A topology is often introduced as a device for saying which subsets of a set count as open. The surprise is that the axioms allow a space in which almost no subset is open. If $X$ is a set, the collection consisting only of $\varnothing$ and $X$ itself satisfies the axioms for open sets, so it defines a [topological space](/page/Topological%20Space) in which the topology remembers only whether a subset is empty or the whole space. This is the indiscrete topology. It is not important because it resembles Euclidean space. It is important because it is the coarsest possible topology, and therefore it tests how much a topological argument really uses. If a statement is meant to hold in all topological spaces, it has to survive this extreme case. If it fails here, the missing hypothesis is often some separation axiom, such as $T_0$, $T_1$, or Hausdorffness. [example: A Two-Point Space With No Separation] Let $X=\{a,b\}$ and let $\tau=\{\varnothing,X\}$. An [open set](/page/Open%20Set) containing $a$ must be an element of $\tau$ and must contain $a$; $\varnothing$ does not contain $a$, while $X$ does, so the only open neighbourhood of $a$ is $X$. The same argument shows that the only open neighbourhood of $b$ is $X$. Hence there are no disjoint open sets $U,V\in\tau$ with $a\in U$ and $b\in V$, because then necessarily $U=X$ and $V=X$, and $U\cap V=X\neq\varnothing$. The closed sets are the complements of the open sets: \begin{align*} X\setminus \varnothing=X \end{align*} and \begin{align*} X\setminus X=\varnothing. \end{align*} Thus the only closed sets are $\varnothing$ and $X$. If $A\subset X$ is nonempty, then $\varnothing$ does not contain $A$, so the only [closed set](/page/Closed%20Set) containing $A$ is $X$, and therefore $\overline{A}=X$. In particular, for $A=\{a\}$ we get $\overline{\{a\}}=X$, so $\{a\}$ is not closed because its closure is larger than itself. Now let $(x_n)_{n\in\mathbb{N}}$ be the constant sequence $x_n=a$. To check convergence to $a$, the only open set containing $a$ is $X$, and $x_n=a\in X$ for every $n\in\mathbb{N}$. Taking $N=1$ satisfies the convergence condition. To check convergence to $b$, the only open set containing $b$ is also $X$, and again $x_n=a\in X$ for every $n\in\mathbb{N}$. Taking $N=1$ works again. Thus the same sequence converges to both points, showing that this topology does not distinguish $a$ from $b$ by separation, closure, or sequential limits. [/example] The example shows the main tension of the subject. The axioms for topology alone speak about open covers, continuity, closure, and convergence, but they do not force points to be distinguishable. The indiscrete topology isolates this phenomenon in its purest form. ## Definition Now ask for the smallest possible amount of topological information. The topology axioms force $\varnothing$ and $X$ to be open in any topology on $X$. If we add no other open sets, the axioms still hold. This gives a space in which open sets do not distinguish locations inside $X$, and this is the page's central object. [definition: Indiscrete Topology] Let $X$ be a set. The indiscrete topology on $X$ is the topology \begin{align*} \tau_{\mathrm{ind}} = \{\varnothing, X\}. \end{align*} A topological space $(X,\tau_{\mathrm{ind}})$ is called an indiscrete space. [/definition] The definition uses the fewest open sets possible. Some texts call it the topology with no proper open sets; this page uses "indiscrete" because it emphasizes the failure to distinguish points. [remark: Empty and Singleton Cases] If $X=\varnothing$, then $\tau_{\mathrm{ind}}=\{\varnothing\}$. If $X$ has one point, then the indiscrete topology is also the [discrete topology](/page/Discrete%20Topology). For sets with at least two points, the indiscrete topology is genuinely different from the discrete topology. [/remark] ## Minimality Among Topologies ### Topologies on a Set To prove that the definition really is minimal, we need the ambient language in which it lives. The point of a topology is to select a family of subsets stable under the operations needed for local arguments: arbitrary unions and finite intersections. These two closure properties are the minimum required to make the phrase "open neighbourhood" behave coherently, so the first task is to name the structure that records them. [definition: Topology] Let $X$ be a set. A topology on $X$ is a collection $\tau \subset \mathcal{P}(X)$ such that $\varnothing \in \tau$, $X \in \tau$, arbitrary unions of elements of $\tau$ belong to $\tau$, and finite intersections of elements of $\tau$ belong to $\tau$. [/definition] The next question is what the object of study should be once such a collection has been chosen. Since the same set can carry many different topologies, the mathematical object must remember both the points and the selected open subsets. [definition: Topological Space] A topological space is a pair $(X,\tau)$ where $X$ is a set and $\tau$ is a topology on $X$. [/definition] ### Coarser and Finer Structures To turn the phrase "fewest open sets" into mathematics, topologies on the same set must be compared by inclusion. Fewer open sets means fewer tests for continuity and fewer open neighbourhoods with which to separate points. This comparison is the language in which the universal minimality of the indiscrete topology can be stated. [definition: Coarser and Finer Topologies] Let $X$ be a set, and let $\tau_1$ and $\tau_2$ be topologies on $X$. The topology $\tau_1$ is coarser than $\tau_2$ if $\tau_1 \subset \tau_2$. The topology $\tau_2$ is finer than $\tau_1$ if $\tau_1 \subset \tau_2$. [/definition] The comparison just defined gives a partial order on all topologies on $X$. The natural question is whether this order has a smallest element, and the answer is exactly the indiscrete topology. [quotetheorem:8867] This result explains why the indiscrete topology appears whenever a construction asks for the weakest topology satisfying some condition. If the condition imposes no open-set requirements, the answer is often the indiscrete topology. ## Continuity as an Extremal Property Continuity is defined by pulling open sets back to open sets. Since the indiscrete topology has almost no open sets to pull back, maps into an indiscrete space face almost no continuity tests. Maps out of an indiscrete space face the opposite problem: every open set in the codomain has a preimage, and the domain allows only two possible open preimages. ### Maps Into Indiscrete Spaces To make that asymmetry precise, recall the topological definition of continuity. It replaces epsilon-delta estimates by a condition on preimages of open sets, which is the right language for seeing how the codomain topology controls the number of tests a function must pass. [definition: Continuous Map Between Topological Spaces] Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces. A function $f:X\to Y$ is continuous if $f^{-1}(V) \in \tau_X$ for every $V \in \tau_Y$. [/definition] For maps into an indiscrete space, the codomain supplies only two open sets. Their preimages are forced to be open in every topology: the preimage of $\varnothing$ is $\varnothing$, and the preimage of the whole codomain is the whole domain. This leaves no further obstruction to continuity. [quotetheorem:4980] This theorem is the first useful diagnostic. If a proposed topological construction is defined by mapping into a space, then putting the indiscrete topology on the codomain may erase all continuity restrictions. [example: Erasing Discontinuities by Changing the Codomain Topology] Let $X=\mathbb{R}$ with its usual topology and let $Y=\mathbb{R}$ with the indiscrete topology $\tau_{\mathrm{ind}}=\{\varnothing,\mathbb{R}\}$. Define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=0$ if $x<0$ and $f(x)=1$ if $x\geq 0$. To check continuity into $(\mathbb{R},\tau_{\mathrm{ind}})$, it is enough to check the two open sets in the codomain. Their preimages are \begin{align*} f^{-1}(\varnothing)=\varnothing \end{align*} and \begin{align*} f^{-1}(\mathbb{R})=\mathbb{R}. \end{align*} Both $\varnothing$ and $\mathbb{R}$ are open in the usual topology on $\mathbb{R}$, so $f$ is continuous as a map from $X$ to the indiscrete space $Y$. With the usual topology on the codomain, the same function is not continuous. The interval $(1/2,3/2)$ is open in the usual topology, and since $f(x)=1$ exactly when $x\geq 0$, its preimage is \begin{align*} f^{-1}((1/2,3/2))=[0,\infty). \end{align*} This subset is not open in the usual topology on $\mathbb{R}$: it contains $0$, but for every $\varepsilon>0$ the point $-\varepsilon/2$ lies in $(-\varepsilon,\varepsilon)$ and does not lie in $[0,\infty)$. Thus the usual codomain topology detects the jump at $0$, while the indiscrete codomain topology has only the tests $\varnothing$ and $\mathbb{R}$ and therefore cannot detect it. [/example] ### Maps Out of Indiscrete Spaces The previous theorem might suggest that the indiscrete topology makes continuity automatic in every direction, but that is false. A nonconstant map out of an indiscrete domain can create proper nonempty preimages of open sets in the codomain. Since the domain has no proper nonempty open sets, such a preimage prevents continuity whenever the codomain can distinguish the two image points. The relevant separation condition is the $T_0$ axiom: for any two distinct points, at least one of them has an open neighborhood that does not contain the other. This is exactly the amount of topological distinguishability needed to force maps from an indiscrete space to collapse all points together. [quotetheorem:8868] The $T_0$ hypothesis is not decorative. It is exactly the condition that distinct points in the codomain can be told apart by at least one open set. Without it, nonconstant continuous maps may exist. [example: Nonconstant Maps Between Indiscrete Spaces] Let $X=\{a,b\}$ and $Y=\{0,1\}$, and give both sets the indiscrete topology: \begin{align*} \tau_X=\{\varnothing,X\} \end{align*} and \begin{align*} \tau_Y=\{\varnothing,Y\}. \end{align*} Define $f:X\to Y$ by $f(a)=0$ and $f(b)=1$. The function is not constant, since $a,b\in X$ and \begin{align*} f(a)=0\neq 1=f(b). \end{align*} To check continuity, let $V\in\tau_Y$. Since $\tau_Y=\{\varnothing,Y\}$, there are only two cases. If $V=\varnothing$, then \begin{align*} f^{-1}(V)=f^{-1}(\varnothing)=\varnothing, \end{align*} and $\varnothing\in\tau_X$. If $V=Y$, then every value of $f$ lies in $Y$, so \begin{align*} f^{-1}(V)=f^{-1}(Y)=X, \end{align*} and $X\in\tau_X$. Thus the preimage of every open subset of $Y$ is open in $X$, so $f$ is continuous by the definition of continuity. This example shows that continuity from an indiscrete space does not force constancy unless the codomain has enough open sets to distinguish the two image points. [/example] This continuity behavior is why the indiscrete topology is a useful stress test. It separates statements about continuity itself from statements that also rely on separation properties of the spaces involved. ## Closure, Interior, and Dense Subsets Open sets in an indiscrete space are scarce, so closed sets are scarce as well. This makes closure very large and interior very small. The resulting formulas are simple, but they have important consequences: every nonempty subset is dense, and most subsets have empty interior. ### Closure and Interior The closure of a set is usually the smallest closed set containing it. In the indiscrete topology, there are only two closed sets to choose from. We need the general definition before calculating what it becomes in this extreme setting. [definition: Closure] Let $(X,\tau)$ be a topological space and let $A\subset X$. The closure of $A$, denoted $\overline{A}$, is the intersection of all closed subsets of $X$ that contain $A$. [/definition] Closure records the points that cannot be separated from $A$ by open neighbourhoods. The dual operation asks which points of $A$ have an open neighbourhood still inside $A$, and this is the right operation for measuring how much open room a subset contains. [definition: Interior] Let $(X,\tau)$ be a topological space and let $A\subset X$. The interior of $A$, denoted $A^\circ$, is the union of all open subsets of $X$ contained in $A$. [/definition] The next task is to compute these two operations in the smallest topology. Since the only open sets are $\varnothing$ and $X$, and the only closed sets are the same two sets, the calculation should expose exactly how much information has been lost. [quotetheorem:8869] The theorem says that closure loses all internal shape. A singleton, a finite nonempty set, and a large proper subset all have the same closure: the whole space. ### Density and Boundary Density is usually interpreted as a set being spread through the whole space. In the indiscrete topology, "spread through" cannot mean metric nearness, because the topology has no small neighbourhoods. The definition therefore collapses to a test of nonemptiness. [definition: Dense Subset] Let $(X,\tau)$ be a topological space. A subset $A\subset X$ is dense in $X$ if $\overline{A}=X$. [/definition] With density phrased through closure, the previous computation immediately raises a question: which subsets have closure equal to all of $X$? In an indiscrete space, every nonempty subset does. [quotetheorem:8870] This is one of the most striking contrasts with metric spaces. In a [metric space](/page/Metric%20Space), dense sets are often large in a limiting sense. In an indiscrete space, density means only that the set is not empty. [example: A Dense Singleton] Let $X=\mathbb{R}$ with the indiscrete topology and let $A=\{0\}$. Since $0\in A$, we have $A\neq \varnothing$, so *Nonempty Subsets Are Dense in an Indiscrete Space* gives \begin{align*} \overline{A}=X=\mathbb{R}. \end{align*} By the definition of density, this means that $\{0\}$ is dense in $\mathbb{R}$ with the indiscrete topology. With the usual topology on $\mathbb{R}$, the same subset has closure $\{0\}$. Indeed, $\mathbb{R}\setminus\{0\}=(-\infty,0)\cup(0,\infty)$ is open in the usual topology, so $\{0\}$ is closed. Since $\{0\}$ is a closed set containing $A$, the definition of closure gives $\overline{A}\subset \{0\}$. Also $A\subset \overline{A}$ because every closed set used in the intersection defining $\overline{A}$ contains $A$. Therefore \begin{align*} \{0\}\subset \overline{A}\subset \{0\}. \end{align*} Hence $\overline{A}=\{0\}\neq \mathbb{R}$ in the usual topology, so $A$ is not dense there. The underlying subset has not changed; only the topology has changed the closure and therefore the density. [/example] Closure alone tells us which points cannot be separated from a set by closed-set tests, but it does not distinguish interior behavior from edge behavior. To describe the edge of a subset, we need to look simultaneously at the set and at its complement: a point should count as boundary when the topology forces it to remain close to both sides. This is exactly the obstruction that becomes extreme in the indiscrete topology, where nonempty sets and their nonempty complements close up to all of $X$. [definition: Boundary] Let $(X,\tau)$ be a topological space and let $A\subset X$. The boundary of $A$ is $\partial A = \overline{A} \cap \overline{X\setminus A}$. [/definition] The next calculation asks which subsets have both themselves and their complements dense. In the indiscrete topology this is controlled only by whether both pieces are nonempty, so proper nonempty subsets have maximal boundary. [quotetheorem:8871] The boundary formula reinforces the same lesson: the indiscrete topology collapses geometric distinctions unless they are forced by emptiness or totality. ## Convergence and Separation Sequences and nets are meant to express limiting behavior. In an indiscrete space, every open neighbourhood of a point is the whole space, so every sequence eventually lies in every neighbourhood of every point. This makes limits maximally nonunique. ### Nonunique Limits For sequences, the usual topological definition asks that every neighbourhood of the proposed limit eventually contain the sequence. The definition does not itself require uniqueness, so it is important to state it before seeing how badly uniqueness can fail. [definition: Sequence Convergence in a Topological Space] Let $(X,\tau)$ be a topological space, let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $X$, and let $x\in X$. The sequence $(x_n)_{n\in\mathbb{N}}$ converges to $x$ if for every open set $U\in\tau$ with $x\in U$, there exists $N\in\mathbb{N}$ such that $x_n\in U$ for all $n\ge N$. [/definition] This definition raises a failure mode: can a sequence have every point as a limit? The indiscrete topology answers yes, because every point has exactly the same open-neighbourhood test, and that test is only membership in the whole space. [quotetheorem:8872] This theorem is often the first warning that [uniqueness of limits](/theorems/625) is not a purely topological fact. It requires separation assumptions. ### Separation Axioms The most familiar way to recover [uniqueness of limits](/theorems/742) is to require distinct points to have disjoint neighbourhoods. That requirement is called Hausdorffness, and the indiscrete topology gives the minimal example showing why it is a real hypothesis rather than a formality. [definition: Hausdorff Space] A topological space $(X,\tau)$ is Hausdorff if for every pair of distinct points $x,y\in X$, there exist open sets $U,V\in\tau$ such that $x\in U$, $y\in V$, and $U\cap V=\varnothing$. [/definition] Hausdorffness asks for two disjoint open neighbourhoods whenever two points are different. In an indiscrete space with at least two points, the only nonempty open set is the whole space, so the definition has no room to operate. [quotetheorem:8873] The same phenomenon appears at weaker separation levels. Before asking for disjoint neighbourhoods, one can ask only that two distinct points be distinguished by some open set. The indiscrete topology cannot do even that when there are two or more points. [definition: $T_0$ Space] A topological space $(X,\tau)$ is a $T_0$ space if for every pair of distinct points $x,y\in X$, there exists an open set $U\in\tau$ such that either $x\in U$ and $y\notin U$, or $y\in U$ and $x\notin U$. [/definition] The next question is whether the weakest standard point-separation axiom survives in an indiscrete space. Since every nonempty open set is the whole space, no open set can contain exactly one of two distinct points. [quotetheorem:8874] These classifications explain why the indiscrete topology belongs in any discussion of [topological spaces](/page/Topology) and separation axioms. It is the model case where open sets carry almost no point-level information. ## Compactness and Connectedness Some global properties become automatic in the indiscrete topology. Compactness asks whether open covers have finite subcovers. Connectedness asks whether the space can be split into two separated open pieces. Since there are so few open sets, both properties become examples of how topology controls global structure. ### Compactness Compactness is a covering condition. The smaller the topology, the fewer open covers exist, and therefore the easier compactness becomes to satisfy. The indiscrete topology makes this mechanism visible because any cover of a nonempty space must contain the whole space. [definition: Compact Space] A topological space $(X,\tau)$ is compact if every [open cover](/page/Open%20Cover) of $X$ has a [finite subcover](/page/Finite%20Subcover). [/definition] The next question is what open covers look like when the only nonempty open set is $X$. In the indiscrete topology, any open cover of a nonempty space contains $X$ itself, so a one-set subcover is already present. [quotetheorem:8875] This result is a useful counterweight to metric intuition. A very large set can be compact in the indiscrete topology, even when the same underlying set is far from compact in a familiar topology. [example: An Uncountable Compact Indiscrete Space] Let $X=\mathbb{R}$ with the indiscrete topology $\tau=\{\varnothing,\mathbb{R}\}$. To prove compactness, let $\mathcal{U}\subset\tau$ be an open cover of $X$. Since $0\in\mathbb{R}$ and $\mathcal{U}$ covers $\mathbb{R}$, there is some $U_0\in\mathcal{U}$ with $0\in U_0$. The only elements of $\tau$ are $\varnothing$ and $\mathbb{R}$, and $\varnothing$ does not contain $0$, so $U_0=\mathbb{R}$. Hence $\{U_0\}$ is a finite subcover of $\mathcal{U}$, so $X$ is compact. With the usual Euclidean topology on $\mathbb{R}$, the open cover \begin{align*} \mathcal{V}=\{(-n,n): n\in\mathbb{Z}_{\ge 1}\} \end{align*} has no finite subcover. Indeed, if $(-n_1,n_1),\ldots,(-n_k,n_k)$ are finitely many members of $\mathcal{V}$ and $m=\max\{n_1,\ldots,n_k\}$, then their union is contained in $(-m,m)$, and the point $m$ is not in $(-m,m)$. Thus compactness has changed because the topology has changed: it is a property of the topological space $(X,\tau)$, not of the underlying set $X$ alone. [/example] ### Connectedness and Paths Connectedness also becomes automatic. A separation requires two nonempty disjoint open sets whose union is the whole space. The indiscrete topology has no nonempty proper open set from which such a split could begin, so the definition is worth isolating before stating the resulting extreme behavior. [definition: Connected Space] A topological space $(X,\tau)$ is connected if there do not exist nonempty open sets $U,V\in\tau$ such that $U\cap V=\varnothing$ and $U\cup V=X$. [/definition] The next test is whether an indiscrete space admits such a separation. Since the only nonempty open set is $X$, two nonempty open sets cannot be disjoint, so no open separation can exist. [quotetheorem:8876] Connectedness should not be read as geometric cohesion here. It says that the topology lacks the open sets needed to split the space. The same underlying set may become disconnected under a finer topology. [example: Same Set, Different Connectedness] Let $X=\{0,1\}$. Give $X$ first the indiscrete topology $\tau_{\mathrm{ind}}=\{\varnothing,X\}$. If $U,V\in\tau_{\mathrm{ind}}$ are both nonempty, then neither can be $\varnothing$, so $U=X$ and $V=X$. Hence \begin{align*} U\cap V=X\cap X=X\neq\varnothing. \end{align*} Therefore there do not exist nonempty disjoint open sets whose union is $X$, so $(X,\tau_{\mathrm{ind}})$ is connected. Now give the same set the discrete topology $\tau_{\mathrm{disc}}=\mathcal{P}(X)$. Since $\{0\}\subset X$ and $\{1\}\subset X$, both $\{0\}$ and $\{1\}$ are open in the discrete topology. They are nonempty, and \begin{align*} \{0\}\cap \{1\}=\varnothing. \end{align*} Also every element of $X=\{0,1\}$ lies in at least one of the two singletons, so \begin{align*} \{0\}\cup \{1\}=X. \end{align*} Thus the discrete topology admits a separation of $X$, so $(X,\tau_{\mathrm{disc}})$ is disconnected. The underlying set has not changed; connectedness changes because the topology changes. [/example] A stronger-looking version of connectedness asks for paths between points. To test whether the indiscrete topology also forces that property, we need the definition that encodes paths as continuous functions from the unit interval. [definition: Path Connected Space] A topological space $(X,\tau)$ is path connected if for every $x,y\in X$, there exists a [continuous function](/page/Continuous%20Function) $\gamma:[0,1]\to X$ such that $\gamma(0)=x$ and $\gamma(1)=y$, where $[0,1]$ has the usual topology. [/definition] The next question is whether arbitrary set-theoretic paths become continuous when the target is indiscrete. Since every map into an indiscrete space is continuous, even a path that jumps from one point to another passes the open-preimage test. [quotetheorem:8877] This provides a compact example of how [path connectedness](/page/Path%20Connectedness) can become weak when the target topology is too coarse. The usual intuition that paths vary continuously in a geometric sense depends on the codomain topology being able to detect variation. ## Quotients, Products, and Constructions The indiscrete topology is not only an isolated example. It appears naturally from standard topological constructions, especially when the construction forgets distinctions. Quotients can collapse open sets, and products of indiscrete spaces stay indiscrete. ### Quotient Topologies A [quotient topology](/page/Quotient%20Topology) is designed to make a surjection continuous while using as few open sets as possible on the target. If the surjection is so coarse that no proper subset of the target has open preimage, the quotient topology becomes indiscrete. [definition: Quotient Topology] Let $(X,\tau_X)$ be a topological space, let $Y$ be a set, and let $q:X\to Y$ be a surjective function. The quotient topology on $Y$ induced by $q$ is the collection of subsets $V\subset Y$ such that $q^{-1}(V)\in\tau_X$. [/definition] This definition is a source of many non-Hausdorff spaces. The indiscrete topology is the most compressed possibility, and it can appear even when the quotient map is a bijection if the domain topology is already indiscrete. [example: A Quotient That Produces the Indiscrete Topology] Let $X=\mathbb{R}$ with its usual topology, let $Y=\{0,1\}$, and define $q:X\to Y$ by \begin{align*} q(x)=0 \text{ if } x\in\mathbb{Q}, \quad q(x)=1 \text{ if } x\in\mathbb{R}\setminus\mathbb{Q}. \end{align*} This map is surjective because $q(0)=0$ and $q(\sqrt{2})=1$. The subsets of $Y$ are \begin{align*} \varnothing,\quad \{0\},\quad \{1\},\quad Y. \end{align*} Their preimages under $q$ are \begin{align*} q^{-1}(\varnothing)=\varnothing. \end{align*} Also, \begin{align*} q^{-1}(\{0\})=\{x\in\mathbb{R}:q(x)=0\}=\mathbb{Q}. \end{align*} Similarly, \begin{align*} q^{-1}(\{1\})=\{x\in\mathbb{R}:q(x)=1\}=\mathbb{R}\setminus\mathbb{Q}. \end{align*} Finally, \begin{align*} q^{-1}(Y)=\mathbb{R}. \end{align*} The set $\mathbb{Q}$ is not open in the usual topology on $\mathbb{R}$. Indeed, if $r\in\mathbb{Q}$ and $\varepsilon>0$, choose an integer $n$ with \begin{align*} n>\frac{\sqrt{2}}{\varepsilon}. \end{align*} Then $\sqrt{2}/n<\varepsilon$, so \begin{align*} r+\frac{\sqrt{2}}{n}\in(r-\varepsilon,r+\varepsilon). \end{align*} The number $r+\sqrt{2}/n$ is irrational, because if it were rational, then $\sqrt{2}=n((r+\sqrt{2}/n)-r)$ would be rational. Thus every usual open interval around $r$ contains a point outside $\mathbb{Q}$, so no rational point has a usual open interval contained in $\mathbb{Q}$. The set $\mathbb{R}\setminus\mathbb{Q}$ is not open either. If $s\in\mathbb{R}\setminus\mathbb{Q}$ and $\varepsilon>0$, choose an integer $n$ with $n>1/\varepsilon$. By the [Archimedean property](/theorems/737), there is an integer $m$ such that \begin{align*} ns-1<m<ns+1. \end{align*} Then $m/n\in\mathbb{Q}$ and \begin{align*} \left|s-\frac{m}{n}\right|<\frac{1}{n}<\varepsilon. \end{align*} So every usual open interval around $s$ contains a rational point, and no irrational point has a usual open interval contained in $\mathbb{R}\setminus\mathbb{Q}$. Therefore the subsets of $Y$ whose preimages are open in the usual topology on $\mathbb{R}$ are exactly $\varnothing$ and $Y$. By the definition of the quotient topology, the quotient topology on $Y$ induced by $q$ is \begin{align*} \{\varnothing,Y\}. \end{align*} Thus this quotient produces the indiscrete topology on the two-point set $Y$. [/example] ### Products Products show a different stability property. Product topologies are generated by coordinate projections, so the open sets available in the factors determine the initial open tests in the product. If each factor is indiscrete, those initial tests carry no proper information. [definition: Product Topology] Let $(X_i,\tau_i)_{i\in I}$ be a family of topological spaces. The [product topology](/page/Product%20Topology) on $\prod_{i\in I} X_i$ is the topology generated by all sets of the form $\pi_i^{-1}(U_i)$, where $i\in I$, $U_i\in\tau_i$, and $\pi_i:\prod_{j\in I}X_j\to X_i$ is the coordinate projection. [/definition] The next question is whether the product construction can create new open sets out of indiscrete factors. For each coordinate projection, the preimage of an open set is either empty or the whole product, so the subbasis supplies no proper nonempty open set. [quotetheorem:8878] This stability under products is one reason the indiscrete topology appears in categorical treatments of topology. It behaves like a minimal structure that is preserved by many forgetful constructions. A related categorical phrase is that the indiscrete topology is the weakest topology on $X$ making every function from a topological space into $X$ continuous. Dually, the discrete topology is the strongest topology making every function from $X$ into a topological space continuous. This initial-versus-final viewpoint is often the cleanest way to remember which direction of mapping becomes automatic. ## Beyond and Connected Topics The [indiscrete topology is the coarsest topology](/theorems/8867) on a fixed set, so its natural counterpart is the discrete topology, the finest topology. Comparing the two gives a useful axis for thinking about topological structure: finer topologies make more maps out of the space continuous, while coarser topologies make more maps into the space continuous. Separation axioms are the main next topic. The indiscrete topology fails $T_0$, $T_1$, and Hausdorff separation on every set with at least two points, so it provides a compact test case for why these hypotheses appear in uniqueness-of-limits theorems, compactness results, and quotient-space arguments. In preorder language, all points in an indiscrete space specialize to all other points, so the specialization preorder collapses every point into a single indistinguishable class. The notes [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology) are a natural continuation for these point-set ideas. Quotient spaces are another direction. Many quotient constructions identify points so aggressively that the resulting topology becomes non-Hausdorff, and the indiscrete topology is the extreme model of this behavior. This connects with constructions in [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology), where quotient spaces and gluing operations become central. In algebraic topology, the indiscrete topology is a warning that homotopy-theoretic invariants depend on the topology, not just the underlying set. Since [maps into an indiscrete space are continuous](/theorems/4980), homotopies into such spaces are abundant. The more advanced viewpoint belongs naturally with [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology). The subject also connects back to elementary analysis. The contrast between usual topology on $\mathbb{R}$ and the indiscrete topology on the same set helps separate metric facts from genuinely topological facts. For the analysis background leading into open sets, closed sets, and convergence, see [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). ## References Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology). Androma, [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology). Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Munkres, *Topology* (2000). Willard, *General Topology* (1970). Kelley, *General Topology* (1955).