From Distances to Open Sets
The preceding tutorials studied connectedness and related ideas in metric spaces. A metric supplies distances, and distances supply open balls. But many arguments in analysis use only the open sets those balls determine. For example, continuity can be expressed by saying that inverse images of open sets are open; compactness is expressed in terms of open covers; and connectedness can be described by the absence of a separation into disjoint nonempty open pieces. In these definitions, the exact numerical distances do not appear.
General topology takes this observation seriously. Instead of beginning with distances and deriving open sets, it begins with a specified family of open sets. This makes it possible to keep the notions that depend on openness while dispensing with distances that may be irrelevant—or may not exist at all. The next tutorial will develop the formal language of topological spaces. Here we establish why that language is useful and see a first example of what it can do.
The axioms capture basic behavior already familiar from metric spaces. The empty set and the whole space are open; combining open sets by any union or by a finite intersection produces another open set. The axioms do not mention points being close, distances, or balls. They retain the structure needed to speak of openness without requiring a metric to generate it.
Every Metric Gives a Topology
For a metric space \((X,d)\), call a set \(U\subseteq X\) open if every \(x\in U\) has some radius \(r>0\) such that \(B_r(x)\subseteq U\). The collection of all such sets is the metric topology determined by \(d\). The following result confirms that this collection satisfies the topology axioms.
Proof. The empty set is open because it has no points at which the definition must be checked. The whole space is open because \(B_1(x)\subseteq X\) for every \(x\in X\).
Let \(\{U_\lambda:\lambda\in\Lambda\}\) be any family of open sets, and let \(x\in\bigcup_{\lambda\in\Lambda}U_\lambda\). Then \(x\in U_{\lambda_0}\) for some index \(\lambda_0\). Since \(U_{\lambda_0}\) is open, there is \(r>0\) with \(B_r(x)\subseteq U_{\lambda_0}\). Therefore \(B_r(x)\subseteq\bigcup_{\lambda\in\Lambda}U_\lambda\), so the union is open. If the family is empty, its union is \(\varnothing\), already shown to be open.
Now let \(U_1,\ldots,U_n\) be finitely many open sets. If their intersection is empty, it is open. Otherwise, take \(x\in\bigcap_{j=1}^n U_j\). For each \(j\), choose \(r_j>0\) such that \(B_{r_j}(x)\subseteq U_j\), and set \(r=\min\{r_1,\ldots,r_n\}\). Since the minimum of finitely many positive numbers is positive, \(r>0\). Also \(B_r(x)\subseteq B_{r_j}(x)\subseteq U_j\) for every \(j\), so \(B_r(x)\subseteq\bigcap_{j=1}^n U_j\). Thus the intersection is open, proving all the topology axioms. \(\square\)
Worked Example: Changing Distances Without Changing Open Sets
On \(\mathbb{R}\), consider the usual metric \(d(x,y)=|x-y|\) and the scaled metric \(\rho(x,y)=2|x-y|\). The open ball for \(\rho\) satisfies $$ B_r^\rho(x)=\{y:2|x-y|<r\}=B_{r/2}^d(x). $$ Conversely, \(B_s^d(x)=B_{2s}^\rho(x)\). Thus every ball for one metric is a ball for the other, and the two metrics have exactly the same open sets.
The metrics assign different numerical distances to distinct points, but they describe the same topology. Consequently, concepts defined solely using open sets cannot distinguish these two metric structures. This is one reason topology is useful: it separates the qualitative notion of closeness relevant to continuity from the particular scale used to measure distance.
What the Open Sets Preserve
Once a topology is specified, familiar definitions can be stated without referring to a metric. A function is continuous when the inverse image of each open set is open. A space is compact when every open cover has a finite subcover. A space is connected when it cannot be written as the union of two disjoint nonempty open subsets. In a metric space these descriptions agree with the concepts developed earlier in the course; they depend on the open sets, not on the choice of radii or numerical distances.
This makes the move to topology a controlled generalization rather than a change of subject. Results proved using only open sets can be applied to any topological space. Results that depend essentially on distances—such as estimates involving \(\varepsilon\) and \(\delta\), Cauchy sequences, or completeness—still require additional structure. General topology identifies which parts of an argument are about openness and which parts genuinely need a metric.
Worked Example: The Discrete Topology Comes from a Metric
Let \(X\) be any set and define \(d(x,y)=0\) when \(x=y\), and \(d(x,y)=1\) when \(x\ne y\). This is the discrete metric. For any \(x\in X\), the ball \(B_{1/2}(x)\) is \(\{x\}\): the distance from \(x\) to itself is \(0<1/2\), while the distance from \(x\) to every other point is \(1\), which is not less than \(1/2\).
Every subset \(A\subseteq X\) is open. Indeed, for each \(x\in A\), the ball \(B_{1/2}(x)=\{x\}\) is contained in \(A\). Thus this metric produces the topology consisting of all subsets of \(X\), called the discrete topology. The example shows that topologies can be much finer than the familiar topology of the real line: in the discrete topology, even a single point is open.
Metric spaces also have a separation property that is not built into the topology axioms: distinct points can be placed in disjoint open neighborhoods. This property is called the Hausdorff property. It helps explain why not every topology can come from a metric.
Proof. Since \(x\ne y\), the metric axiom gives \(d(x,y)>0\). Set \(r=d(x,y)/3\), so \(r>0\). The balls \(B_r(x)\) and \(B_r(y)\) are open neighborhoods of \(x\) and \(y\). If a point \(z\) belonged to both, the triangle inequality would give $$ d(x,y)\leq d(x,z)+d(z,y)<r+r=\frac{2d(x,y)}{3}<d(x,y), $$ a contradiction. Hence the two neighborhoods are disjoint. \(\square\)
A Compact Space That No Metric Can Produce
To see why abstract topologies enlarge the subject, consider the cofinite topology on an infinite set \(X\). Its open sets are \(\varnothing\) together with all subsets whose complements in \(X\) are finite. This topology may seem unusual, but it satisfies the axioms and has useful properties.
Proof. First verify the topology axioms. Both \(\varnothing\) and \(X\) are open. For any family of open sets, if every member is empty, its union is empty. Otherwise the union contains a nonempty cofinite open set \(U\); its complement is a subset of the finite set \(X\setminus U\), so the union is cofinite and open.
For a finite intersection, if one of the sets is empty, the intersection is empty. Otherwise each set has finite complement. The complement of their intersection is the union of finitely many finite complements, hence finite. The intersection is therefore cofinite and open. This verifies that the specified family is a topology.
Now let \(\mathcal{U}\) be an open cover of \(X\). Since \(X\) is nonempty, at least one member \(U_0\in\mathcal{U}\) is nonempty. It is cofinite, so \(X\setminus U_0\) is finite. For each point of \(X\setminus U_0\), choose a member of \(\mathcal{U}\) containing that point. These finitely many chosen sets, together with \(U_0\), cover \(X\). Thus every open cover has a finite subcover, and \(X\) is compact.
Finally, take distinct \(x,y\in X\). Any open neighborhood of \(x\) is nonempty and cofinite, and any open neighborhood of \(y\) is also nonempty and cofinite. The intersection of two such sets is nonempty: its complement is contained in the union of two finite sets, and an infinite set cannot be exhausted by that finite union. Therefore \(x\) and \(y\) cannot have disjoint open neighborhoods. The space is not Hausdorff. By the theorem that every metric space is Hausdorff, this topology cannot be induced by any metric. \(\square\)
Worked Example: The Cofinite Topology on the Integers
Give \(\mathbb{Z}\) the cofinite topology. The set \(U=\mathbb{Z}\setminus\{-1,4\}\) is open because its complement is finite. The sets \(U\) and \(V=\mathbb{Z}\setminus\{2,4,7\}\) intersect in $$ U\cap V=\mathbb{Z}\setminus\{-1,2,4,7\}, $$ which is again open. They cannot be disjoint: only finitely many integers are excluded from their intersection, while \(\mathbb{Z}\) is infinite.
Every open cover of \(\mathbb{Z}\) has a finite subcover by the compactness argument above. Yet no pair of distinct integers has disjoint open neighborhoods, so this topology is not generated by a metric. Compactness here is defined entirely through open covers; there is no metric available to use the metric-space compactness results from earlier tutorials.
Why the General Viewpoint Matters
The cofinite example illustrates the central motivation. General topology is not merely a new vocabulary for metric spaces: it allows the study of open-set behavior in settings where no distance function can encode that behavior. At the same time, when a metric is available, its open sets provide the bridge back to the analysis developed earlier. Theorems stated using only open sets can then be recognized as results about the underlying topology, while metric-specific theorems retain their additional hypotheses.
A useful habit is to ask what a proof actually uses. If it uses open neighborhoods, open covers, or relative openness, it may have a topological formulation. If it compares distances, constructs a Cauchy sequence, or relies on a numerical bound, it may need metric structure. Keeping these roles distinct prevents a common mistake: assuming that every topological space has distances, or that a theorem proved for metric spaces automatically applies without its metric hypotheses.
The next step is to develop topological spaces systematically: how to work with their open sets, subspaces, and continuous maps. The examples here provide the reason for doing so. The open-set viewpoint includes every metric space, preserves many central ideas of analysis, and reaches spaces that metric language alone cannot describe.
Check Your Understanding
Use the relationship between metrics and open sets to answer the following questions.
- Which parts of the definition of a topology are verified when proving that metric-open sets form a topology?
- Why do the metrics \(d(x,y)=|x-y|\) and \(\rho(x,y)=2|x-y|\) on \(\mathbb{R}\) determine the same open sets?
- Why is every subset open in a space with the discrete topology generated by the discrete metric?
- Where does the triangle inequality enter the proof that every metric space is Hausdorff?
- Why does compactness of an infinite cofinite space not imply that its topology comes from a metric?