Complete metric space

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

A metrizable space is topologically complete when some metric inducing its topology is complete. The given metric need not be complete: the open interval $(0,1)$ is an example. A Polish space additionally requires separability.

A subspace of a complete metric space admits a compatible complete metric if and only if it is a G-delta set. For the reverse direction, adjoining the coordinates $1/d(x,X\setminus U_n)$ to the ambient metric prevents Cauchy sequences from approaching any excluded closed boundary. For the forward direction, small-diameter relative open covers in a compatible complete metric produce the ambient open sets.

A Polish space is a topological space whose topology can be induced by a complete separable metric. The completeness requirement concerns some compatible metric; it need not hold for every metric inducing the topology. Euclidean spaces are examples.

A metric space is complete if and only if every decreasing sequence of nonempty closed sets whose diameters tend to zero has nonempty intersection. In that case the intersection contains exactly one point.

Every complete metric space is a Baire space: every countable intersection of open dense subsets is dense. Equivalently, no nonempty open subset is a countable union of nowhere-dense subsets.

To prove the dense-intersection form, let $V$ be nonempty and open and let $G_1,G_2,\dots$ be open dense sets. Successively choose closed ballsstarting with a ball contained in $V\cap G_1$. The centres form a Cauchy sequence. Its limit belongs to every closed ball, and hence to $V\cap {\bigcap }_n{G}_n$.

A set containing a countable intersection of open dense subsets. The intersection itself is also called a residual set. In a Baire space it is dense. This topological notion of genericity does not by itself assert full measure.

A dense additive subgroup of a complete metrizable topological group cannot be proper if it is comeagre. Every translate is comeagre and meets the subgroup, so every translating element is a difference of two subgroup elements. Applied to a topologically complete normed space inside its completion, this proves that its original norm is complete.

No infinite-dimensional Banach space has a countable Hamel basis. If $(e_n)$ were such a basis, the space would be the countable union of the finite-dimensional subspaces $\mathrm{span}(e_1,\dots ,e_n)$. Each is closed and has empty interior, contradicting the Baire category theorem.

A nowhere differentiable continuous function is continuous at every point and has no finite derivative anywhere. A Baire-category argument shows that such functions form a dense subset of $C([0,1])$ in the uniform norm.

A Baire space is a topological space in which every countable intersection of open dense sets is dense, equivalently one in which no nonempty open set is a countable union of nowhere-dense sets. Every compact Hausdorff space and every complete metric space is a Baire space.

A Baire function belongs to the smallest hierarchy of real-valued functions containing the continuous functions and closed under pointwise sequential limits.

A Baire class one function is a pointwise limit of continuous functions. On a complete metric domain, its set of discontinuities is meagre; in particular, it has a point of continuity on every nonempty perfect closed subset.

The indicator of the rational numbers in the real numbers is discontinuous everywhere because both the rationals and irrationals are dense. The discontinuity theorem for Baire class one functions therefore shows that it is not Baire class one.

Let $S$ be a nonempty closed, convex, symmetric subset of a Banach space and suppose ${\bigcup }_{n\ge 1}nS=X$. Baire's theorem puts a ball $B(x,r)$ inside some $nS$. Symmetry also puts $B(-x,r)$ there, and convexity puts the midpoint of $x+z$ and $-x+z$, namely $z$, in $nS$ whenever $\Vert z\Vert <r$. Hence $B(0,r/n)\subset S$.

Convexity is essential. In $\mathbb{R}$, letThis set is closed, symmetric, and has gaps arbitrarily near zero. Given $x\ne 0$, choose $k$ so small-scale that the interval
$[|x|/(2\cdot 4^{-k}),|x|/4^{-k}]$ has length at least one; it contains an integer $n$, and then $x/n\in S$. Thus ${\bigcup }_{n}nS=\mathbb{R}$, although $S$ is not a neighbourhood of zero.

Every nonempty countable complete metric space has an isolated point. Otherwise it is the countable union of its singleton subsets, each of which is nowhere dense, contradicting the Baire category theorem.

If such a space has infinitely many points, it has infinitely many isolated points. Indeed, after deleting any finite collection of isolated points, the remaining set is closed, complete, countable, and nonempty. It therefore has a point isolated in the remaining space; because the deleted set is finite, that point is also isolated in the original space.

A subset $A$ of a topological space is nowhere dense when the interior of its closure is empty. Equivalently, every nonempty open set contains a nonempty open subset disjoint from $A$.

A meagre, or first-category, set is a countable union of nowhere dense sets.

An uncountable subset of the real numbers with countable intersection with every meagre set. This is the category counterpart of a Sierpiński set.

The complement of a comeagre set is a meagre set, a countable union of nowhere dense sets. A dense G-delta set is comeagre. The Baire category theorem ensures that two comeagre subsets of a nonempty complete metric space cannot be disjoint.

A Lusin set is an uncountable subset $A$ of the real numbers such that $A\cap M$ is countable for every meagre set $M$. Every meagre subset has a meagre $F_{\sigma }$ cover, so testing those covers suffices. The Lusin set construction under CH gives an example. A Lusin set has no uncountable meagre subset of itself.

Under the Continuum hypothesis, enumerate the meagre $F_{\sigma }$ covers by ${\omega }_1$. At stage $\alpha$ choose a new real outside all earlier covers and the previous choices. The excluded union is meagre, so the Baire category theorem makes the choice possible. Each cover then contains only countably many selected points. Arbitrary meagre sets need not themselves have only continuum many possible choices; the coded covers are what is enumerated.

In $C([0,1])$ with the uniform norm, the continuous functions that are monotone on some interval of positive length form a meagre set. Indeed, it is enough to use intervals with rational endpoints. For each such interval, the nondecreasing and nonincreasing functions form closed sets with empty interior: a sufficiently small local triangular perturbation breaks the relevant inequality. The Baire category theorem therefore shows that a dense set of continuous functions is monotone on no nontrivial interval.

Let $f\in C^{\infty }(\mathbb{R})$ and suppose that for every $x$ there is an $n=n(x)\ge 0$ such that $f^{(n)}(x)=0$. Then $f$ is a polynomial.

If an entire function $f$ is not a polynomial, then there is a point $z_0\in \mathbb{C}$ for which $f^{(n)}(z_0)\ne 0$ for every nonnegative integer $n$. Consequently every coefficient in the Taylor series of $f$ about $z_0$ is nonzero.

For each $n$, the function $f^{(n)}$ is not identically zero, since otherwise $f$ would be a polynomial. Its zero set is closed and has empty interior by the identity theorem, hence is a nowhere dense set. The Baire category theorem says that the complete metric space $\mathbb{C}$ cannot be the union of these countably many zero sets.

Set $S_n=\{x:f^{(n)}(x)=0\}$ and $E_n=\mathrm{int}{S}_n$. The sets $S_n$ are closed, $E_n\subset E_{n+1}$, and Baire's theorem on each interval shows that $\Omega ={\bigcup }_n{E}_n$ is dense. On every connected component of $\Omega$, compactness and the increasing cover by the $E_n$ show that one derivative vanishes locally with a uniform finite order, so $f$ agrees there with one polynomial.

If $F=\mathbb{R}\setminus \Omega$ were nonempty, it would be closed and have no isolated points: polynomial pieces on both sides of an isolated point would have matching derivatives of every order and hence join into one polynomial piece. Applying Baire's theorem to the cover $F={\bigcup }_n(F\cap S_n)$ gives an interval $U$ and an index $k$ for which $F\cap U$ is nonempty and contained in $S_k$. Because $F$ has no isolated points, difference quotients show that every derivative of order at least $k$ vanishes on $F\cap U$. Each polynomial component of $\Omega$ meeting $U$ has an endpoint in $F\cap U$, so its degree is less than $k$. Thus $f^{(k)}=0$ throughout $U$, contradicting $F\cap U\ne \varnothing$. Hence $\Omega =\mathbb{R}$, and its sole connected component carries one polynomial.

The space $C^{\infty }([0,1])$ becomes a complete metric space underConvergence in this metric is exactly uniform convergence of every derivative separately.

If $(f_n)$ is Cauchy in the smooth-function metric, each derivative sequence $(f_n^{(r)})$ converges uniformly to a continuous function $g_r$. Passing to the limit inshows that $g_r^{\prime }=g_{r+1}$. Hence $g_0$ is smooth and $f_n\to g_0$ in the metric.

Outside a meagre set in $C^{\infty }([0,1])$, every rational $q\in (0,1)$ and every positive integer $M$ admit $m\ge M$ such thatFor fixed $q,M$, the union over $m\ge M$ of these strict-inequality sets is open. It is dense because a perturbation $\delta \cos (K(x-q)-m\pi /2)$ can make the $m$th derivative arbitrarily large while keeping any prescribed finite collection of lower derivatives arbitrarily small. The Baire category theorem applied over the countable pairs $(q,M)$ gives the claim.

The generic derivative bound implies that at every rational $q\in (0,1)$,The Cauchy-Hadamard theorem therefore gives radius zero to the Taylor series at every such $q$. If a Taylor series represented $f$ on a neighborhood of any point, that neighborhood would contain a rational point at which $f$ was real analytic, a contradiction.

A closed subspace of a complete metric space is complete. A Cauchy sequence in the subspace converges in the ambient space, and closedness keeps its limit in the subspace.