Topological bases and subbases

Overview §

Bases are convenient for defining and describing topologies without having to specify a large number of sets.

Definition: Basis for a topology

Given a set $X$, a basis for a topology on $X$ is a collection $\mathcal{B}\subseteq \mathcal{P}(X)$ of subsets of $X$ such that:

• (B1) For each $x\in X$, there exists at least one basis element $B$ containing $x$ (that is, $\mathcal{B}$ covers $X$);
• (B2) Given two basis elements $B_1,B_2$ and a point in their intersection $x\in B_1\cap B_2$, there exists a basis element $B_3$ such that $x\in B_3$ and $B_3\subseteq B_1\cap B_2$.

Equivalently, condition (B1) says that ${\bigcup }_{B\in \mathcal{B}}B=X$, while (B2) says that if $B_1,B_2\in \mathcal{B}$, then $B_1\cap B_2$ is a union of elements (sets) in $\mathcal{B}$.

Note that (B2) says every point in the intersection has an element $B\in \mathcal{B}$ that contains that point, but the intersection itself does not have to be back in $\mathcal{B}$!

Lemma (Munkres 13.3): Using bases to determine which topology is finer

Let $\mathcal{B},\mathcal{B}’$ be bases for topologies $\mathcal{T},\mathcal{T}’$ on $X$. The following are equivalent:

1. $\mathcal{T}’$ is finer than $\mathcal{T}$ (i.e., $\mathcal{T}\subseteq \mathcal{T}’$);
2. For all $U\in \mathcal{B}$ and all $x\in U$, there exists $V\in \mathcal{B}’$ such that $x\in V\subseteq U$.

Related: Basis for a subspace topology

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Bases to topologies §

There are two ways of going from a basis to a topology.

Definition: Topology generated by a basis

Given a collection $\mathcal{B}$ which satisfies the two conditions of a basis, a subset $U\subset X$ is said to be open in $X$ if for all $x\in U$, there exists a basis element $B\in \mathcal{B}$ such that $x\in B$ and $B\subset U$.

The topology $\mathcal{T}$ generated by $\mathcal{B}$ is defined as the collection of all subsets $U$ that are open in $X$:

$$\mathcal{T}=\{U\subset X|\forall x\in U\exists B\in \mathcal{B}\text{ such that }x\in B\subset U\}.$$

Equivalently, $U\in \mathcal{T}$ if and only if $U$ is a union of basis elements $B\in \mathcal{B}$ (see Munkres, Lemma 13.1).

Lemma (Munkres 13.1): Every open set can be expressed as a union of basis elements

Let $\mathcal{B}$ be a basis for a topology $\mathcal{T}$ on a set $X$. Then $\mathcal{T}$ is equal to the collection of all unions of basis elements $B\in \mathcal{B}$.

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Topologies to bases §

Lemma (Munkres 13.2): Obtaining a basis for a given topology

Let $(X,\mathcal{T})$ be a topological space and let $\mathcal{C}\subseteq \mathcal{T}$ be a collection of open sets of $X$ such that for each open set $U\subseteq X$ and each $x\in U$, there exists $C\in \mathcal{C}$ such that $x\in C\subseteq U$. Then (1) $\mathcal{C}$ is a basis and (2) $\mathcal{C}$ generates $\mathcal{T}$.

Link to original

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Subbases §

Definition: Subbasis, topology generated by a subbasis

A subbasis for a topology on $X$ is a collection of subsets $\mathcal{S}\subseteq \mathcal{P}(X)$ such that $\mathcal{S}$ satisfies the (B1) for bases. In other words:

• For all $x\in X$, there exists $S\in \mathcal{S}$ such that $x\in S$.
• $\mathcal{S}$ covers $X$, so we have precisely ${\bigcup }_{S\in \mathcal{S}}S=X$.

The topology generated by the subbasis $\mathcal{S}$ is the topology generated by the basis

$${\mathcal{B}}_{\mathcal{S}}=\{V_1\cap \cdots \cap V_n|V_1,...,V_n\in \mathcal{S},n\ge 1\},$$

the collection of all finite intersections of elements in $\mathcal{S}$. That is, the topology generated by $\mathcal{S}$ is the union of all finite intersections of elements of $\mathcal{S}$. It is the coarsest topology in which every element of $\mathcal{S}$ is open.

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Review §

Definitions §

• Basis
• Topology generated by a basis
• Statement of (Lemma) Obtaining a basis from a topology
• Subbasis
• Topology generated by a subbasis

Exercises §

• Check that the topology generated by a basis is indeed a topology. ⭐
• Prove that $U$ is open in $X$ if and only if $U$ is a union of elements in $\mathcal{B}$ (check that the topology $\mathcal{T}$ generated by the basis and the collection of all unions of elements of $\mathcal{B}$ are equal sets).
• Prove (Lemma) Obtaining a basis from a topology. (Hint: to prove $\mathcal{C}$ generates $\mathcal{T}$, use double containment to show $\mathcal{T}$ is the same as the topology generated by $\mathcal{C}$.) ⭐
• (Topology HW1, P4) Show that the countable collection $\{(a,b)|a<b,a,b\in \mathbb{Q}\}$ is a basis that generates the standard topology on $\mathbb{R}$. ⭐
• Check the basis ${\mathcal{B}}_{\mathcal{S}}$ given by a subbasis $\mathcal{S}$ is indeed a basis; this implies that the topology generated by $\mathcal{S}$ is indeed a topology. ⭐