Base (topology)

Given a topological space ${\displaystyle (X,\tau )}$, a base[2] (or basis[3]) for the topology ${\displaystyle \tau }$ (also called a base for ${\displaystyle X}$ if the topology is understood) is a family ${\displaystyle {\mathcal {B}}\subseteq \tau }$ of open sets such that every open set of the topology can be represented as the union of some subfamily of ${\displaystyle {\mathcal {B}}}$.[note 1] The elements of ${\displaystyle {\mathcal {B}}}$ are called basic open sets. Equivalently, a family ${\displaystyle {\mathcal {B}}}$ of subsets of ${\displaystyle X}$ is a base for the topology ${\displaystyle \tau }$ if and only if ${\displaystyle {\mathcal {B}}\subseteq \tau }$ and for every open set ${\displaystyle U}$ in ${\displaystyle X}$ and point ${\displaystyle x\in U}$ there is some basic open set ${\displaystyle B\in {\mathcal {B}}}$ such that ${\displaystyle x\in B\subseteq U}$.

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 ${\displaystyle M}$ the collection of all open balls about points of ${\displaystyle M}$ forms a base for the topology.

In general, a topological space ${\displaystyle (X,\tau )}$ can have many bases. The whole topology ${\displaystyle \tau }$ is always a base for itself (that is, ${\displaystyle \tau }$ is a base for ${\displaystyle \tau }$). 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 ${\displaystyle X}$ is the minimum cardinality of a base for its topology, called the weight of ${\displaystyle X}$ and denoted ${\displaystyle w(X)}$. From the examples above, the real line has countable weight.

If ${\displaystyle {\mathcal {B}}}$ is a base for the topology ${\displaystyle \tau }$ of a space ${\displaystyle X}$, it satisfies the following properties:[4]

(B1) The elements of ${\displaystyle {\mathcal {B}}}$ cover ${\displaystyle X}$, i.e., every point ${\displaystyle x\in X}$ belongs to some element of ${\displaystyle {\mathcal {B}}}$.

(B2) For every ${\displaystyle B_{1},B_{2}\in {\mathcal {B}}}$ and every point ${\displaystyle x\in B_{1}\cap B_{2}}$, there exists some ${\displaystyle B_{3}\in {\mathcal {B}}}$ such that ${\displaystyle x\in B_{3}\subseteq B_{1}\cap B_{2}}$.

Property (B1) corresponds to the fact that ${\displaystyle X}$ is an open set; property (B2) corresponds to the fact that ${\displaystyle B_{1}\cap B_{2}}$ is an open set.

Conversely, suppose ${\displaystyle X}$ is just a set without any topology and ${\displaystyle {\mathcal {B}}}$ is a family of subsets of ${\displaystyle X}$ satisfying properties (B1) and (B2). Then ${\displaystyle {\mathcal {B}}}$ is a base for the topology that it generates. More precisely, let ${\displaystyle \tau }$ be the family of all subsets of ${\displaystyle X}$ that are unions of subfamilies of ${\displaystyle {\mathcal {B}}.}$ Then ${\displaystyle \tau }$ is a topology on ${\displaystyle X}$ and ${\displaystyle {\mathcal {B}}}$ is a base for ${\displaystyle \tau }$.[5] (Sketch: ${\displaystyle \tau }$ defines a topology because it is stable under arbitrary unions by construction, it is stable under finite intersections by (B2), it contains ${\displaystyle X}$ by (B1), and it contains the empty set as the union of the empty subfamily of ${\displaystyle {\mathcal {B}}}$. The family ${\displaystyle {\mathcal {B}}}$ is then a base for ${\displaystyle \tau }$ by construction.) Such families of sets are a very common way of defining a topology.

In general, if ${\displaystyle X}$ is a set and ${\displaystyle {\mathcal {B}}}$ is an arbitrary collection of subsets of ${\displaystyle X}$, there is a (unique) smallest topology ${\displaystyle \tau }$ on ${\displaystyle X}$ containing ${\displaystyle {\mathcal {B}}}$. (This topology is the intersection of all topologies on ${\displaystyle X}$ containing ${\displaystyle {\mathcal {B}}}$.) The topology ${\displaystyle \tau }$ is called the topology generated by ${\displaystyle {\mathcal {B}}}$, and ${\displaystyle {\mathcal {B}}}$ is called a subbase for ${\displaystyle \tau }$. The topology ${\displaystyle \tau }$ can also be characterized as the set of all arbitrary unions of finite intersections of elements of ${\displaystyle {\mathcal {B}}}$. (See the article about subbase.) Now, if ${\displaystyle {\mathcal {B}}}$ also satisfies properties (B1) and (B2), the topology generated by ${\displaystyle {\mathcal {B}}}$ can be described in a simpler way without having to take intersections: ${\displaystyle \tau }$ is the set of all unions of elements of ${\displaystyle {\mathcal {B}}}$ (and ${\displaystyle {\mathcal {B}}}$ is base for ${\displaystyle \tau }$ in that case).

There is often an easy way to check condition (B2). If the intersection of any two elements of ${\displaystyle {\mathcal {B}}}$ is itself an element of ${\displaystyle {\mathcal {B}}}$ or is empty, then condition (B2) is automatically satisfied (by taking ${\displaystyle B_{3}=B_{1}\cap B_{2}}$). 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 ${\displaystyle S}$ of all semi-infinite intervals of the forms ${\displaystyle (-\infty ,a)}$ and ${\displaystyle (a,\infty )}$ with ${\displaystyle a\in \mathbb {R} }$. The topology generated by ${\displaystyle S}$ contains all open intervals ${\displaystyle (a,b)=(-\infty ,b)\cap (a,\infty )}$, hence ${\displaystyle S}$ generates the standard topology on the real line. But ${\displaystyle S}$ is only a subbase for the topology, not a base: a finite open interval ${\displaystyle (a,b)}$ does not contain any element of ${\displaystyle S}$ (equivalently, property (B2) does not hold).

The set Γ of all open intervals in ${\displaystyle \mathbb {R} }$ forms a basis for the Euclidean topology on ${\displaystyle \mathbb {R} }$.

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 π-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 π-system and so also a base for a topology. In fact, if Γ is a filter on X then { ∅ } ∪ Γ is a topology on X and Γ 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 ${\displaystyle \mathbb {R} }$ is closed under finite intersections and so each forms a basis for some topology on ${\displaystyle \mathbb {R} }$:

• The set Γ of all bounded open intervals in ${\displaystyle \mathbb {R} }$ generates the usual Euclidean topology on ${\displaystyle \mathbb {R} }$.
• The set Σ of all bounded closed intervals in ${\displaystyle \mathbb {R} }$ generates the discrete topology on ${\displaystyle \mathbb {R} }$ and so the Euclidean topology is a subset of this topology. This is despite the fact that Γ is not a subset of Σ. Consequently, the topology generated by Γ, which is the Euclidean topology on ${\displaystyle \mathbb {R} }$, is coarser than the topology generated by Σ. In fact, it is strictly coarser because Σ contains non-empty compact sets which are never open in the Euclidean topology.
• The set Γ_{${\displaystyle \mathbb {Q} }$} of all intervals in Γ such that both endpoints of the interval are rational numbers generates the same topology as Γ. This remains true if each instance of the symbol Γ is replaced by Σ.
• Σ_∞ = { [r, ∞) : r ∈ ${\displaystyle \mathbb {R} }$ } generates a topology that is strictly coarser than the topology generated by Σ. No element of Σ_∞ is open in the Euclidean topology on ${\displaystyle \mathbb {R} }$.
• Γ_∞ = { (r, ∞) : r ∈ ${\displaystyle \mathbb {R} }$ } generates a topology that is strictly coarser than both the Euclidean topology and the topology generated by Σ_∞. The sets Σ_∞ and Γ_∞ are disjoint, but nevertheless Γ_∞ is a subset of the topology generated by Σ_∞.

• A topology ${\displaystyle \tau _{2}}$ is finer than a topology ${\displaystyle \tau _{1}}$ if and only if for each ${\displaystyle x\in X}$ and each basic open set ${\displaystyle B}$ of ${\displaystyle \tau _{1}}$ containing ${\displaystyle x}$, there is a basic open set of ${\displaystyle \tau _{2}}$ containing ${\displaystyle x}$ and contained in ${\displaystyle B}$.
• If ${\displaystyle {\mathcal {B}}_{1},\ldots ,{\mathcal {B}}_{n}}$ are bases for the topologies ${\displaystyle \tau _{1},\ldots ,\tau _{n}}$ then the collection of all set products ${\displaystyle B_{1}\times \cdots \times B_{n}}$ with each ${\displaystyle B_{i}\in {\mathcal {B}}_{i}}$ is a base for the product topology ${\displaystyle \tau _{1}\times \cdots \times \tau _{n}.}$ 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 ${\displaystyle {\mathcal {B}}}$ be a base for ${\displaystyle X}$ and let ${\displaystyle Y}$ be a subspace of ${\displaystyle X}$. Then if we intersect each element of ${\displaystyle {\mathcal {B}}}$ with ${\displaystyle Y}$, the resulting collection of sets is a base for the subspace ${\displaystyle Y}$.
• If a function ${\displaystyle f:X\to Y}$ maps every basic open set of ${\displaystyle X}$ into an open set of ${\displaystyle Y}$, it is an open map. Similarly, if every preimage of a basic open set of ${\displaystyle Y}$ is open in ${\displaystyle X}$, then ${\displaystyle f}$ is continuous.
• ${\displaystyle {\mathcal {B}}}$ is a base for a topological space ${\displaystyle X}$ if and only if the subcollection of elements of ${\displaystyle {\mathcal {B}}}$ which contain ${\displaystyle x}$ form a local base at ${\displaystyle x}$, for any point ${\displaystyle x\in X}$.

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 ${\displaystyle X,}$ a family ${\displaystyle {\mathcal {C}}}$ of closed sets forms a base for the closed sets if and only if for each closed set ${\displaystyle A}$ and each point ${\displaystyle x}$ not in ${\displaystyle A}$ there exists an element of ${\displaystyle {\mathcal {C}}}$ containing ${\displaystyle A}$ but not containing ${\displaystyle x.}$ A family ${\displaystyle {\mathcal {C}}}$ is a base for the closed sets of ${\displaystyle X}$ if and only if its dual in ${\displaystyle X,}$ that is the family ${\displaystyle \{X\setminus C:C\in {\mathcal {C}}\}}$ of complements of members of ${\displaystyle {\mathcal {C}}}$, is a base for the open sets of ${\displaystyle X.}$

Let ${\displaystyle {\mathcal {C}}}$ be a base for the closed sets of ${\displaystyle X.}$ Then

1. ${\displaystyle \bigcap {\mathcal {C}}=\varnothing }$
2. For each ${\displaystyle C_{1},C_{2}\in {\mathcal {C}}}$ the union ${\displaystyle C_{1}\cup C_{2}}$ is the intersection of some subfamily of ${\displaystyle {\mathcal {C}}}$ (that is, for any ${\displaystyle x\in X}$ not in ${\displaystyle C_{1}{\text{ or }}C_{2}}$ there is some ${\displaystyle C_{3}\in {\mathcal {C}}}$ containing ${\displaystyle C_{1}\cup C_{2}}$ and not containing ${\displaystyle x}$).

Any collection of subsets of a set ${\displaystyle X}$ satisfying these properties forms a base for the closed sets of a topology on ${\displaystyle X.}$ The closed sets of this topology are precisely the intersections of members of ${\displaystyle {\mathcal {C}}.}$

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 ${\displaystyle X,}$ the zero sets form the base for the closed sets of some topology on ${\displaystyle X.}$ This topology will be the finest completely regular topology on ${\displaystyle X}$ coarser than the original one. In a similar vein, the Zariski topology on Aⁿ is defined by taking the zero sets of polynomial functions as a base for the closed sets.

We shall work with notions established in (Engelking 1989, p. 12, pp. 127-128).

Fix X a topological space. Here, a network is a family ${\displaystyle {\mathcal {N}}}$ of sets, for which, for all points x and open neighbourhoods U containing x, there exists B in ${\displaystyle {\mathcal {N}}}$ for which ${\displaystyle x\in B\subseteq U.}$ 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, ${\displaystyle \chi (x,X),}$ as the minimum cardinality of a neighbourhood basis for x in X; and the character of X to be

$${\displaystyle \chi (X)\triangleq \sup\{\chi (x,X):x\in X\}.}$$

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) ≤ 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 ${\displaystyle B'\subseteq B}$ of size ${\displaystyle |B'|\leq w(X).}$
• if N a neighbourhood basis for x in X then there is a neighbourhood basis ${\displaystyle N'\subseteq N}$ of size ${\displaystyle |N'|\leq \chi (x,X).}$
• if ${\displaystyle f:X\to Y}$ is a continuous surjection, then nw(Y) ≤ w(X). (Simply consider the Y-network ${\displaystyle f'''B\triangleq \{f''U:U\in B\}}$ for each basis B of X.)
• if ${\displaystyle (X,\tau )}$ is Hausdorff, then there exists a weaker Hausdorff topology ${\displaystyle (X,\tau ')}$ so that ${\displaystyle w(X,\tau ')\leq nw(X,\tau ).}$ 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 ${\displaystyle f:X\to Y}$ 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 ${\displaystyle nw(f(X))=w(f(X))\leq w(X)\leq \aleph _{0}}$ (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) ≤ κ some infinite cardinal. Then there does not exist a strictly increasing sequence of open sets (equivalently strictly decreasing sequence of closed sets) of length ≥ κ⁺.

To see this (without the axiom of choice), fix

$${\displaystyle \left\{U_{\xi }\right\}_{\xi \in \kappa },}$$

as a basis of open sets. And suppose per contra, that

$${\displaystyle \left\{V_{\xi }\right\}_{\xi \in \kappa ^{+}}}$$

were a strictly increasing sequence of open sets. This means

$${\displaystyle \forall \alpha <\kappa ^{+}:\qquad V_{\alpha }\setminus \bigcup _{\xi <\alpha }V_{\xi }\neq \varnothing .}$$

For

$${\displaystyle x\in V_{\alpha }\setminus \bigcup _{\xi <\alpha }V_{\xi },}$$

we may use the basis to find some U_γ with x in U_γ ⊆ V_α. In this way we may well-define a map, f : κ⁺ → κ mapping each α to the least γ for which U_γ ⊆ V_α and meets

$${\displaystyle V_{\alpha }\setminus \bigcup _{\xi <\alpha }V_{\xi }.}$$

This map is injective, otherwise there would be α < β with f(α) = f(β) = γ, which would further imply U_γ ⊆ V_α but also meets

$${\displaystyle V_{\beta }\setminus \bigcup _{\xi <\alpha }V_{\xi }\subseteq V_{\beta }\setminus V_{\alpha },}$$

which is a contradiction. But this would go to show that κ⁺ ≤ κ, a contradiction.

• Esenin-Volpin's theorem
• Gluing axiom
• Neighbourhood system