Discrete and Indiscrete Topologies

Two Extremes on the Same Set

A set alone does not determine which of its subsets are open. As seen in Examples of Topologies, a partition into one block gives the topology $\{\varnothing,X\}$, while a partition into singleton blocks gives the topology consisting of every subset of $X$. These two choices are the extremes: the first has as few open sets as possible, and the second has as many as possible.

This tutorial gives these constructions standard names and studies what they imply. The difference is not merely a matter of listing open sets. It affects whether points have small open neighborhoods, which identity maps are continuous, and how restrictive continuity is for functions whose domain or target has one of these topologies. Throughout, a topology on $X$ means a collection of subsets satisfying the topology axioms from Topology Axioms.

Definition: The discrete topology on a set $X$ is $\mathcal{T}_{\mathrm{dis}}=\mathcal{P}(X)$, the collection of all subsets of $X$. The indiscrete topology on $X$ is $\mathcal{T}_{\mathrm{ind}}=\{\varnothing,X\}$.

The names describe the amount of open-set information: in a discrete space, every subset is open; in an indiscrete space, there are no open sets other than the two required by the axioms. The discrete topology is the topology of the partition into singleton blocks, and the indiscrete topology comes from the partition with the single block $X$.

Why Both Collections Are Topologies

The two collections satisfy the topology axioms directly. For the discrete topology, arbitrary unions and finite intersections of subsets of $X$ are still subsets of $X$, so they belong to $\mathcal{P}(X)$. It contains $\varnothing$ and $X$. For the indiscrete topology, any union of members is either $\varnothing$ or $X$, and any finite intersection of members is also either $\varnothing$ or $X$. The empty union is $\varnothing$, and the intersection of no sets is $X$, so the conventions for empty families are respected as well.

Proposition: The discrete and indiscrete collections are topologies on $X$. If $X=\varnothing$, they coincide. If $X$ has exactly one point, they also coincide; if $X$ has at least two points, they are different.

Proof. The preceding checks verify the topology axioms for both collections. If $X=\varnothing$, then $\mathcal{P}(X)=\{\varnothing\}$, which is also $\{\varnothing,X\}$. If $X=\{x\}$, its only subsets are $\varnothing$ and $X=\{x\}$, so again the two collections agree. If $X$ has at least two points, choose $x\in X$. Then $\{x\}$ is a subset of $X$, so it belongs to $\mathcal{T}_{\mathrm{dis}}$, but it is neither $\varnothing$ nor $X$. Hence it does not belong to $\mathcal{T}_{\mathrm{ind}}$, and the topologies differ. $\square$

Worked Example: Listing the Extremes on a Finite Set

Let $X=\{a,b,c\}$. The discrete topology is $$ \mathcal{T}_{\mathrm{dis}}= \{\varnothing,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},X\}. $$ It has eight members, one for each subset of $X$. The indiscrete topology is $$ \mathcal{T}_{\mathrm{ind}}=\{\varnothing,\{a,b,c\}\}. $$ For example, $\{a\}$ is open in the discrete topology but not in the indiscrete topology, while $\varnothing$ and $X$ are open in both. Since $X$ has more than one point, the proposition predicts that these topologies are different.

Every Topology Lies Between Them

The terms “finest” and “coarsest” refer to inclusion of collections of open sets. A topology with more open sets is finer; one with fewer open sets is coarser. The discrete topology is finest because it includes every subset, while the indiscrete topology is coarsest because every topology must include $\varnothing$ and $X$.

Theorem: If $\mathcal{T}$ is any topology on $X$, then $$ \mathcal{T}_{\mathrm{ind}}\subseteq\mathcal{T}\subseteq\mathcal{T}_{\mathrm{dis}}. $$ Thus the indiscrete topology is the coarsest topology on $X$, and the discrete topology is the finest.

Proof. By the topology axioms, $\varnothing\in\mathcal{T}$ and $X\in\mathcal{T}$. Therefore $\{\varnothing,X\}\subseteq\mathcal{T}$, which is the first inclusion. Every member of $\mathcal{T}$ is, by definition, a subset of $X$. Since $\mathcal{T}_{\mathrm{dis}}$ contains every subset of $X$, each member of $\mathcal{T}$ belongs to $\mathcal{T}_{\mathrm{dis}}$. This gives the second inclusion. $\square$

This ordering also describes identity maps. For topologies $\mathcal{T}_1$ and $\mathcal{T}_2$ on the same set, consider the identity function from $(X,\mathcal{T}_1)$ to $(X,\mathcal{T}_2)$. By the open-set characterization of continuity, it is continuous exactly when the inverse image of every $\mathcal{T}_2$-open set is $\mathcal{T}_1$-open. The inverse image of a set under the identity is the set itself.

Proposition: The identity map $\operatorname{id}_X:(X,\mathcal{T}_1)\to(X,\mathcal{T}_2)$ is continuous if and only if $\mathcal{T}_2\subseteq\mathcal{T}_1$. It is a homeomorphism if and only if $\mathcal{T}_1=\mathcal{T}_2$.

Proof. Suppose the identity map is continuous. For every $U\in\mathcal{T}_2$, continuity requires $\operatorname{id}_X^{-1}(U)\in\mathcal{T}_1$. Since $\operatorname{id}_X^{-1}(U)=U$, every member of $\mathcal{T}_2$ belongs to $\mathcal{T}_1$. Thus $\mathcal{T}_2\subseteq\mathcal{T}_1$. Conversely, if $\mathcal{T}_2\subseteq\mathcal{T}_1$, then for every $\mathcal{T}_2$-open $U$, its inverse image $U$ is $\mathcal{T}_1$-open, so the identity is continuous. The identity is bijective, and its inverse is the identity map in the opposite direction. Both maps are continuous exactly when $\mathcal{T}_2\subseteq\mathcal{T}_1$ and $\mathcal{T}_1\subseteq\mathcal{T}_2$, which is equivalent to equality. $\square$

Worked Example: The Direction of Continuity

Take $X=\{u,v,w\}$, with its discrete and indiscrete topologies. The identity map $$ \operatorname{id}_X:(X,\mathcal{T}_{\mathrm{dis}})\longrightarrow(X,\mathcal{T}_{\mathrm{ind}}) $$ is continuous: the target has only $\varnothing$ and $X$ as open sets, and both are open in the discrete domain. In the reverse direction, the identity is not continuous. The set $\{u\}$ is open in the discrete target, but its inverse image is $\{u\}$, which is not open in the indiscrete domain. Thus the direction matters: the domain having more open sets makes continuity of this identity possible, not the reverse.

Continuity at the Extremes

Continuity is defined by inverse images of open sets. That criterion makes some maps automatic. A discrete domain has every subset open, so no inverse image can fail to be open. An indiscrete target has so few open sets that every inverse image is one of the open sets already present.

Theorem: Let $X$ have the discrete topology and $Y$ any topological space. Every function $f:X\to Y$ is continuous. Also, every function $g:Z\to Y$ is continuous when $Y$ has the indiscrete topology.

Proof. For the first claim, let $V$ be open in $Y$. Its inverse image $f^{-1}(V)$ is a subset of $X$. Every subset of a discrete space is open, so $f^{-1}(V)$ is open in $X$. This holds for every open $V$, proving continuity. For the second claim, the only open subsets of $Y$ are $\varnothing$ and $Y$. Their inverse images under $g$ are $\varnothing$ and $Z$, respectively, both open in $Z$. Hence $g$ is continuous. $\square$

The converse patterns are more restrictive. If the domain is indiscrete, a nonempty proper subset of the domain is not open, so inverse images of target open sets cannot be arbitrary. If the target is discrete, every singleton is open, and continuity forces the inverse image of each singleton to be open.

Theorem: Let $f:X\to Y$, where $Y$ has the discrete topology. Then $f$ is continuous if and only if every fiber $f^{-1}(\{y\})$, for $y\in Y$, is open in $X$.

Proof. If $f$ is continuous, then $\{y\}$ is open in the discrete space $Y$ for each $y\in Y$. Therefore $f^{-1}(\{y\})$ is open in $X$. Conversely, suppose every such fiber is open. Each open set $V\subseteq Y$ is a union of singletons, and $$ f^{-1}(V)=\bigcup_{y\in V}f^{-1}(\{y\}). $$ An arbitrary union of open sets is open, including the empty union when $V=\varnothing$. Thus the inverse image of every open set in $Y$ is open, so $f$ is continuous. $\square$

Worked Example: A Map into a Discrete Space

Let $X=\{p,q,r,s\}$ have the topology generated by the partition $\{\{p,q\},\{r,s\}\}$, so its open sets are unions of these two blocks. Give $Y=\{0,1\}$ the discrete topology. Define $f(p)=f(q)=0$ and $f(r)=f(s)=1$. Its fibers are $f^{-1}(\{0\})=\{p,q\}$ and $f^{-1}(\{1\})=\{r,s\}$, both open blocks. The theorem shows that $f$ is continuous.

Now define $h(p)=0$, $h(q)=1$, $h(r)=h(s)=1$. The fiber $h^{-1}(\{0\})=\{p\}$ is not a union of the partition blocks, so it is not open in $X$. Since $\{0\}$ is open in $Y$, this inverse image witnesses that $h$ is not continuous. A map into a discrete space must respect the open-set structure of its domain through all of its fibers.

Indiscrete Domains and Hausdorff Targets

The Hausdorff condition provides a useful contrast. In a Hausdorff space, two distinct points have disjoint open neighborhoods. An indiscrete domain cannot have a nonempty proper open subset, so a continuous map from it cannot send two points to distinct points in a Hausdorff target.

Theorem: Let $X$ be a nonempty indiscrete space and let $Y$ be Hausdorff. A function $f:X\to Y$ is continuous if and only if it is constant.

Proof. If $f$ is constant, then the inverse image of any subset of $Y$ is either $\varnothing$ or $X$. Both sets are open in the indiscrete domain, so $f$ is continuous. Conversely, suppose $f$ is continuous and is not constant. Then there are $x_1,x_2\in X$ such that $f(x_1)\ne f(x_2)$. Since $Y$ is Hausdorff, there are disjoint open sets $U,V\subseteq Y$ with $f(x_1)\in U$ and $f(x_2)\in V$. The inverse image $f^{-1}(U)$ contains $x_1$, so it is nonempty. It does not contain $x_2$, because $f(x_2)\in V$ and $U\cap V=\varnothing$. Thus $f^{-1}(U)$ is a nonempty proper subset of $X$, which is not open in the indiscrete topology. This contradicts continuity. Therefore $f$ is constant. $\square$

Worked Example: A Nonconstant Map from an Indiscrete Space

Let $X=\{m,n\}$ have the indiscrete topology, and let $Y=\mathbb{R}$ have its usual topology. Define $f(m)=0$ and $f(n)=2$. The open interval $(-1,1)$ contains $f(m)$ but not $f(n)$, so $$ f^{-1}((-1,1))=\{m\}. $$ The set $\{m\}$ is nonempty and proper, hence not open in $X$. This single inverse image proves that $f$ is not continuous. By contrast, any constant map from this $X$ to $\mathbb{R}$ is continuous, as the theorem predicts.

What These Extremes Do and Do Not Say

The discrete and indiscrete topologies give quick tests, but the conclusions depend on which side of a function carries the extreme topology. A discrete domain makes every function continuous; a discrete target instead makes continuity depend on whether the fibers are open. An indiscrete target makes every function into it continuous; an indiscrete domain can severely restrict continuous maps, especially when the target is Hausdorff.

There is also a useful distinction between a topology and a metric. A discrete topology is a collection of open sets, not a distance function. The discrete metric discussed in Examples of Topologies is one way to produce this topology, but the definition of a discrete topological space does not require choosing any metric. Conversely, if $X$ has at least two points, the indiscrete topology cannot be induced by a metric: every metric topology is Hausdorff, while the only nonempty open set in the indiscrete topology is $X$, so distinct points have no disjoint open neighborhoods.

Finally, the extremal descriptions are relative to a fixed underlying set. The theorem says that every topology on that same $X$ lies between $\mathcal{T}_{\mathrm{ind}}$ and $\mathcal{T}_{\mathrm{dis}}$; it does not say that a topology is determined by the size of $X$, or that every intermediate collection is automatically a topology. Intermediate choices must still satisfy the topology axioms. The following study of open sets in topological spaces develops how those axioms govern such choices.

Check Your Understanding

Use the definitions and continuity criteria in this tutorial to answer the following questions.

For a set with at least two points, give a subset that is open in the discrete topology but not the indiscrete topology.
Why does every topology $\mathcal{T}$ on $X$ satisfy $\mathcal{T}_{\mathrm{ind}}\subseteq\mathcal{T}\subseteq\mathcal{T}_{\mathrm{dis}}$?
For which inclusion between $\mathcal{T}_1$ and $\mathcal{T}_2$ is the identity map from $(X,\mathcal{T}_1)$ to $(X,\mathcal{T}_2)$ continuous?
What must be checked to determine whether a map into a discrete space is continuous?
Why is a nonconstant continuous map from a nonempty indiscrete space into a Hausdorff space impossible?
When do the discrete and indiscrete topologies on $X$ coincide?