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Base (topology)

In mathematics, a base for the topology of a topological space is a family of open subsets of such that every open set of the topology is equal to the union of some sub-family of . For example, the set of all open intervals in the real number line is a basis for the Euclidean topology on because every open interval is an open set, and also every open subset of can be written as a union of some family of open intervals.

Definition and basic properties
Given a topological space (X,\tau), a base (or basis) for the topology \tau (also called a base for X if the topology is understood) is a family \mathcal{B}\subseteq\tau of open sets such that every open set of the topology can be represented as the union of some subfamily of \mathcal{B}. The elements of \mathcal{B} are called basic open sets. Equivalently, a family \mathcal{B} of subsets of X is a base for the topology \tau if and only if \mathcal{B}\subseteq\tau and for every open set U in X and point x\in U there is some basic open set B\in\mathcal{B} such that x\in B\subseteq U. For example, the collection of all open intervals in the real line forms a base for the standard topology on the real numbers. More generally, in a metric space M the collection of all open balls about points of M forms a base for the topology. In general, a topological space (X,\tau) can have many bases. The whole topology \tau is always a base for itself (that is, \tau is a base for \tau). For the real line, the collection of all open intervals is a base for the topology. So is the collection of all open intervals with rational endpoints, or the collection of all open intervals with irrational endpoints, for example. Note that two different bases need not have any basic open set in common. One of the topological properties of a space X is the minimum cardinality of a base for its topology, called the weight of X and denoted w(X). From the examples above, the real line has countable weight. If \mathcal{B} is a base for the topology \tau of a space X, it satisfies the following properties: :(B1) The elements of \mathcal{B} cover X, i.e., every point x\in X belongs to some element of \mathcal{B}. :(B2) For every B_1,B_2\in\mathcal{B} and every point x\in B_1\cap B_2, there exists some B_3\in\mathcal{B} such that x\in B_3\subseteq B_1\cap B_2. Property (B1) corresponds to the fact that X is an open set; property (B2) corresponds to the fact that B_1\cap B_2 is an open set. Conversely, suppose X is just a set without any topology and \mathcal{B} is a family of subsets of X satisfying properties (B1) and (B2). Then \mathcal{B} is a base for the topology that it generates. More precisely, let \tau be the family of all subsets of X that are unions of subfamilies of \mathcal{B}. Then \tau is a topology on X and \mathcal{B} is a base for \tau. (Sketch: \tau defines a topology because it is stable under arbitrary unions by construction, it is stable under finite intersections by (B2), it contains X by (B1), and it contains the empty set as the union of the empty subfamily of \mathcal{B}. The family \mathcal{B} is then a base for \tau by construction.) Such families of sets are a very common way of defining a topology. In general, if X is a set and \mathcal{B} is an arbitrary collection of subsets of X, there is a (unique) smallest topology \tau on X containing \mathcal{B}. (This topology is the intersection of all topologies on X containing \mathcal{B}.) The topology \tau is called the topology generated by \mathcal{B}, and \mathcal{B} is called a subbase for \tau. The topology \tau consists of X together with all arbitrary unions of finite intersections of elements of \mathcal{B} (see the article about subbase.) Now, if \mathcal{B} also satisfies properties (B1) and (B2), the topology generated by \mathcal{B} can be described in a simpler way without having to take intersections: \tau is the set of all unions of elements of \mathcal{B} (and \mathcal{B} is a base for \tau in that case). There is often an easy way to check condition (B2). If the intersection of any two elements of \mathcal{B} is itself an element of \mathcal{B} or is empty, then condition (B2) is automatically satisfied (by taking B_3=B_1\cap B_2). For example, the Euclidean topology on the plane admits as a base the set of all open rectangles with horizontal and vertical sides, and a nonempty intersection of two such basic open sets is also a basic open set. But another base for the same topology is the collection of all open disks; and here the full (B2) condition is necessary. An example of a collection of open sets that is not a base is the set S of all semi-infinite intervals of the forms (-\infty,a) and (a,\infty) with a\in\mathbb{R}. The topology generated by S contains all open intervals (a,b)=(-\infty,b)\cap(a,\infty), hence S generates the standard topology on the real line. But S is only a subbase for the topology, not a base: a finite open interval (a,b) does not contain any element of S (equivalently, property (B2) does not hold). ==Examples==
Examples
The set \Gamma of all open intervals in \mathbb{R} forms a basis for the Euclidean topology on \mathbb{R}. A non-empty family of subsets of a set X that is closed under finite intersections of two or more sets, which is called a \pi-system on X, is necessarily a base for a topology on X if and only if it covers X. By definition, every σ-algebra, every filter (and so in particular, every neighborhood filter), and every topology is a covering \pi-system and so also a base for a topology. In fact, if \Gamma is a filter on X then \{\varnothing\}\cup\Gamma is a topology on X and \Gamma is a basis for it. A base for a topology does not have to be closed under finite intersections and many are not. But nevertheless, many topologies are defined by bases that are also closed under finite intersections. For example, each of the following families of subset of \mathbb{R} is closed under finite intersections and so each forms a basis for some topology on \mathbb{R}: • The set \Gamma of all bounded open intervals in \mathbb{R} generates the usual Euclidean topology on \mathbb{R}. • The set \Sigma of all bounded closed intervals in \mathbb{R} generates the discrete topology on \mathbb{R} and so the Euclidean topology is a subset of this topology. This is despite the fact that \Gamma is not a subset of \Sigma. Consequently, the topology generated by \Gamma, which is the Euclidean topology on \mathbb{R}, is coarser than the topology generated by \Sigma. In fact, it is strictly coarser because \Sigma contains non-empty compact sets which are never open in the Euclidean topology. • The set \Gamma_\Q of all intervals in \Gamma such that both endpoints of the interval are rational numbers generates the same topology as \Gamma. This remains true if each instance of the symbol \Gamma is replaced by \Sigma. • \Sigma_\infty = \{[r,\infty): r\in\R\} generates a topology that is strictly coarser than the topology generated by \Sigma. No element of \Sigma_\infty is open in the Euclidean topology on \mathbb{R}. • \Gamma_\infty = \{[r,\infty): r\in\R\} generates a topology that is strictly coarser than both the Euclidean topology and the topology generated by \Sigma_\infty. The sets \Sigma_\infty and \Gamma_\infty are disjoint, but nevertheless \Gamma_\infty is a subset of the topology generated by \Sigma_\infty. Objects defined in terms of bases • The order topology on a totally ordered set admits a collection of open-interval-like sets as a base. • In a metric space the collection of all open balls forms a base for the topology. • The discrete topology has the collection of all singletons as a base. • A second-countable space is one that has a countable base. The Zariski topology on the spectrum of a ring has a base consisting of open sets that have specific useful properties. For the usual base for this topology, every finite intersection of basic open sets is a basic open set. • The Zariski topology of \C^n is the topology that has the algebraic sets as closed sets. It has a base formed by the set complements of algebraic hypersurfaces. • The Zariski topology of the spectrum of a ring (the set of the prime ideals) has a base such that each element consists of all prime ideals that do not contain a given element of the ring. ==Theorems==
Theorems
• A topology \tau_2 is finer than a topology \tau_1 if and only if for each x\in X and each basic open set B of \tau_1 containing x, there is a basic open set of \tau_2 containing x and contained in B. • If \mathcal{B}_1, \ldots, \mathcal{B}_n are bases for the topologies \tau_1, \ldots, \tau_n then the collection of all set products B_1 \times \cdots \times B_n with each B_i\in\mathcal{B}_i is a base for the product topology \tau_1 \times \cdots \times \tau_n. In the case of an infinite product, this still applies, except that all but finitely many of the base elements must be the entire space. • Let \mathcal{B} be a base for X and let Y be a subspace of X. Then if we intersect each element of \mathcal{B} with Y, the resulting collection of sets is a base for the subspace Y. • If a function f : X \to Y maps every basic open set of X into an open set of Y, it is an open map. Similarly, if every preimage of a basic open set of Y is open in X, then f is continuous. • \mathcal{B} is a base for a topological space X if and only if the subcollection of elements of \mathcal{B} which contain x form a local base at x, for any point x\in X. ==Base for the closed sets==
Base for the closed sets
Closed sets are equally adept at describing the topology of a space. There is, therefore, a dual notion of a base for the closed sets of a topological space. Given a topological space X, a family \mathcal{C} of closed sets forms a base for the closed sets if and only if for each closed set A and each point x not in A there exists an element of \mathcal{C} containing A but not containing x. A family \mathcal{C} is a base for the closed sets of X if and only if its in X, that is the family \{X\setminus C: C\in \mathcal{C}\} of complements of members of \mathcal{C}, is a base for the open sets of X. Let \mathcal{C} be a base for the closed sets of X. Then • \bigcap \mathcal{C} = \varnothing • For each C_1, C_2 \in \mathcal{C} the union C_1 \cup C_2 is the intersection of some subfamily of \mathcal{C} (that is, for any x \in X not in C_1 \text{ or } C_2 there is some C_3 \in \mathcal{C} containing C_1 \cup C_2 and not containing x). Any collection of subsets of a set X satisfying these properties forms a base for the closed sets of a topology on X. The closed sets of this topology are precisely the intersections of members of \mathcal{C}. In some cases it is more convenient to use a base for the closed sets rather than the open ones. For example, a space is completely regular if and only if the zero sets form a base for the closed sets. Given any topological space X, the zero sets form the base for the closed sets of some topology on X. This topology will be the finest completely regular topology on X coarser than the original one. In a similar vein, the Zariski topology on A^n is defined by taking the zero sets of polynomial functions as a base for the closed sets. ==Weight and character==
Weight and character
We shall work with notions established in . Fix X a topological space. Here, a network is a family \mathcal{N} of sets, for which, for all points x and open neighbourhoods U containing x, there exists B in \mathcal{N} for which x \in B \subseteq U. Note that, unlike a basis, the sets in a network need not be open. We define the weight, w(X), as the minimum cardinality of a basis; we define the network weight, nw(X), as the minimum cardinality of a network; the character of a point, \chi(x,X), as the minimum cardinality of a neighbourhood basis for x in X; and the character of X to be \chi(X)\triangleq\sup\{\chi(x,X):x\in X\}. The point of computing the character and weight is to be able to tell what sort of bases and local bases can exist. We have the following facts: • nw(X) \leq w(X). • if X is discrete, then w(X) = nw(X) = |X|. • if X is Hausdorff, then nw(X) is finite if and only if X is finite discrete. • if B is a basis of X then there is a basis B'\subseteq B of size |B'|\leq w(X). • if N is a neighbourhood basis for x in X then there is a neighbourhood basis N'\subseteq N of size |N'|\leq \chi(x,X). • if f : X \to Y is a continuous surjection, then nw(Y) \leq w(X). (Simply consider the Y-network fB \triangleq \{f(U) : U \in B\} for each basis B of X.) • if (X,\tau) is Hausdorff, then there exists a weaker Hausdorff topology (X,\tau') so that w(X,\tau')\leq nw(X,\tau). So a fortiori, if X is also compact, then such topologies coincide and hence we have, combined with the first fact, nw(X) = w(X). • if f : X \to Y a continuous surjective map from a compact metrizable space to an Hausdorff space, then Y is compact metrizable. The last fact follows from f(X) being compact Hausdorff, and hence nw(f(X))=w(f(X))\leq w(X)\leq\aleph_0 (since compact metrizable spaces are necessarily second countable); as well as the fact that compact Hausdorff spaces are metrizable exactly in case they are second countable. (An application of this, for instance, is that every path in a Hausdorff space is compact metrizable.) Increasing chains of open sets Using the above notation, suppose that w(X) \leq \kappa some infinite cardinal. Then there does not exist a strictly increasing sequence of open sets (equivalently strictly decreasing sequence of closed sets) of length \leq \kappa^+\!. To see this (without the axiom of choice), fix \left \{ U_{\xi} \right \}_{\xi\in\kappa}, as a basis of open sets. And suppose per contra, that \left \{ V_{\xi}\right \}_{\xi\in\kappa^{+}} were a strictly increasing sequence of open sets. This means \forall \alpha For x\in V_{\alpha}\setminus\bigcup_{\xi we may use the basis to find some U_\gamma with x in U_\gamma \subseteq V_\alpha. In this way we may well-define a map, f : \kappa^+ \!\to \kappa mapping each \alpha to the least \gamma for which U_\gamma \subseteq V_\alpha and meets V_{\alpha} \setminus \bigcup_{\xi This map is injective, otherwise there would be \alpha with f(\alpha) = f(\beta) = \gamma, which would further imply U_\gamma \subseteq V_\alpha but also meets V_{\beta} \setminus \bigcup_{\xi which is a contradiction. But this would go to show that \kappa^+ \!\leq \kappa, a contradiction. ==See also==
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