained in $G_n$ and in the interior of the previous ball, with the first also contained in $O$. Density provides a point at each stage, and openness permits the smaller closed ball. The centers are a Cauchy sequence; completeness supplies a limit lying in every closed ball, hence in $O\cap {\bigcap }_n{G}_n$. The proof also works at isolated points, where sufficiently small balls are singletons.

The incomplete metric space $\mathbb{Q}$ is a counterexample: it is the countable union of its closed nowhere dense singletons, so the complements are open dense with empty intersection.

For the sequence in the problem, putEach $E_N$ is closed by continuity. Pointwise convergence makes each point's values a Cauchy sequence, so $[0,1]={\bigcup }_N{E}_N$. By the Baire category theorem, one $E_N$ has nonempty relative interior, which contains a nonempty open interval $J$ in $(0,1)$. This proves the required common tail bound.

If continuous $g_n$ converged pointwise to the rationality indicator, take $ϵ=1/4$ and let $m\to \infty$ in this bound. On $J$, $|g_n-g|\le 1/4$ for $n\ge N$. Hence a fixed continuous $g_N$ is at least $3/4$ at every rational point and at most $1/4$ at every irrational point of $J$. The density of both sets contradicts continuity. Thus the rationality indicator is not a pointwise limit of continuous functions.

Nevertheless $h_n(x)=x^n$ are continuous and converge pointwise to $h(x)=0$ for $x<1$, $h(1)=1$. This limit is discontinuous at $1$: a pointwise limit of continuous functions can be discontinuous.

A metric space metric $d:X\times X\to [0,\infty )$ satisfies positivity, $d(x,y)=0$ exactly when $x=y$, symmetry, and the triangle inequality. A subset $U\subseteq X$ is an open set when every $x\in U$ has some $\epsilon >0$ with the open ball $B_d(x,\epsilon )\subseteq U$.

Both $\varnothing$ and $X$ are open. An arbitrary union of open sets is open because a point lies in one member supplying a ball. A finite intersection is open because the minimum of the finitely many available radii supplies a ball. Thus the metric-open sets satisfy the topology axioms.

For $C[0,1]$, the Cauchy-Schwarz inequality gives $d_1(f,g)\le d_2(f,g)$, so the $d_2$ topology is at least as fine as the $d_1$ topology. To see strictness, define the continuous triangular spikesThenThus $f_n$ converges to zero in the L1 norm but not in the L2 norm. The two metrics induce different topologies.

A metric space is complete when every Cauchy sequence converges to a point of the space.

Let $x^{(1)},x^{(2)},\dots$ be a Cauchy sequence in $X$, where $x^{(j)}=(x_1^{(j)},x_2^{(j)},\dots )$. For each positive integer $N$, choose $J_N$ such thatThis inequality says that the first $N$ coordinates of $x^{(j)}$ and $x^{(k)}$ agree. Consequently each coordinate $x_n^{(j)}$ is eventually constant. Let $x_n$ be its eventual value and put $x=(x_1,x_2,\dots )\in X$.

For $j\ge J_N$, the first $N$ coordinates of $x^{(j)}$ agree with those of $x$, soAs $N$ is arbitrary, $x^{(j)}\to x$. Therefore

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 compact real tree is a compact metric space $(T,d)$ such that any $x,y\in T$ are joined by a unique arc, and that arc is isometric to $[0,d(x,y)]$. The multiplicity of a point in a real tree $a\in T$ is the number of connected components of $T\setminus \{a\}$.

For nonempty compact subsets $A,B$ of a metric space $(Z,d)$, the Hausdorff distance isFor compact metric spaces $X,Y$, the Gromov-Hausdorff distance iswhere $\phi$ and $\psi$ range over isometric embeddings into a common metric space $Z$.

The collection of compact real trees is not compact in the Gromov-Hausdorff topology. Indeed, the intervals $T_n=[0,n]$ are compact real trees andTheir diameters are unbounded, so $(T_n)$ has no convergent subsequence in the Gromov-Hausdorff topology.

Let $A,B$ be disjoint closed subsets of the metric space $(X,d)$; the cases where one is empty are immediate. The distance functionsare continuous, because each is 1-Lipschitz. Their sum is strictly positive: if both distances vanished, closedness would put $x$ in $A\cap B$. Thereforeis continuous, equals zero on $A$, and equals one on $B$. The setsare disjoint open neighbourhoods of $A$ and $B$. Thus every metric space is a normal topological space; this is the normality of every metric space.

Weak convergence of probability measures ${\mu }_n\to \mu$ on a metric space means thatfor every bounded continuous real function $f$. Weak convergence of random variables means weak convergence of their probability distributions; equivalently,for every bounded continuous $f$.

A computational problem in the SCI hierarchy is a quadrupleHere $\Omega$ is the primary set of inputs, $\Lambda$ is the set of permitted evaluation functions, $(\mathcal{M},d)$ is the output metric space, and $\Xi :\Omega \to \mathcal{M}$ is the problem function. The Solvability complexity index is defined from the minimum height of a tower of algorithms that computes $\Xi$ from finite subsets of $\Lambda$.

For the classical computational spectral problem, takeandSince the spectrum of a bounded operator is a nonempty compact subset of the complex numbers, one may take $\mathcal{M}$ to be the nonempty compact subsets of $\mathbb{C}$ with the Hausdorff distance. The Attouch--Wets topology gives an equivalent convenient formulation on bounded spectral sets and also extends naturally to unbounded closed sets. This choice of $\mathcal{M}$ makes convergence mean convergence of the whole spectrum as a set, including both the absence of persistent spectral pollution and the approximation of every genuine spectral point.

In a compact metric space with a continuous map $T$, points $x,y$ are proximal ifThis property is independent of the compatible metric. If $T$ is injective and $x\ne y$, arbitrarily small proximal distances must occur at arbitrarily large times, since each finite collection of distances is strictly positive. Proximal points need not be equal or have dense orbits.

A real tree is a metric space $(T,d)$ in which every pair $a,b\in T$ is joined by a unique arc and that arc is isometric to the real interval $[0,d(a,b)]$. Equivalently, it is a geodesic metric space with no nontrivial simple loops.

In a metric space, applying the triangle inequality in both directions gives $|d(x,z)-d(y,z)|\le d(x,y)$. In a normed vector space, its corresponding form is $|\Vert x\Vert -\Vert y\Vert |\le \Vert x-y\Vert$. In particular, distance to a fixed point is a Lipschitz continuous real function.

A topological space is separable when it has a countable dense subset. The topology is part of the assertion: a dual may be separable in its weak-star topology and not in its norm topology. A compact metric space is separable by taking the union of finite nets at scales tending to zero.

A sequentially closed set is one that contains every limit of its convergent sequences. This is weaker than being a closed set in a general topological space, but the two conditions agree in metric spaces.

Topological dynamics studies continuous maps and their orbits on topological spaces. For a compact metric space $X$ and a continuous map $T:X\to X$, the pair $(X,T)$ is a compact Hausdorff topological dynamical system.

For a uniquely ergodic continuous map $T$ of a compact metric space, every continuous $f$ satisfies $N^{-1}{\sum }_{j=0}^{N-1}f(T^{j}x)\to \int fd\mu$ uniformly in $x$. Otherwise choose increasingly long orbit empirical measures with a fixed discrepancy. Compactness provides a weak limit; the telescoping identity makes it invariant, so it must be the unique measure $\mu$, contradicting that discrepancy. Compactness and continuity are essential hypotheses.

The supremum of pointwise distances defines a metric on maps when it is finite, for example if the target metric space is bounded. Convergence in this metric is uniform convergence. If the target is complete, Cauchy sequences have pointwise limits and the Cauchy estimate makes convergence uniform.

A continuous self-map of a compact metric space is uniquely ergodic when it has exactly one invariant Borel probability measure. An irrational rotation of the circle is uniquely ergodic: invariance multiplies its nonzero-index Fourier coefficients by nontrivial phases, forcing those coefficients to vanish. The Stone-Weierstrass theorem then identifies the invariant measure as normalized Lebesgue measure.