Examples of Topologies

One Set, Different Choices of Open Sets

A topology is a collection of subsets, not an extra property attached to the underlying set alone. The same set can carry very different topologies: one may have only two open sets, while another may declare every subset open. Between these extremes are topologies that come from a metric, a partition, or a condition on the size of a complement.

Recall the topology axioms from Topology Axioms: the empty set and the whole set are open, arbitrary unions of open sets are open, and finite intersections of open sets are open. The examples in this tutorial illustrate several ways to satisfy those requirements. In particular, checking arbitrary unions requires care: the empty union is $\varnothing$, and a family of open sets may contain only empty sets.

Topologies from Metrics

Every metric gives a topology, as established in Why General Topology?. In the metric topology, a set $U$ is open when every $x\in U$ has some open ball $B_r(x)$ contained in $U$. This familiar construction supplies many examples on sets used throughout analysis.

Worked Example: The Usual Topology on the Real Line

Give $\mathbb{R}$ its usual metric $d(x,y)=|x-y|$. The open ball of radius $r>0$ centered at $x$ is $$ B_r(x)=\{y\in\mathbb{R}:|y-x|<r\}=(x-r,x+r). $$ Thus the metric topology consists exactly of the sets $U\subseteq\mathbb{R}$ such that every $x\in U$ lies in some open interval $(x-r,x+r)\subseteq U$. In particular, every open interval is open: if $x\in(a,b)$, choose $r=\min\{x-a,b-x\}/2$. Both quantities in the minimum are positive, and $x-r>a$ and $x+r<b$, so $(x-r,x+r)\subseteq(a,b)$.

The criterion is not that an open set must be one interval. For instance, $(-2,-1)\cup(1,3)$ is open because each of its points lies in a sufficiently small interval contained in the same component. The metric definition tests openness point by point.

Worked Example: A Metric Whose Topology Is Discrete

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 a metric: it is nonnegative and symmetric, it vanishes exactly when its arguments agree, and the triangle inequality holds. Indeed, if $x\ne z$, then at least one of $x\ne y$ or $y\ne z$ holds, so $d(x,y)+d(y,z)\geq1=d(x,z)$. If $x=z$, the triangle inequality follows from nonnegativity.

For any $x\in X$, the open ball $B_{1/2}(x)$ is $\{x\}$, since $d(x,y)<1/2$ holds exactly when $y=x$. Every subset $A\subseteq X$ is therefore open: it is the union $\bigcup_{x\in A}\{x\}$, with the empty case giving $\varnothing$. This is one route to the discrete topology, in which all subsets are open.

The Cofinite Topology

A topology need not come from a metric. One construction uses the sizes of complements. A subset is called finite if it has only finitely many elements; this includes the empty set. The cofinite topology declares a set open whenever its complement is finite, together with the empty set. Including $\varnothing$ separately matters when the underlying set is infinite: its complement is then not finite.

Definition: Let $X$ be a set. The cofinite topology on $X$ is $$ \mathcal{T}_{\mathrm{cof}}=\{\varnothing\}\cup\{U\subseteq X:X\setminus U\text{ is finite}\}. $$

Theorem: The cofinite topology is a topology on $X$.

Proof. By definition, $\varnothing\in\mathcal{T}_{\mathrm{cof}}$, and $X\in\mathcal{T}_{\mathrm{cof}}$ because $X\setminus X=\varnothing$ is finite.

Consider any family $\{U_i:i\in I\}$ of cofinite-open sets. If $I$ is empty, its union is $\varnothing$, which is open. If every $U_i$ is empty, the union is also $\varnothing$. Otherwise, the family contains some nonempty $U_j$. Since $U_j$ is cofinite-open and nonempty, its complement $X\setminus U_j$ is finite. The union contains $U_j$, so $$ X\setminus\bigcup_{i\in I}U_i\subseteq X\setminus U_j. $$ A subset of a finite set is finite. Hence the union has finite complement and is open. This covers arbitrary families, including those that contain empty members.

Now take finitely many cofinite-open sets $U_1,\ldots,U_n$. If any $U_k=\varnothing$, their intersection is empty and therefore open. Otherwise each complement $X\setminus U_k$ is finite, and $$ X\setminus\bigcap_{k=1}^{n}U_k=\bigcup_{k=1}^{n}(X\setminus U_k). $$ A finite union of finite sets is finite, so the intersection is open. For $n=0$, the intersection is $X$, already shown to be open. All topology axioms hold. $\square$

Worked Example: Open Sets in the Cofinite Topology

Let $X=\mathbb{Z}$, and consider $U=\mathbb{Z}\setminus\{-2,0,5\}$. Its complement has three elements, so $U$ is open in the cofinite topology. The intersection of $U$ with $V=\mathbb{Z}\setminus\{1,5\}$ is $$ U\cap V=\mathbb{Z}\setminus\{-2,0,1,5\}, $$ which is also open: the complement of the intersection is the finite union $\{-2,0,5\}\cup\{1,5\}=\{-2,0,1,5\}$.

By contrast, a union of open sets should not be checked by claiming that its complement is always finite merely because every member has finite complement. If the family consists only of empty sets, the union is empty and its complement may be infinite. The proof above handles this case directly; if the union contains a nonempty member, its complement is contained in that member’s finite complement.

When $X$ is finite, every subset has finite complement, so the cofinite topology is the discrete topology. When $X$ is infinite, it is a different example: the theorem in Why General Topology? establishes that the cofinite topology is compact and not Hausdorff.

Topologies from Partitions

A partition of $X$ is a collection of nonempty, pairwise disjoint subsets whose union is $X$. Its members are called blocks. Declare an open set to be any union of blocks, allowing the union of no blocks to be empty. This construction groups points into blocks that the topology cannot separate using open sets.

Theorem: If $\mathcal{P}$ is a partition of $X$, the collection of all unions of members of $\mathcal{P}$ is a topology on $X$.

Proof. The empty set is the union of no blocks, and $X$ is the union of all blocks, so both belong to the collection. Let $\{U_i:i\in I\}$ be any family of unions of blocks. Their union is still a union of blocks: a block is included in the union exactly when it occurs in at least one $U_i$. If $I$ is empty, this union is $\varnothing$, also allowed.

For two unions of blocks $U$ and $V$, their intersection is a union of blocks as well. To see this, suppose a block $P\in\mathcal{P}$ meets $U\cap V$, and choose $x\in P\cap U\cap V$. Since $U$ is a union of blocks, it contains the entire block $P$; likewise, $V$ contains $P$. Therefore $P\subseteq U\cap V$. It follows that the intersection consists exactly of the blocks it meets. Repeated application proves closure under nonempty finite intersections, and the intersection of no sets is $X$. Thus the collection is a topology. $\square$

Worked Example: A Topology from Three Blocks

Let $X=\{a,b,c,d,e\}$ and partition it into $P_1=\{a,b\}$, $P_2=\{c\}$, and $P_3=\{d,e\}$. The open sets are precisely the unions of these three blocks. For example, $$ P_1\cup P_3=\{a,b,d,e\} $$ is open, and $P_2\cup P_3=\{c,d,e\}$ is open. Their intersection is $P_3=\{d,e\}$, again open. The singleton $\{a\}$ is not open: a union of blocks that contains $a$ must contain all of $P_1$, including $b$.

At one extreme, the partition with the single block $X$ gives only $\varnothing$ and $X$, the indiscrete topology. At the other extreme, partitioning $X$ into singleton blocks gives every subset as a union of blocks, hence the discrete topology. These are useful boundary examples; the next tutorial examines them in detail.

The Cocountable Topology

On an uncountable set, one can replace “finite complement” by “countable complement.” Here countable includes finite. As with the cofinite construction, the empty set must be included separately: on an uncountable $X$, its complement $X$ is not countable.

Definition: If $X$ is uncountable, its cocountable topology is $$ \mathcal{T}_{\mathrm{cc}}=\{\varnothing\}\cup\{U\subseteq X:X\setminus U\text{ is countable}\}. $$

Worked Example: Open Sets on an Uncountable Set

Take $X=\mathbb{R}$. The set of irrational numbers $\mathbb{R}\setminus\mathbb{Q}$ is open in the cocountable topology because its complement $\mathbb{Q}$ is countable. For any $x\in\mathbb{R}$, the set $\mathbb{R}\setminus\{x\}$ is also open because a singleton is countable. But $\{x\}$ is not open: its complement is uncountable, and it is not the empty set.

Any two nonempty cocountable-open subsets $U$ and $V$ intersect. If they were disjoint, then $X= (X\setminus U)\cup(X\setminus V)$, a union of two countable sets, which would make $X$ countable, contrary to the hypothesis. Thus this topology cannot provide disjoint open neighborhoods for distinct points.

The axiom check for this construction follows the same careful pattern as for the cofinite topology. The empty set and $X$ are included. For an arbitrary family of cocountable-open sets, the union is empty if all members are empty. Otherwise choose a nonempty member $U_j$; the complement of the full union is contained in $X\setminus U_j$, hence is countable. For a finite intersection, if one member is empty the intersection is empty; otherwise its complement is a finite union of countable sets and is countable. Consequently $\mathcal{T}_{\mathrm{cc}}$ is a topology.

How to Read an Example

An example of a topology is more than a list of sets. It specifies which subsets are open, and that choice controls notions such as neighborhoods, continuity, and separation. When checking a proposed collection, the most reliable approach is to test the topology axioms in order and explicitly address empty families and empty sets. These boundary cases are not technical distractions: they are where a superficially plausible argument about arbitrary unions can fail.

The constructions also emphasize that topology depends on both the underlying set and the chosen collection of open sets. A metric can produce the usual topology on $\mathbb{R}$, while a complement-size rule produces a different topology on the same set. A partition gives still another choice by forcing points in one block to occur together in every open set. The open-set axioms make all of these constructions legitimate, but their topological behavior need not be alike.

Check Your Understanding

Use the constructions and arguments in this tutorial to answer the following questions.

Why does the cofinite-topology proof distinguish a union that is empty from a union containing a nonempty member?
For the metric $d(x,y)=1$ when $x\ne y$, what is $B_{1/2}(x)$, and what does this imply about open subsets?
In a topology defined by a partition, why can an open set not contain only part of a block?
What are the open sets in the topology determined by the partition $\{\{a,b\},\{c\}\}$ of $\{a,b,c\}$?
Why must two nonempty open sets in the cocountable topology on an uncountable set intersect?
When $X$ is finite, why does the cofinite topology agree with the discrete topology?

What You'll Learn