The Timpson Lyceum · General Topology (Point Set Topology)

Basis for a Topology – Lecture 7 (Topology)

Overview

This lecture introduces the idea of a basis for a topology — a small, manageable collection of "building-block" sets from which an entire topology can be generated, much the way a finite set of Lego pieces generates infinitely many structures. Rather than describing every open set directly (often impossibly tedious), we specify a basis and let arbitrary unions of basis elements fill out the topology. We give the precise definition of a basis, define the topology it generates, and work through the discrete and standard-on-$\mathbb{R}$ examples. The technical heart of the lecture is proving that the collection generated by a basis really is a topology, which requires a preliminary lemma (proved by induction) about finite intersections of basis elements. We close with a useful characterization of open sets in terms of basis elements.

Key Concepts

Recap — what a topology is. A collection $\tau$ of subsets of $X$ is a topology on $X$ iff:

• $\varnothing \in \tau$ and $X \in \tau$;
• $\tau$ is closed under finite intersections: if $U_1, \dots, U_n \in \tau$ then $\bigcap_{i=1}^{n} U_i \in \tau$;
• $\tau$ is closed under arbitrary (finite or infinite) unions.

The pair $(X, \tau)$ is a topological space; members of $\tau$ are open sets. We loosely call $X$ "a topological space" with $\tau$ understood.

Basis for a topology. A collection $\mathcal{B}$ of subsets of $X$ is a basis for a topology on $X$ if it satisfies two conditions:

• (B1) Covering. For every $x \in X$ there exists $B \in \mathcal{B}$ with $x \in B$. (Every point lies in at least one basis element.)
• (B2) Intersection condition. If $B_1, B_2 \in \mathcal{B}$ and $x \in B_1 \cap B_2$, then there exists $B_3 \in \mathcal{B}$ with $x \in B_3 \subseteq B_1 \cap B_2$.

The members of $\mathcal{B}$ are called basis elements. (B2) does not require the intersection itself to be a basis element — only that each point of the overlap sits inside some basis element contained in the overlap. This is a deliberately weak requirement: it is often easy to fit a small, easy-to-describe set (e.g. a tiny ball) around a point even when the overlap itself is awkward to describe.

Topology generated by a basis. Given a basis $\mathcal{B}$ on $X$, the topology generated by $\mathcal{B}$ is $\tau = \{\varnothing\} \cup \{\, U : U \text{ is a union of basis elements} \,\}.$ A set is open precisely when it can be written as a union of basis elements. Every basis element is itself open (union it with itself). A given basis generates exactly one topology.

Example — singletons generate the discrete topology. Let $X$ be any set and $\mathcal{B} = \{\, \{x\} : x \in X \,\}$, the collection of all singletons. Any subset of $X$ is a union of singletons, so the generated topology is the power set $\mathcal{P}(X)$ — the discrete topology. For $X = \{a,b,c,d\}$ this produces all $2^4 = 16$ subsets.

Example — the standard topology on $\mathbb{R}$. Let $\mathcal{B} = \{\, (a,b) : a, b \in \mathbb{R},\ a < b \,\}$, all bounded open intervals. The topology it generates is the standard topology on $\mathbb{R}$ — the one used in calculus and analysis. Its open sets are exactly the unions of open intervals.

• $\mathbb{R}$ itself is open: e.g. $\mathbb{R} = \bigcup_{x \in \mathbb{R}} (x-1,\, x+1)$.
• A set like $\{1\}$ is not open: any interval containing $1$ also contains infinitely many points on either side, so no basis element equals $\{1\}$. (Note $(1,1) = \varnothing$.) Likewise $\mathbb{Z}$ and $[1,2]$ are not open.
• The integers cannot be isolated even by an infinite intersection: $\bigcap_{n=1}^{\infty}\left(-\tfrac1n, \tfrac1n\right) = \{0\}$, but this is an infinite intersection, which a topology need not contain — only finite intersections of open sets are guaranteed open. This is exactly why condition (B2)/finite intersections matter.

Finiteness vs. infinity, summarized. Unions may be arbitrary; intersections must be finite. The would-be "trick" of intersecting infinitely many shrinking intervals to capture a single point fails precisely because the topology axioms only promise finite intersections.

Proof by induction (the engine for the lemma). To show a statement $P(n)$ holds for all integers $n \ge n_0$: prove the base case $P(n_0)$, then prove the inductive step $P(k) \Rightarrow P(k+1)$. The truth then cascades to $n_0+1, n_0+2, \dots$

Lemma (finite intersections of basis elements). If $\mathcal{B}$ is a basis on $X$ and $B_1, \dots, B_n \in \mathcal{B}$ with $x \in \bigcap_{i=1}^{n} B_i$, then there exists $B' \in \mathcal{B}$ with $x \in B' \subseteq \bigcap_{i=1}^{n} B_i$. (This extends (B2) from two basis elements to any finite number; it is the key step in the main theorem below.)

Characterization of open sets via the basis. Let $\mathcal{B}$ generate $\tau$ and let $U \subseteq X$. Then $U$ is open iff for every $x \in U$ there exists a basis element $B$ with $x \in B \subseteq U$. (This mirrors the earlier theorem stated with arbitrary open sets, specialized to basis elements.)

Worked Example

Main Theorem. The collection $\tau$ generated by a basis $\mathcal{B}$ on $X$ is a topology on $X$.

We verify the three axioms.

(1) Contains $\varnothing$ and $X$. $\varnothing \in \tau$ by definition. By (B1) every point of $X$ lies in some basis element, so $X = \bigcup_{B \in \mathcal{B}} B \in \tau$.

(3) Arbitrary unions. Each open set is a union of basis elements, so a union of open sets is again a union of basis elements, hence open.

(2) Finite intersections. Let $V = \bigcap_{i=1}^{n} U_i$ with each $U_i \in \tau$. If $V = \varnothing$, it is open. Otherwise take any $x \in V$, so $x \in U_i$ for every $i$. Since each $U_i$ is a union of basis elements, for each $i$ there is a basis element $B_i$ with $x \in B_i \subseteq U_i$. Then $x \in \bigcap_{i=1}^{n} B_i$, and applying the lemma gives a basis element $B'_x$ with $x \in B'_x \subseteq \bigcap_i B_i \subseteq \bigcap_i U_i = V$. As $x$ ranged over all of $V$, taking the union of these $B'_x$ recovers $V$ as a union of basis elements, so $V \in \tau$. $\blacksquare$

Proof of the lemma, by induction on $n$. State the rule first, then carry out the step:

The step applies (B2) to the two basis elements $B^{*}$ and $B_{k+1}$, then chains the containments. By induction the result holds for every finite $n \ge 2$. $\blacksquare$

Summary

• A basis $\mathcal{B}$ is a collection of subsets satisfying (B1) every point lies in some basis element, and (B2) each point of an overlap $B_1 \cap B_2$ lies in some basis element contained in that overlap.
• The topology generated by $\mathcal{B}$ consists of $\varnothing$ together with all unions of basis elements; each basis element is open, and a basis determines exactly one topology.
• Examples: all singletons generate the discrete topology; all bounded open intervals generate the standard topology on $\mathbb{R}$, whose open sets are unions of open intervals.
• Single points (and sets like $\mathbb{Z}$) are not open in the standard topology — and they cannot be rescued by infinite intersections, since topologies guarantee only finite intersections.
• A lemma proved by induction extends the two-set intersection condition to any finite intersection of basis elements; this is the crux of proving the generated collection satisfies the finite-intersection axiom.
• An open set can be characterized intrinsically: $U$ is open iff every point of $U$ sits inside some basis element contained in $U$.