Base (topology)

Examples

The set ${\displaystyle \mathrm{\Gamma }}$ 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 ${\displaystyle X}$ that is closed under finite intersections of two or more sets, which is called a ${\displaystyle \pi }$-system on ${\displaystyle X}$, is necessarily a base for a topology on ${\displaystyle X}$ if and only if it covers ${\displaystyle X}$. By definition, every σ-algebra, every filter (and so in particular, every neighborhood filter), and every topology is a covering ${\displaystyle \pi }$-system and so also a base for a topology. In fact, if ${\displaystyle \mathrm{\Gamma }}$ is a filter on ${\displaystyle X}$ then ${\displaystyle \{\varnothing \}\cup \mathrm{\Gamma }}$ is a topology on ${\displaystyle X}$ and ${\displaystyle \mathrm{\Gamma }}$ 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 ${\displaystyle \mathbb{R}}$ is closed under finite intersections and so each forms a basis for some topology on ${\displaystyle \mathbb{R}}$:

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

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 ${\displaystyle {\mathbb{C}}^n}$ 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

• 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 ${\displaystyle {\mathcal{B}}_1,\dots ,{\mathcal{B}}_n}$ are bases for the topologies ${\displaystyle {\tau }_1,\dots ,{\tau }_n}$ 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 ${\displaystyle x\in X}$.

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 ${\displaystyle X,}$ 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 ${\displaystyle A^n}$ is defined by taking the zero sets of polynomial functions as a base for the closed sets.

Weight and character

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

Fix ${\displaystyle X}$ a topological space. Here, a network is a family ${\displaystyle \mathcal{N}}$ of sets, for which, for all points ${\displaystyle x}$ and open neighbourhoods U containing ${\displaystyle x}$, there exists ${\displaystyle B}$ 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, ${\displaystyle w(X)}$, as the minimum cardinality of a basis; we define the network weight, ${\displaystyle nw(X)}$, 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 ${\displaystyle x}$ in ${\displaystyle X}$; and the character of ${\displaystyle X}$ 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:

• ${\displaystyle nw(X)\le w(X)}$.
• if ${\displaystyle X}$ is discrete, then ${\displaystyle w(X)=nw(X)=|X|}$.
• if ${\displaystyle X}$ is Hausdorff, then ${\displaystyle nw(X)}$ is finite if and only if ${\displaystyle X}$ is finite discrete.
• if ${\displaystyle B}$ is a basis of ${\displaystyle X}$ then there is a basis ${\displaystyle B^{\prime }\subseteq B}$ of size ${\displaystyle |B^{\prime }|\le w(X).}$
• if ${\displaystyle N}$ is a neighbourhood basis for ${\displaystyle x}$ in ${\displaystyle X}$ then there is a neighbourhood basis ${\displaystyle N^{\prime }\subseteq N}$ of size ${\displaystyle |N^{\prime }|\le \chi (x,X).}$
• if ${\displaystyle f:X\to Y}$ is a continuous surjection, then ${\displaystyle nw(Y)\le w(X)}$. (Simply consider the ${\displaystyle Y}$-network ${\displaystyle fB\triangleq \{f(U):U\in B\}}$ for each basis ${\displaystyle B}$ of ${\displaystyle X}$.)
• if ${\displaystyle (X,\tau )}$ is Hausdorff, then there exists a weaker Hausdorff topology ${\displaystyle (X,{\tau }^{\prime })}$ so that ${\displaystyle w(X,{\tau }^{\prime })\le nw(X,\tau ).}$ So a fortiori, if ${\displaystyle X}$ is also compact, then such topologies coincide and hence we have, combined with the first fact, ${\displaystyle nw(X)=w(X)}$.
• if ${\displaystyle f:X\to Y}$ a continuous surjective map from a compact metrizable space to an Hausdorff space, then ${\displaystyle Y}$ is compact metrizable.

The last fact follows from ${\displaystyle f(X)}$ being compact Hausdorff, and hence ${\displaystyle nw(f(X))=w(f(X))\le w(X)\le {\mathrm{\aleph }}_0}$ (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 ${\displaystyle w(X)\le \kappa }$ some infinite cardinal. Then there does not exist a strictly increasing sequence of open sets (equivalently strictly decreasing sequence of closed sets) of length ${\displaystyle \le {\kappa }^{+}}$.

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 $${\displaystyle \mathrm{\forall }\alpha <{\kappa }^{+}:V_{\alpha }\setminus \bigcup _{\xi <\alpha }{V}_{\xi }\ne \varnothing .}$$

For $${\displaystyle x\in V_{\alpha }\setminus \bigcup _{\xi <\alpha }{V}_{\xi },}$$ we may use the basis to find some ${\displaystyle U_{\gamma }}$ with ${\displaystyle x}$ in ${\displaystyle U_{\gamma }\subseteq V_{\alpha }}$. In this way we may well-define a map, ${\displaystyle f:{\kappa }^{+}\to \kappa }$ mapping each ${\displaystyle \alpha }$ to the least ${\displaystyle \gamma }$ for which ${\displaystyle U_{\gamma }\subseteq V_{\alpha }}$ and meets $${\displaystyle V_{\alpha }\setminus \bigcup _{\xi <\alpha }{V}_{\xi }.}$$

This map is injective, otherwise there would be ${\displaystyle \alpha <\beta }$ with ${\displaystyle f(\alpha )=f(\beta )=\gamma }$, which would further imply ${\displaystyle U_{\gamma }\subseteq V_{\alpha }}$ 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 ${\displaystyle {\kappa }^{+}\le \kappa }$, a contradiction.