Topological Spaces and Open Sets

The open-set axioms are the entry point to essentially all of modern geometry and analysis whenever "distance" is either unavailable or the wrong tool. In functional analysis, the weak topology on an infinite-dimensional Banach space and the weak-\ topology on its dual are defined directly by declaring certain sets open (via seminorms), and are typically not metrizable on infinite-dimensional spaces, yet remain essential for compactness arguments (Banach–Alaoglu theorem). In <strong>algebraic geometry</strong>, the Zariski topology on the solution set of polynomial equations declares closed sets to be common zero loci of polynomials; this topology is almost always non-Hausdorff and never arises from a metric, yet it is the correct language for schemes and varieties. In <strong>computer science and logic</strong>, domain theory and denotational semantics use topological (Scott) topologies on posets to model computation and partial information. In <strong>statistics and probability</strong>, topologies on spaces of probability measures (e.g. the topology of weak convergence) underlie convergence theorems that have no natural metric a priori, though metrizations (e.g. the Lévy–Prokhorov metric) are sometimes constructed after the fact. Even within pure analysis, recognizing that pointwise convergence, uniform convergence, and $L^p$ convergence correspond to genuinely different* topologies on function spaces — rather than different metrics on the "same" underlying geometric object — is what the open-set formalism was built to clarify.

Ask the tutor: How does the definition of a basis let us generate an entire topology from a smaller, more manageable collection of "basic" open sets?

The metric-space definition of an open set says: $U$ is open if every point of $U$ has some open ball around it, entirely contained in $U$. This is a local condition, checked point by point, and it is built directly from distances. The topological axioms instead treat "openness" as a primitive, global bookkeeping device: we simply declare, once and for all, a family of subsets to be the open ones, and demand that this family behaves sensibly under the two set-theoretic operations that matter most — union and (finite) intersection. Why finite intersection but arbitrary union? This asymmetry is not an accident; it is exactly the pattern that holds for open balls in a metric space. An arbitrary union of open balls is open (a point in the union lies in one ball, which already witnesses openness). But an infinite intersection of open balls can collapse to a single point or even become empty of interior — for instance $\bigcap_{n=1}^\infty (-\tfrac1n, \tfrac1n) = \{0\}$ in $\mathbb{R}$, and a single point is not open in $\mathbb{R}$ with its usual topology. So the topological axioms are reverse-engineered from what is provably true of metric open sets, and nothing more. Once we accept these three axioms as the definition, a topological space carries no residual notion of "distance," only a notion of which sets are open. This is deliberately less information than a metric — many metrics (even very different-looking ones) can produce the same collection of open sets, and some topologies arise with no metric behind them whatsoever. The picture to keep in mind: a topology is a coarse net thrown over a set, remembering only "which regions are open neighborhoods," discarding numerical distance but preserving enough structure to talk about continuity and convergence.

Example 1: The cofinite topology on an infinite set, and why it is a genuine topology. Let $X$ be an infinite set (say $X = \mathbb{Z}$). Define

We check this satisfies (T1)–(T3). (T1): $\emptyset \in \mathcal{T}_{cf}$ by definition. $X \in \mathcal{T}_{cf}$ because $X \setminus X = \emptyset$ is finite. (T2): Let $\{U_\alpha\}_{\alpha \in A}$ be a family in $\mathcal{T}_{cf}$, not all empty (if all are empty the union is $\emptyset$, already handled). Pick some $\alpha_0$ with $U_{\alpha_0} \neq \emptyset$, so $X \setminus U_{\alpha_0}$ is finite. By De Morgan,

and a subset of a finite set is finite, so $X \setminus \bigcup_\alpha U_\alpha$ is finite. Hence $\bigcup_\alpha U_\alpha \in \mathcal{T}_{cf}$. (T3): Let $U_1, \dots, U_n \in \mathcal{T}_{cf}$ (if any $U_i = \emptyset$ the intersection is $\emptyset \in \mathcal{T}_{cf}$; so assume all nonempty, hence cofinite). By De Morgan,

a finite union of finite sets, hence finite. So $\bigcap_i U_i \in \mathcal{T}_{cf}$. Concretely, on $X = \mathbb{Z}$: the set $U = \mathbb{Z} \setminus \{0, 7, -3\}$ is open (its complement $\{0,7,-3\}$ has 3 elements, finite), but $E = \{0, 1, 2, 3, \dots\} = \mathbb{Z}_{\geq 0}$ is not open, since $\mathbb{Z} \setminus E = \{-1,-2,-3,\dots\}$ is infinite. Note also this topology has no metric behind it in general: in the cofinite topology on an infinite set, any two nonempty open sets intersect (since their complements are finite and $X$ is infinite, they cannot be disjoint — this is worth checking: if $U_1 \cap U_2 = \emptyset$ then $X = (X\setminus U_1) \cup (X \setminus U_2)$, a union of two finite sets, forcing $X$ finite, contradiction). This "any two nonempty opens meet" property is impossible in a metric space with at least two points (take disjoint balls of small enough radius around two distinct points), so $\mathcal{T}_{cf}$ is a topology that is genuinely not a metric topology.

Example 2: Two different metrics on $\mathbb{R}^2$ producing the same topology. Let $d_2(x,y) = \sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$ (Euclidean) and $d_\infty(x,y) = \max(|x_1-y_1|, |x_2-y_2|)$ (sup metric) on $X = \mathbb{R}^2$. We show $\mathcal{T}_{d_2} = \mathcal{T}_{d_\infty}$, i.e. the two metrics generate the identical collection of open sets, despite having visibly different balls (disks vs. squares). Key inequality: for all $x, y \in \mathbb{R}^2$,

The left inequality holds since $\max(|x_1-y_1|,|x_2-y_2|) \le \sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$ (the max of two numbers is at most the Euclidean norm of the vector they form, as $a \le \sqrt{a^2+b^2}$ for $a,b\ge0$). The right inequality holds since $(x_1-y_1)^2 + (x_2-y_2)^2 \le 2\max(|x_1-y_1|,|x_2-y_2|)^2$, taking square roots. Now let $U \in \mathcal{T}_{d_2}$ and let $x \in U$. There is $r>0$ with $B_{d_2}(x,r) \subseteq U$. Using the left inequality, if $d_\infty(x,y) < r/\sqrt2$ then $d_2(x,y) \le \sqrt2\, d_\infty(x,y) < r$, so $B_{d_\infty}(x, r/\sqrt2) \subseteq B_{d_2}(x,r) \subseteq U$. Hence $U \in \mathcal{T}_{d_\infty}$. Conversely let $U \in \mathcal{T}_{d_\infty}$ and $x \in U$, with $B_{d_\infty}(x,r) \subseteq U$. Using the right inequality's reverse direction, if $d_2(x,y) < r$ then $d_\infty(x,y) \le d_2(x,y) < r$, so $B_{d_2}(x,r) \subseteq B_{d_\infty}(x,r) \subseteq U$. Hence $U \in \mathcal{T}_{d_2}$. So $\mathcal{T}_{d_2} = \mathcal{T}_{d_\infty}$: the same topology on $\mathbb{R}^2$ arises from two metrics that are not equal as functions (e.g. $d_2((0,0),(1,1)) = \sqrt2$ while $d_\infty((0,0),(1,1)) = 1$). This is the phenomenon of equivalent metrics: distances differ, but the family of open sets — and hence everything topological built from it — is identical. This is precisely why topology, not metric, is the "right" invariant to study when only continuity/convergence-type questions are at stake.