The Discrete and Indiscrete Topologies as Extremes

How Topologies on One Set Are Bounded

A topology specifies which subsets of a set are to count as open. On a fixed set, one topology may have fewer open sets than another, but there are definite limits: every topology must include the empty set and the whole set, and no topology can include a set that is not a subset of the underlying set. These simple facts give two extreme topologies. The indiscrete topology has only the open sets forced to be present, while the discrete topology declares every subset open.

Throughout, $X$ is a set, and a topology on $X$ is a family $\mathcal{T}\subseteq\mathcal{P}(X)$ satisfying three axioms: $\varnothing,X\in\mathcal{T}$; the union of any family of members of $\mathcal{T}$ belongs to $\mathcal{T}$; and the intersection of any finite family of members of $\mathcal{T}$ belongs to $\mathcal{T}$. The members of $\mathcal{T}$ are the open sets. The topology is written $(X,\mathcal{T})$ when specifying the topology matters.

Definition: On a set $X$, the indiscrete topology is $\mathcal{T}_{\mathrm{ind}}=\{\varnothing,X\}$, and the discrete topology is $\mathcal{T}_{\mathrm{dis}}=\mathcal{P}(X)$. For two topologies $\mathcal{T}_1,\mathcal{T}_2$ on the same set, $\mathcal{T}_1\subseteq\mathcal{T}_2$ means every set open in $\mathcal{T}_1$ is also open in $\mathcal{T}_2$.

The comparison $\mathcal{T}_1\subseteq\mathcal{T}_2$ is about inclusion of families of subsets, not inclusion of the underlying spaces: both topologies are on the same $X$. Inclusion is reflexive, since every family contains all its own members; antisymmetric, since $\mathcal{T}_1\subseteq\mathcal{T}_2$ and $\mathcal{T}_2\subseteq\mathcal{T}_1$ imply equality; and transitive, since membership in $\mathcal{T}_1$ then implies membership in $\mathcal{T}_2$, and then in $\mathcal{T}_3$. Thus inclusion gives a natural way to compare topologies on one set.

The Two Extreme Topologies

First, both families in the definition really are topologies. For $\mathcal{T}_{\mathrm{ind}}$, any union of members is either empty or $X$: if any member of the family is $X$, the union is $X$; otherwise all members are empty. Any finite intersection is also empty or $X$: if one of the sets is empty, the intersection is empty; if all are $X$, it is $X$. The required sets $\varnothing$ and $X$ are present, including when $X=\varnothing$, in which case they coincide.

For $\mathcal{T}_{\mathrm{dis}}=\mathcal{P}(X)$, every union of subsets of $X$ is a subset of $X$, and every finite intersection of subsets of $X$ is a subset of $X$. Thus both operations produce members of $\mathcal{P}(X)$; also $\varnothing,X\in\mathcal{P}(X)$. This verifies the topology axioms directly.

Theorem: For every set $X$ and every topology $\mathcal{T}$ on $X$, $$ \mathcal{T}_{\mathrm{ind}}\subseteq\mathcal{T}\subseteq\mathcal{T}_{\mathrm{dis}}. $$ In particular, the indiscrete topology is contained in every topology on $X$, and every topology on $X$ is contained in the discrete topology.

Proof. Since $\mathcal{T}$ is a topology, it contains $\varnothing$ and $X$. These are exactly the members of $\mathcal{T}_{\mathrm{ind}}$, so $\mathcal{T}_{\mathrm{ind}}\subseteq\mathcal{T}$. For the other inclusion, every member $U\in\mathcal{T}$ is, by the definition of a topology on $X$, a subset of $X$. Hence $U\in\mathcal{P}(X)=\mathcal{T}_{\mathrm{dis}}$. This holds for every $U\in\mathcal{T}$, giving $\mathcal{T}\subseteq\mathcal{T}_{\mathrm{dis}}$. $\square$

The theorem says that the indiscrete topology has the fewest open sets among topologies on $X$, and the discrete topology has the most. It does not say that every subset other than $\varnothing$ and $X$ must be open or must be closed. Whether those additional sets are open depends on the particular topology.

Worked Example: A Topology Between the Extremes

Let $X=\{u,v\}$ and set $\mathcal{T}=\{\varnothing,\{u\},X\}$. Check that this is a topology and compare it with the two extremes.

The required sets are present: $\varnothing\in\mathcal{T}$ and $X\in\mathcal{T}$. To check unions, take any family of members of $\mathcal{T}$. If one member is $X$, the union is $X$. If no member is $X$, every member is either $\varnothing$ or $\{u\}$, so the union is either $\varnothing$ or $\{u\}$. In every case the union belongs to $\mathcal{T}$.

For a nonempty finite family, if one member is $\varnothing$, its intersection is $\varnothing$. Otherwise each member is either $\{u\}$ or $X$, and their intersection is $\{u\}$ if at least one member is $\{u\}$, or $X$ if all members are $X$. These outcomes belong to $\mathcal{T}$. The intersection of an empty family, when included in the convention for finite intersections, is $X$, which is also in $\mathcal{T}$. Thus all topology axioms hold.

Here $\mathcal{T}_{\mathrm{ind}}=\{\varnothing,X\}$, so $\mathcal{T}_{\mathrm{ind}}\subseteq\mathcal{T}$, with $\{u\}$ as an additional member. The discrete topology is $\mathcal{P}(X)=\{\varnothing,\{u\},\{v\},X\}$. Thus $\mathcal{T}\subseteq\mathcal{T}_{\mathrm{dis}}$, and $\{v\}$ is a subset of $X$ that is not open in $\mathcal{T}$.

Check: the list defining $\mathcal{T}$ has exactly the three sets $\varnothing,\{u\},X$; the union and intersection checks show that no operation required by the axioms produces a set outside that list. The omitted subset $\{v\}$ confirms that this topology is not discrete, and the presence of $\{u\}$ confirms that it is not indiscrete.

When Is There Only One Topology?

The extreme topologies can coincide. This happens when there are no subsets of $X$ other than $\varnothing$ and $X$. For a set with at most one element, those are all its subsets. If $X$ has at least two elements, a nonempty proper subset exists, and there is room for a topology strictly between the two extremes.

Theorem: A set $X$ has exactly one topology if and only if $X$ has at most one element. Equivalently, either extreme is the only possible topology precisely when $X$ has at most one element; in that case the two extremes are the same topology.

Proof. Suppose first that $X$ has at most one element. If $X=\varnothing$, then $\mathcal{P}(X)=\{\varnothing\}$, and the requirement that a topology contain both $\varnothing$ and $X$ forces it to be $\{\varnothing\}$. If $X$ has one element, its only subsets are $\varnothing$ and $X$. Again, every topology must contain both, and there are no other subsets to include. In either case the only topology is $\{\varnothing,X\}$, which is both $\mathcal{T}_{\mathrm{ind}}$ and $\mathcal{T}_{\mathrm{dis}}$.

Now suppose $X$ has at least two elements. Choose $a\in X$. Then $\{a\}$ is nonempty and is not equal to $X$, so $\mathcal{T}_a=\{\varnothing,\{a\},X\}$ has three distinct members. It is a topology: a union of members is $X$ if any member is $X$, and otherwise is $\varnothing$ or $\{a\}$; a finite intersection is $\varnothing$ if any member is empty, and otherwise is $X$ or $\{a\}$. It contains both required sets, so these checks verify all the axioms. Since it contains $\{a\}$, it is not indiscrete. Since $\{a\}\ne X$, the subset $\{a\}$ is not open in $\mathcal{T}_{\mathrm{ind}}$, and since $X$ has at least two elements there is some $b\in X$ with $b\ne a$; the subset $\{b\}$ is not a member of $\mathcal{T}_a$, so $\mathcal{T}_a$ is not discrete. Therefore at least one topology differs from each extreme, and neither extreme is the only possibility. $\square$

Worked Example: A Set with at Least Two Elements

Let $X=\{a,b,c\}$, and consider the two families $\mathcal{T}_a=\{\varnothing,\{a\},X\}$ and $\mathcal{T}_b=\{\varnothing,\{b\},X\}$. Verify that both are topologies and compare them.

For $\mathcal{T}_a$, a union that includes $X$ equals $X$; if it does not include $X$, its members are drawn from $\varnothing,\{a\}$, so its union is $\varnothing$ or $\{a\}$. A nonempty finite intersection containing $\varnothing$ is $\varnothing$; otherwise its members are $\{a\}$ or $X$, and the intersection is $\{a\}$ or $X$. The empty-family intersection, if considered, is $X$. The required sets are present, so $\mathcal{T}_a$ is a topology.

For $\mathcal{T}_b$, the same checks with the listed sets give unions in $\{\varnothing,\{b\},X\}$ and finite intersections in $\{\varnothing,\{b\},X\}$; it also contains $\varnothing$ and $X$. Thus $\mathcal{T}_b$ is a topology.

Both contain $\mathcal{T}_{\mathrm{ind}}=\{\varnothing,X\}$ and are contained in $\mathcal{T}_{\mathrm{dis}}=\mathcal{P}(X)$. But neither contains the other: $\{a\}\in\mathcal{T}_a$ and $\{a\}\notin\mathcal{T}_b$, while $\{b\}\in\mathcal{T}_b$ and $\{b\}\notin\mathcal{T}_a$.

Check: each family has only the displayed three members, and each includes the empty and whole sets. The union and intersection checks above stay within the respective lists. The two missing-set tests establish both failures of inclusion, without affecting the theorem: each topology still lies between the same two extremes.

A Common Mistake and Why the Bounds Matter

Common pitfall. The statement $\mathcal{T}_{\mathrm{ind}}\subseteq\mathcal{T}\subseteq\mathcal{T}_{\mathrm{dis}}$ is sometimes reversed by confusing “a topology contains open sets” with “an open set contains points.” For instance, in the topology $\{\varnothing,\{u\},\{u,v\}\}$ on $X=\{u,v\}$, it is wrong to say that the topology is contained in the indiscrete topology because every topology contains $\varnothing$ and $X$. The set $\{u\}$ belongs to this topology but not to $\{\varnothing,X\}$, so the claimed inclusion fails. The correct inclusion is $\{\varnothing,X\}\subseteq\{\varnothing,\{u\},X\}$. The discrete topology goes on the other side because it contains every subset, including every open set of the chosen topology.

These bounds are useful as a first check when a family of subsets is proposed as a topology: it must contain $\varnothing$ and $X$, and every listed member must be a subset of $X$. Those checks alone do not establish that the family is a topology; closure under arbitrary unions and finite intersections must still be verified. The bounds also clarify the role of the underlying set: on a set with at least two elements the extremes differ, and the construction $\{\varnothing,\{a\},X\}$ supplies an intermediate topology. On a set with at most one element there is no additional subset to include, so the two extremes coincide and there can be no other topology.

Topology	Open sets	Position among topologies on $X$
Indiscrete	Only $\varnothing$ and $X$	Contained in every topology
Any topology $\mathcal{T}$	A family of subsets satisfying the topology axioms	$\mathcal{T}_{\mathrm{ind}}\subseteq\mathcal{T}\subseteq\mathcal{T}_{\mathrm{dis}}$
Discrete	Every subset of $X$	Contains every topology on $X$

Check Your Understanding

Use the topology axioms and the inclusion theorem to justify your answers.

For $X=\{r,s,t\}$, explain why every topology on $X$ must contain $\varnothing$ and $X$, and why it cannot contain a set that is not a subset of $X$.
Verify directly that $\{\varnothing,\{r,s\},X\}$ is a topology on $X=\{r,s,t\}$.
Show that if a topology on $X$ equals the indiscrete topology, then it contains no open sets other than $\varnothing$ and $X$.
What is the unique topology on $X=\varnothing$? Explain why the requirements to include $\varnothing$ and $X$ do not produce two different sets in this case.
If $X$ has at least two elements, choose $a\in X$ and explain why $\{\varnothing,\{a\},X\}$ is different from both extreme topologies.

Topology	Open sets	Position among topologies on \(X\)
Indiscrete	Only \(\varnothing\) and \(X\)	Contained in every topology
Any topology \(\mathcal{T}\)	A family of subsets satisfying the topology axioms	\(\mathcal{T}_{\mathrm{ind}}\subseteq\mathcal{T}\subseteq\mathcal{T}_{\mathrm{dis}}\)
Discrete	Every subset of \(X\)	Contains every topology on \(X\)

The Discrete and Indiscrete Topologies as Extremes

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