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The closure of a subset $A$ of a topological space $X$ is the smallest closed set containing $A$. Equivalently, it is the intersection of all closed sets containing $A$, or the set of points whose every neighbourhood in $X$ meets $A$. The notation ${\mathrm{cl}}_{X}A$ makes the ambient topological space explicit. This definition works without a metric or any appeal to sequences. ^[ref]

For $A=[0,1)$ in the usual real line, $\overline{A}=[0,1]$: every neighbourhood of $1$ meets $A$, while a point outside $[0,1]$ has an open ball disjoint from $A$. Passing to the closure adds the missing endpoint and retains all original points, including isolated points when present. In general, the closure is $A$ together with its limit points; its individual points are the adherent points of $A$.

The closure turns an arbitrary subset into a closed set with no unnecessary added points. It satisfies $A\subseteq \overline{A}$ and $\overline{\overline{A}}=\overline{A}$; a subset is closed exactly when $A=\overline{A}$. In a metric space, $x\in \overline{A}$ exactly when some sequence of points of $A$ has limit of a sequence $x$: choose its $n$th point in $A\cap B(x,1/n)$. For a general topological space, keep the neighbourhood definition; sequences need not detect every point of the closure.

The cofinite topology on a set $X$ declares the empty set and all cofinite subsets to be open sets: a nonempty open set may exclude only a finite set of points. There is no assumed metric. On an infinite set, this gives a particularly small collection of open sets while still making every singleton a closed set. On a finite set, every subset has finite complement, so the cofinite topology is the discrete topology.^[ref]

For an original example, use $X={\mathbb{N}}_{>0}$ and chooseBoth are open sets, and their intersection is $U\cap V=X\setminus \{1,2,3,5,7,8\}$. Thus $4,6,9,10,\dots$ are in both. Their finite exclusions cannot separate the infinite remainder. The finite intersection of any finite collection of nonempty open sets is again a nonempty open set, since its complement is a finite union of finite sets. An arbitrary nonempty union is open because its complement is contained in the finite complement of any one nonempty member.

The cofinite topology is useful for separating hypotheses that familiar metric spaces package together. An infinite cofinite topological space is a T1 space, because its singletons are closed, but is not a Hausdorff space, because two nonempty open sets cannot be disjoint sets. It is a hyperconnected space and hence a connected space. Moreover, every subset of a cofinite space is compact, and sequence convergence in the cofinite topology can have many limits. These are compatible properties, rather than contradictions: the usual uniqueness and closedness results use the Hausdorff space hypothesis.^[ref]

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 metric is compatible with a given topology when its open balls generate exactly that topology. Compatible metrics on a compact topological space are uniformly equivalent: the identity maps between the resulting compact metric spaces are uniformly continuous. Consequently they define the same proximality and the same asymptotic-pair relation for a fixed continuous map.

Here $C^{\infty }([0,1])$ means the real- or complex-valued functions continuous on $[0,1]$, smooth on $(0,1)$, with every interior derivative having a continuous extension to the endpoints; endpoint values of derivatives are these extensions. On this space of smooth functions on a compact interval, defineThese finite-valued seminorms increase with $k$ and separate points. In fact every $p_k$ is a norm, since it includes $p_0(f)=\Vert f{\Vert }_{\infty }$. The terminology seminorm includes norms; it does not assert that every measurement has a nontrivial kernel of a seminorm. Smooth convergence means $p_k(f_n-f)\to 0$ for every fixed $k$, equivalently uniform convergence of every derivative separately. It does not demand one tolerance or one convergence index that works for all orders simultaneously. The maximum convention is equivalent to the derivative supremum norm convention using a finite sum, since $p_k(f)\le {\sum }_{j=0}^k\Vert f^{(j)}{\Vert }_{\infty }\le (k+1)p_k(f)$.

For a worked example, $f(t)=t^3$ gives $p_0(f)=1$, $p_1(f)=3$, $p_2(f)=6$, and $p_k(f)=6$ for $k\ge 3$. The linear functional $\ell (f)=f^{\prime \prime }(1/2)$ obeys $|\ell (f)|\le p_2(f)$ and is continuous. Similarly, differentiation is a continuous linear operator $D:f\mapsto f^{\prime }$ for smooth convergence, because $p_k(Df)\le p_{k+1}(f)$.

The space of smooth functions on a compact interval is complete for this seminorm topology. A Cauchy sequence has uniform limits $g_j$ for its derivatives of each order. Passing to the limit in the fundamental theorem of calculus identitygives $g_j(t)-g_j(0)={\int }_0^t{g}_{j+1}(s)ds$, so $g_j^{\prime }=g_{j+1}$ and $g_0$ belongs to the same space. This is the completeness of the smooth-function metric argument; the metric there and the increasing-$p_k$ metric above induce the same smooth topology and the same Cauchy sequences.

The set distance is the infimum of pairwise metric distances. Two nonempty disjoint compact sets have positive set distance: continuity attains the minimum on their Cartesian product, and a zero minimum would give a common point. Disjoint closed sets without compactness need not have positive separation.

A distortion measure assigns a nonnegative cost to replacing a source symbol $x$ by a reconstruction $\hat{x}$. It need not be a metric: the two alphabets may differ, and neither symmetry nor the triangle inequality is required. Choosing the cost is part of the model. For numerical symbols, $d(x,\hat{x})=(x-\hat{x}{)}^2$ gives mean squared error after averaging; for exact symbol agreement, use Hamming distortion.

For $0<p<1$, write $Q_p(f)=(\int |f{|}^{p}d\mu {)}^{1/p}$ to emphasize that this is a quasi-norm, as in the Lq quasi-norm branch. On two coordinates with counting measure, $u=(1,0)$ and $v=(0,1)$ give $Q_p(u)=Q_p(v)=1$, but $Q_p(u+v)=2^{1/p}>2$. At $p=1/2$, this is the explicit failure $4>1+1$ in the right panel of the diagram. The midpoint $(1/2,1/2)$ has $Q_p=2^{1/p-1}>1$, so the corresponding unit sublevel set is not a convex set.

A useful replacement survives: for nonnegative real numbers $a,b$, $(a+b{)}^p\le a^p+b^p$. Together with the scalar triangle inequality, this yieldsConsequently $d_p(f,g)=Q_p(f-g{)}^p$ is a metric on the equivalence classes of measurable functions with finite $Q_p$, even though $Q_p(f-g)$ itself need not be a metric. Do not replace the failed Minkowski inequality by a universal reversed bound for signed or complex functions: $g=-f\ne 0$ would give a zero sum but two positive separate sizes.

A Fréchet space is a Hausdorff locally convex space whose topology is induced by a complete translation-invariant metric. Translation invariance means $d(x+z,y+z)=d(x,y)$; completeness means every Cauchy sequence converges in the space. Use a translation-invariant metric for this condition: an arbitrary compatible metric can change which sequences are Cauchy sequences. ^[ref].

Equivalently, the topology comes from a countable separating family of seminorms $(p_k{)}_{k\ge 0}$ and is complete in their common sense: if $p_k(x_n-x_m)\to 0$ as $n,m\to \infty$ for every fixed $k$, then some $x\in E$ satisfies $p_k(x_n-x)\to 0$ for every $k$. A useful compatible translation-invariant metric isIt is finite because the weights sum to one. The separating condition makes it a metric; without that condition the formula gives only a pseudometric. Its convergence in a metric space and Cauchy sequence conditions are exactly the conditions for all the $p_k$ separately: control finitely many terms, then bound the remaining tail by $2^{-(K+1)}$. Replacing $p_k$ by ${\max }_{0\le j\le k}{p}_j$ gives an increasing family with the same topology.

Every Banach space is a Fréchet space, using its norm as the single measurement. The space of smooth functions on a compact interval is a Fréchet space with derivative seminorms but with no single norm inducing its smooth topology. This is why Fréchet spaces are useful in differential equations: one can require accuracy in every order of derivative, while each particular estimate uses only finitely many orders.

The name honors Maurice Fréchet. His doctoral thesis was published as ^[ref]; an ^[ref] is available. This identifies a historical source without attributing the modern locally convex definition to that paper.

The two-sided full shift on a finite alphabet $[k]$ consists of all functions $z:\mathbb{Z}\to [k]$, with the product topology and the left shift. It is a compact metric space. A compatible metric isAgreement on increasingly large finite coordinate sets is equivalent to convergence in this topology. The metric above is compatible but is not invariant under the left shift.

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.