A Topology with Only Two Open Sets
A topology does not have to declare many subsets open. At one end of the simplest possibilities, a set can have just two open sets: the empty set and the whole set. This choice is called the indiscrete topology. Its definition is short, but checking it on different underlying sets helps separate the topology from the elements of the set itself: changing the underlying set changes the whole set, while the rule for which subsets are open remains the same.
Recall that a topology \(\mathcal{T}\) on a set \(X\) is a collection of subsets of \(X\) satisfying three conditions: \(\varnothing\) and \(X\) belong to \(\mathcal{T}\); the union of any collection of members of \(\mathcal{T}\) belongs to \(\mathcal{T}\); and the intersection of finitely many members of \(\mathcal{T}\) belongs to \(\mathcal{T}\). The members of \(\mathcal{T}\) are the open sets.
In particular, “only two open sets” means that no proper nonempty subset of \(X\) is open. The definition does not depend on whether \(X\) is finite or infinite, or on whether its elements have a familiar interpretation.
Why This Collection Is a Topology
Proof. First suppose \(X\ne\varnothing\). The collection contains \(\varnothing\) and \(X\), as required. Consider any indexed collection of its members. If at least one member is \(X\), the union is \(X\); if every member is \(\varnothing\), the union is \(\varnothing\). If the indexed collection has no members, its union is also \(\varnothing\). Thus every such union belongs to \(\{\varnothing,X\}\). For a finite nonempty collection, if every member is \(X\), its intersection is \(X\); if at least one member is \(\varnothing\), its intersection is \(\varnothing\). The intersection of no sets, by convention, is \(X\). Thus the finite-intersection condition holds as well. Now suppose \(X=\varnothing\). The collection is \(\{\varnothing\}\); every union of its members is \(\varnothing\), and every finite intersection is \(\varnothing\), including the empty intersection since the whole set is \(\varnothing\). It therefore satisfies all three topology axioms in this case too. \(\square\)
The empty-set case matters because the symbols \(\varnothing\) and \(X\) do not denote two different sets when \(X=\varnothing\). The topology is still defined, but it has one member rather than two distinct members.
Worked Example: A Three-Point Set
Let \(X=\{a,b,c\}\) and equip it with \(\mathcal{T}_{\mathrm{ind}}=\{\varnothing,X\}\). There are \(2^3=8\) subsets of \(X\): \(\varnothing\), \(\{a\}\), \(\{b\}\), \(\{c\}\), \(\{a,b\}\), \(\{a,c\}\), \(\{b,c\}\), and \(\{a,b,c\}\). Applying the definition, the open sets are exactly \(\varnothing\) and \(\{a,b,c\}=X\). Each of the other six subsets is nonempty and proper, so none is open.
For example, \(\{a,b\}\) is not open, even though it contains two of the three points. Its size does not change the test: it is not equal to \(\varnothing\), and it is not equal to \(X\). Likewise, \(\{a\}\) is not open. This example checks that the topology is determined by the stated collection, not by a rule such as “small subsets are open.”
Check: the collection includes the required empty set and whole set. A union of sets chosen from \(\varnothing\) and \(X\) is \(X\) if any chosen set is \(X\), and otherwise is \(\varnothing\); a finite intersection is \(X\) if all chosen sets are \(X\), and otherwise is \(\varnothing\). So the topology axioms hold in this concrete case.
Worked Example: The Real Line as an Underlying Set
Take the set \(X=\mathbb{R}\), but equip it with the indiscrete topology rather than assuming any other topology. Thus \(\mathcal{T}_{\mathrm{ind}}=\{\varnothing,\mathbb{R}\}\). The open sets are precisely \(\varnothing\) and \(\mathbb{R}\).
The subset \((-1,1)\) is nonempty: it contains \(0\). It is proper: for example, \(2\notin(-1,1)\), since \(2\) is not strictly between \(-1\) and \(1\). Therefore \((-1,1)\) is not open in this topology. The singleton \(\{0\}\) is also nonempty and proper, since \(0\in\{0\}\) and \(1\notin\{0\}\), so it is not open either. These conclusions concern the chosen topology on \(\mathbb{R}\); the fact that these sets are familiar subsets of the real line does not make them open here.
Check: the open-set list has not omitted any additional set: by definition a subset is open only if it equals \(\varnothing\) or the whole underlying set \(\mathbb{R}\). Neither \((-1,1)\) nor \(\{0\}\) equals either of those two sets.
Worked Example: Singleton and Empty Underlying Sets
First let \(X=\{p\}\). Its indiscrete topology is \(\{\varnothing,\{p\}\}\). In this case these are also all subsets of \(X\): the only possible membership choices for a subset are to omit \(p\), giving \(\varnothing\), or include \(p\), giving \(\{p\}\). Hence every subset of this singleton happens to be open.
Now let \(X=\varnothing\). Its power set, meaning the set of all its subsets, is \(\mathcal{P}(X)=\{\varnothing\}\): there is just one subset, the empty set. The indiscrete topology is therefore \(\{\varnothing\}\), not a collection with two distinct members. It contains both required sets because \(\varnothing=X\) in this case.
Check: for the singleton, the displayed topology contains exactly the two subsets \(\varnothing\) and \(\{p\}\). For the empty underlying set, the only subset is \(\varnothing\), so the displayed topology lists every subset once.
Open Sets in Subspaces
If \(A\subseteq X\), the subspace topology on \(A\) consists of the sets \(A\cap U\) as \(U\) ranges over the open sets of \(X\). This definition lets us check what happens when we keep only some of the points of an indiscrete space.
Proof. The only open sets \(U\) of \(X\) are \(\varnothing\) and \(X\). Intersecting \(A\) with these gives \(A\cap\varnothing=\varnothing\) and \(A\cap X=A\), because \(A\subseteq X\). There are no other open sets \(U\) to use in forming intersections, so the subspace topology is exactly \(\{\varnothing,A\}\). When \(A=\varnothing\), both listed intersections are the same set, \(\varnothing\), as stated. \(\square\)
Worked Example: Restricting to Two Points
Let \(X=\{a,b,c\}\) have the indiscrete topology and take \(A=\{a,c\}\). To find the subspace topology, intersect \(A\) with each open set of \(X\). For \(U=\varnothing\), \(A\cap U=\{a,c\}\cap\varnothing=\varnothing\). For \(U=X\), \(A\cap U=\{a,c\}\cap\{a,b,c\}=\{a,c\}=A\). Thus the open sets in the subspace are exactly \(\varnothing\) and \(\{a,c\}\). In particular, \(\{a\}\) is not open in this subspace: it is neither empty nor all of \(A\).
Check: the only open sets available before intersecting were \(\varnothing\) and \(\{a,b,c\}\), and the two intersections just computed produce exactly the claimed list.
Continuous Maps and the Indiscrete Topology
A function \(f:X\to Y\) is continuous if for every open set \(V\) in \(Y\), its preimage \(f^{-1}(V)=\{x\in X:f(x)\in V\}\) is open in \(X\). The open sets in the codomain determine which preimages must be checked. If the codomain is indiscrete, that check has a particularly simple outcome.
Proof. Let \(V\) be any open subset of \(Y\). Since \(Y\) is indiscrete, either \(V=\varnothing\) or \(V=Y\). If \(V=\varnothing\), then \(f^{-1}(V)=\varnothing\). If \(V=Y\), then \(f^{-1}(V)=X\), because every value of \(f\) belongs to \(Y\). Both \(\varnothing\) and \(X\) are open in the topology on \(X\). Thus the preimage of every open set in \(Y\) is open in \(X\), which is the definition of continuity. This reasoning also covers an empty domain or an empty codomain: the possible preimages remain open. \(\square\)
Worked Example: Checking a Map into an Indiscrete Space
Let \(X=\{r,s\}\) with topology \(\mathcal{T}_X=\{\varnothing,\{r\},X\}\), and let \(Y=\{u,v\}\) have the indiscrete topology \(\mathcal{T}_Y=\{\varnothing,Y\}\). Define \(f(r)=u\) and \(f(s)=v\). The open set \(\varnothing\) in \(Y\) has preimage \(\varnothing\) in \(X\). The open set \(Y\) has preimage \(\{r,s\}=X\), since both \(f(r)\) and \(f(s)\) lie in \(Y\). Both preimages are open in \(X\), so \(f\) is continuous.
For contrast, consider the reverse direction with \(X=\{r,s\}\) indiscrete and \(Y=\{u,v\}\) with topology \(\{\varnothing,\{u\},Y\}\). This is a topology: the union of any chosen sets from this collection is one of \(\varnothing,\{u\},Y\), and the intersection of any finite collection is also one of these sets. Define \(g(r)=u\) and \(g(s)=v\). The open set \(\{u\}\subseteq Y\) has preimage \(g^{-1}(\{u\})=\{r\}\). But \(\{r\}\) is nonempty and proper in \(X\), so it is not open in the indiscrete topology on \(X\). Therefore \(g\) is not continuous.
Check: the failed continuity test is an explicit preimage calculation: \(g(r)=u\) puts \(r\) in the preimage, and \(g(s)=v\ne u\) leaves \(s\) out. Hence the preimage is exactly \(\{r\}\), not \(\varnothing\) or \(X\).
A Common Mistake
Common pitfall: treating the name “indiscrete” as if it meant that every subset is open. For \(X=\{a,b\}\), a wrong calculation might list \(\mathcal{T}=\{\varnothing,\{a\},\{b\},X\}\), then conclude that \(\{a\}\) is open. The correction is to apply the definition: the indiscrete topology is \(\{\varnothing,X\}\), so for this \(X\) it is \(\{\varnothing,\{a,b\}\}\). The set \(\{a\}\) is nonempty and is not equal to \(X\), hence it is not open. Always compare a proposed open set with the two sets named in the definition.
This topology is useful as a clean test of definitions. It makes the list of open sets as short as possible while still satisfying the topology axioms. In particular, checking openness here is an equality test: a subset is open precisely when it equals the empty set or the whole underlying set. The same check applies whether the underlying set has no elements, one element, several elements, or infinitely many elements.
Check Your Understanding
For each question, begin by writing down the underlying set and its indiscrete topology.
- List all the open sets when \(X=\{1,2,3,4\}\), and decide whether \(\{1,3\}\) is open.
- What is the indiscrete topology on \(X=\varnothing\)? How many distinct open sets does it have?
- If \(X\) is indiscrete and \(A\subseteq X\), determine the subspace topology on \(A\), including the case \(A=\varnothing\).
- Explain why every function into an indiscrete codomain is continuous by calculating the preimages of its open sets.
- Give a function from an indiscrete domain to a space with an open proper nonempty subset that is not continuous, and identify the failing preimage.