Metric Spaces and Topology

Connectedness in topological spaces

A topological space X is connected if it cannot be written as the union of two disjoint nonempty open sets, and such a pair of sets is called a separation. Equivalently, X is connected exactly when \emptyset and X are the only subsets that are both open and closed. Connectedness depends on which sets are declared open, so one underlying set can be connected in one topology and disconnected in another. In the upper limit topology on \mathbb{R}, generated by half-open intervals (a,b], the sets (-\infty,c] and (c,\infty) are open, disjoint and cover the line, so that space is disconnected. By contrast, in the cofinite topology any two nonempty open sets have finite complements and therefore intersect, which forces connectedness. A continuous image of a connected space is connected, which is the topological reason behind the intermediate value theorem. A union of connected sets that share a common point is connected. If A is connected and A\subseteq C\subseteq\overline{A}, then C is connected. Two sets that merely touch at a point can have disjoint interiors, so the union of the interiors may be disconnected even when the union of the sets is connected. The real line modulo the integers is compact, being the continuous image of [0,1], and removing one point from a circle leaves a connected open arc. A cone with vertex at the origin is connected because each point is joined to the vertex by a segment lying in the cone. A two-sheeted hyperboloid, the level set of a quadratic form at a negative value, splits into two components according to the sign of the last coordinate. For a continuous real function on the circle, the image is a compact connected set, that is, a closed bounded interval. Applying the intermediate value theorem to g(\theta)=f(\cos\theta,\sin\theta)-f(-\cos\theta,-\sin\theta), which satisfies g(\pi)=-g(0), produces a zero of g. An unbounded union of compact pieces is not compact, and a union of disjoint clopen pieces is disconnected. A family of circles whose centres escape to infinity is therefore neither compact nor connected.

Comparison of topologies on R

For two topologies \tau_1 and \tau_2 on the same set, \tau_1 is finer than \tau_2 when \tau_2\subseteq\tau_1, and coarser when the inclusion is reversed. The intersection of two topologies is again a topology, and it can collapse to the indiscrete topology. If one topology requires every nonempty open set to contain a fixed point 1, and another requires every nonempty open set to omit 1, then only \emptyset and the whole set survive. In that case the closure of any nonempty set is the whole space. On \mathbb{R}, the three topologies form the chain cofinite \subseteq standard \subseteq upper limit. Cofinite sets are standard-open because their complements are finite, hence closed. Each standard open interval is a union of half-open intervals (c,x], so it is open in the upper limit topology. Compactness restricts comparisons. If a topology is compact and Hausdorff, no strictly finer topology is compact, and no strictly coarser topology is Hausdorff. The reason is that the identity map is a continuous bijection from a compact space onto a Hausdorff space, and such a map is a homeomorphism, forcing the two topologies to be equal.

Compact sets in metric spaces

A subset of a metric space is compact if every open cover has a finite subcover. In a metric space this is equivalent to being complete and totally bounded, and in \mathbb{R}^n the Heine-Borel theorem says compact means closed and bounded. Closedness in an ambient space does not give compactness when the ambient space is not compact. For example, a set of rational numbers whose squares lie in an open interval is closed in \mathbb{Q} but not compact, because a rational sequence converging to an irrational number has no subsequence converging inside the set. In function spaces, total boundedness is the key test. In C[0,1] with the sup metric, the monomials x^n all have norm 1, and distinct terms with large indices are at distance close to 1. The family is therefore bounded but not totally bounded. Since any limit of a sequence in it would be a discontinuous function, the family is closed in C[0,1]. A useful pattern is the unit sphere of \mathbb{R}^3 with a great circle removed. This set is two disjoint open hemispheres, so it is not closed, hence not compact, and it is disconnected. Such a set arises from a diagonalizability condition: a rank-one matrix uw^{T} is diagonalizable exactly when its trace w^{T}u is nonzero, and when the trace is zero the nonzero matrix is nilpotent.

Metric space axioms

A metric on a set X is a function d:X\times X\to[0,\infty) such that d(x,y)=0 exactly when x=y, d is symmetric, and the triangle inequality d(x,z)\le d(x,y)+d(y,z) holds. Each axiom must be checked separately when a new formula is proposed. A standard device is that if f is injective, then d(x,y)=|f(x)-f(y)| is a metric. The identity axiom follows from injectivity, and the other axioms follow from the absolute value. For example, |\arctan x-\arctan y| is a metric on (0,\infty). Some modifications break an axiom. The squared distance (x-y)^2 fails the triangle inequality, and the formula \max(d,c) destroys the identity axiom because \max(d(x,x),c)=c>0. By contrast, the truncation \min(d,c) for a constant c>0 preserves all axioms, the bounded transform d/(1+d) of a metric is again a metric, and a positive combination such as an average of metrics is a metric.

Cofinite topology on R

The cofinite topology on a set X consists of \emptyset together with all sets whose complements are finite. Its closed sets are the finite sets and X itself, so it is the smallest topology in which every singleton is closed. Any two nonempty open sets intersect, so the space is T_1 but not Hausdorff. Every finite subset is closed. Every subset of this space is compact. Given an open cover of a subset, one member has finite complement, and only finitely many further members are needed to cover the remaining finitely many points. The whole real line with this topology is therefore compact and connected. The space is not first countable. A countable family of neighbourhoods of a point contains only countably many finite complements, so some other point y lies in every member. The open set \mathbb{R}\setminus\{y\} then contains the point but no member of the family. Since second countability implies first countability, the space is not second countable either.

Subspace topology and homeomorphisms

The subspace topology on A\subseteq X consists of all sets U\cap A with U open in X. A set that is open in a subspace need not be open in the ambient space. For example, [0,2) is open in [0,5] with the usual metric, because every small ball about 0 inside [0,5] is a set [0,\varepsilon) lying in the set. But [0,2) is not open in \mathbb{R}, since every ball about 0 contains negative numbers. A homeomorphism preserves compactness, connectedness, path-connectedness and the number of path components. These invariants distinguish spaces that look alike. A unit sphere is compact, while a union of coordinate planes is not, so the two are not homeomorphic. When a set is a union of level sets of a continuous function with values in \mathbb{Q}, any path in it keeps that function constant. The image of a path is a connected subset of \mathbb{Q}, hence a single point. So the path components are the level sets, and an infinite number of them rules out homeomorphism with a path-connected space.

Proper continuous maps

A continuous map f:X\to Y is proper if f^{-1}(K) is compact for every compact K\subseteq Y. Properness prevents compact sets from being pulled back to unbounded or non-compact sets. A linear invertible map of \mathbb{R}^2 is proper, because the preimage of a compact set K is the image of K under the continuous inverse, which is compact. Projections and inclusions of non-compact spaces are typically not proper. The projection \mathbb{R}^2\to\mathbb{R} onto the first coordinate pulls back [0,1] to a strip. The inclusion of \mathbb{R}\setminus\{0\} into \mathbb{R} pulls back [-1,1] to [-1,1]\setminus\{0\}, which is not compact. Quadratic functions show the same distinction. The function (x_1-x_2)^2 vanishes on an unbounded line, so its zero set is not compact. A positive definite quadratic form g satisfies g(x)\ge c\|x\|^2 for some c>0, so preimages of compact sets are bounded and closed, hence compact.

Topology generated by a basis

A collection \mathcal B of subsets of X is a basis for a topology when \mathcal B covers X and the intersection of two members is a union of members. The generated topology consists of all unions of members. Arithmetic progressions give an example on \mathbb{Z}: the sets S(a,b)=\{an+b : n\in\mathbb{Z}\} with a\ne 0 form a basis, because two such progressions meet in an empty set or in another progression, by the Chinese remainder theorem. Each progression is clopen, since its complement is a union of the other residue classes with the same difference, and these are basic open sets. Consequently, for distinct integers x and y with d=x-y, the progression through x with difference 2d is clopen, contains x and omits y. So no connected set contains two points. A second standard basis on \mathbb{R} is the family of half-open intervals [a,b). The open interval (a,b) is the union of the basic sets [c,b) with a<c<b, so it is open. The topology is first countable, because \{[x,x+1/n) : n\ge 1\} is a countable local base at x. It is not second countable, since a base must contain for each real x a member whose least element is x, which forces uncountably many members. A closed interval is not compact. The cover \{[2,x) : 2<x<5\}\cup\{[5,6)\} of [2,5] has no finite subcover.

Separability and metrizability

A space is separable if it has a countable dense subset, and metrizable if some metric induces its topology. A compact metric space is separable. For each n, it is covered by finitely many balls of radius 1/n, and choosing one point from each ball gives a countable set that is 1/n-dense for every n. In a Hausdorff space, each singleton is the intersection of finitely many open neighbourhoods of its point, so a finite Hausdorff space is discrete. The discrete metric induces the discrete topology, so every finite Hausdorff topology is metrizable. The Hausdorff condition cannot be dropped in the second claim. The indiscrete topology on a two-point set is not metrizable, because every metrizable space is Hausdorff.

Quotient maps and open/closed maps

Let p:X\to Y be a surjection. The quotient topology on Y declares V open exactly when p^{-1}(V) is open in X, so it is the finest topology making p continuous. A map is open if it sends open sets to open sets, and closed if it sends closed sets to closed sets. Neither property follows from continuity or surjectivity. To test a map, take a set in the domain and check whether its image is open or closed in the target. For example, consider a map from an interval onto a two-point set given by the indicator of a closed subinterval. An open subset lying entirely in the upper piece maps to a singleton that is not open in the quotient topology, so the map is not open. A closed subset of the lower piece maps to a set whose complement is not open, so the map is not closed.

Product topology and local compactness

The product topology on \prod_n X_n has as basis the sets \prod_n U_n in which U_n=X_n for all but finitely many indices. A basic open set therefore constrains only finitely many coordinates, which makes the product topology coarser than the box topology. In \mathbb{R}^{\mathbb{N}}, the diagonal \Delta=\{(x,x,\dots) : x\in\mathbb{R}\} is closed and has empty interior. It is homeomorphic to \mathbb{R} through the first coordinate projection, which is continuous on \Delta. A space is locally compact if every point has a compact neighbourhood. Since \mathbb{R} is locally compact and connected, the diagonal inherits both properties through the homeomorphism. Being a proper closed subset, it is not dense.

Topology generated by a subbase

A subbase \mathcal S generates the smallest topology containing it. Its basis consists of the finite intersections of members of \mathcal S together with the whole space. On \mathbb{Z}, taking as subbase all residue classes m+n\mathbb{Z} with n\ne 0 gives a topology in which each class is closed, since its complement is a union of other classes of the same modulus, which are open. Such a space is Hausdorff. For distinct integers x,y with d=|x-y|, the classes x+2d\mathbb{Z} and y+2d\mathbb{Z} are disjoint open neighbourhoods. The countable family of classes forms a base, so the space is second countable. It is also regular, and Urysohn's metrization theorem then gives metrizability. The space is not connected. The even integers and the odd integers form two disjoint nonempty open sets covering \mathbb{Z}.