The Timpson Lyceum · General Topology (Point Set Topology)

Basis for a Topology – Lecture 8 (Topology)

Overview

This lecture continues the study of bases for a topology, moving from the abstract definition to the concrete bases that generate the standard topologies on $\mathbb{R}$, $\mathbb{R}^2$, and $\mathbb{R}^n$. After recapping what a topology and a basis are, we build geometric intuition for which subsets of $\mathbb{R}$ and $\mathbb{R}^2$ are open, then rigorously prove that the collection of all open balls really is a basis. We show that the basis of open rectangles generates the same standard topology on $\mathbb{R}^2$ — so a single topology can come from very different-looking bases. The lecture closes with a powerful criterion (Theorem 1.13) for recognizing when a sub-collection of a known topology's open sets forms a basis for it, which justifies generating the standard topology from open balls, rectangles, or even hearts.

Key Concepts

Topology (recap). A collection $\tau$ of subsets of a set $X$ is a topology on $X$ if:

• $\varnothing \in \tau$ 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 unions: for any (finite or infinite) sub-collection of $\tau$, its union is in $\tau$.

The members of $\tau$ are called open sets — "open" means exactly "belongs to the topology."

Basis (recap). A collection $\mathcal{B}$ of subsets of $X$ is a basis if:

• (B1) Covering: for every $x \in X$ there is some $B \in \mathcal{B}$ with $x \in B$ (equivalently, $\bigcup \mathcal{B} = X$);
• (B2) Overlap: for any $B_1, B_2 \in \mathcal{B}$ and any $x \in B_1 \cap B_2$, there exists $B_3 \in \mathcal{B}$ with $x \in B_3 \subseteq B_1 \cap B_2$.

Topology generated by a basis. $\mathcal{B}$ generates exactly one topology $\tau$: the open sets are $\varnothing$, every basis element, and every arbitrary union of basis elements. One topology can be generated by several different bases, but each basis generates only one topology.

• Example (discrete topology). For $X = \{a,b,c\}$ with $\mathcal{B} = \{\{a\},\{b\},\{c\}\}$, unions of singletons give every subset of $X$, so $\mathcal{B}$ generates the discrete topology.

Standard topology on $\mathbb{R}$. Generated by the basis of all open intervals $(a,b)$. Open sets are unions of open intervals. Intuition: a set is open iff, drawn on the number line, it includes no endpoints (no "filled-in" boundary points).

• $(1,3)$, $(1,3)\cup(-4,-1)$, and $(-\infty, 3)$ are open; all of $\mathbb{R}$ is open (union of all intervals).
• $\{0\}\cup(1,5)$, $\{1,2,3\}$, and $[2,7)$ are not open: every open interval around a point contains infinitely many points on both sides, so no finite set or included endpoint can appear in an open set.

Distance (Euclidean metric) in $\mathbb{R}^2$. For $x=(x_1,x_2)$ and $y=(y_1,y_2)$, $d(x,y) = \sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}.$ Here $\mathbb{R}^2 = \mathbb{R}\times\mathbb{R}$ is the set of all ordered pairs of reals.

Open ball. The open ball centered at $x$ of radius $r>0$ is $B(x,r) = \{\, y \in \mathbb{R}^2 : d(x,y) < r \,\}$ — all points strictly inside the circle, boundary excluded. (Radius must be positive; there is no zero or negative radius.)

Standard topology on $\mathbb{R}^2$. Generated by the basis of all open balls $B(x,r)$, $x\in\mathbb{R}^2$, $r>0$. Open sets are unions of open balls; any region without its boundary is open (you can fill it to arbitrary accuracy with infinitely many tiny balls). Including any single boundary point destroys openness.

Triangle inequality. For any points $a,b,c$: $d(a,b) \le d(a,c) + d(c,b)$ — a detour through a third point is never shorter than the direct path. Used in the lemma below.

Open rectangles. $(a,b)\times(c,d) = \{(x,y): a<x<b,\ c<y<d\}$, the boundary excluded. The basis of all open rectangles generates the same standard topology on $\mathbb{R}^2$ as open balls do.

Generalization to $\mathbb{R}^n$. The distance formula extends as $d(x,y)=\sqrt{\sum_{i=1}^{n}(x_i-y_i)^2}$. Open balls in $\mathbb{R}^n$ (e.g. solid spheres in $\mathbb{R}^3$, boundary excluded) form the basis for the standard topology on $\mathbb{R}^n$; open boxes $(a_1,b_1)\times\cdots\times(a_n,b_n)$ work equally well.

Theorem 1.13 (recognizing a basis inside a topology). Let $(X,\tau)$ be a topological space and let $\mathcal{B}\subseteq\tau$ be a collection of open sets. If for every $U\in\tau$ and every $x\in U$ there exists $B_x\in\mathcal{B}$ with $x\in B_x\subseteq U$, then $\mathcal{B}$ is a basis and the topology it generates is exactly $\tau$. This is why open hearts, diamonds, or any shape can serve as a basis for the standard topology — any open set, no matter how close a point sits to its edge, can be locally captured by a small enough copy of the shape.

Worked Example

Lemma (a ball inside a ball). Given $B(x,r)$ and any $y \in B(x,r)$, there is a radius $\varepsilon>0$ with $B(y,\varepsilon)\subseteq B(x,r)$.

Proof. Choose $\varepsilon$ with $0 < \varepsilon < r - d(x,y)$ (positive since $y$ is inside the ball, so $d(x,y)<r$). Let $z \in B(y,\varepsilon)$ be arbitrary; we bound $d(x,z)$ using the triangle inequality and the choice of $\varepsilon$:

Since $d(x,z) < r$, we have $z \in B(x,r)$. As $z$ was arbitrary, $B(y,\varepsilon)\subseteq B(x,r)$. $\blacksquare$

Using the lemma — open balls form a basis. Let $\mathcal{B}=\{B(x,r): x\in\mathbb{R}^2,\ r>0\}$.

• (B1) For any $x\in\mathbb{R}^2$, $x \in B(x,r)$ for any $r>0$ (e.g. $r=7$), so every point is covered.
• (B2) Suppose $x \in B_1 \cap B_2$ for two open balls. By the lemma there are radii $\varepsilon_1,\varepsilon_2>0$ with $B(x,\varepsilon_1)\subseteq B_1$ and $B(x,\varepsilon_2)\subseteq B_2$. Let $\varepsilon=\min(\varepsilon_1,\varepsilon_2)$; then $x \in B(x,\varepsilon) \subseteq B_1 \cap B_2,$ giving the required third basis element. Hence $\mathcal{B}$ is a basis, and the topology it generates is, by definition, the standard topology on $\mathbb{R}^2$.

Summary

• A topology is closed under finite intersections and arbitrary unions and contains $\varnothing$ and $X$; its members are the open sets.
• A basis satisfies the covering property (B1) and the overlap property (B2); it generates exactly one topology, whose open sets are all unions of basis elements.
• The standard topology on $\mathbb{R}$ comes from open intervals; on $\mathbb{R}^2$ from open balls $B(x,r)=\{y: d(x,y)<r\}$; on $\mathbb{R}^n$ from $n$-dimensional open balls via the generalized Euclidean distance.
• The key lemma — every interior point of a ball is the center of a smaller ball still inside it — combined with the triangle inequality proves that open balls satisfy (B1) and (B2).
• Open rectangles generate the same standard topology on $\mathbb{R}^2$ as open balls, showing one topology can arise from very different bases.
• Theorem 1.13: any collection $\mathcal{B}\subseteq\tau$ such that every point of every open set has a member of $\mathcal{B}$ between it and that set is automatically a basis for $\tau$ — so balls, rectangles, or even hearts all work.