When Do Open Sets Tell Two Points Apart?
Topology describes a space through its open sets. This suggests a basic question: can two different points have exactly the same open-set membership? If so, no open set can detect which of the two points is present. Topological indistinguishability gives a precise name to this situation, and the \(T_0\) axiom says exactly that it never happens for distinct points.
Let \((X,\mathcal{T})\) be a topological space. Recall that an open neighborhood of \(x\in X\) is an open set containing \(x\). Two points may have many neighborhoods in common without being indistinguishable; the defining condition compares membership in every open set of the space.
The open set witnessing distinguishability need not contain a specified one of the two points: it may contain \(x\) and not \(y\), or \(y\) and not \(x\). The requirement is not that there be disjoint neighborhoods. It asks only that at least one open set tell the points apart.
Indistinguishability Is an Equivalence Relation
An equivalence relation is a relation that is reflexive, symmetric, and transitive. Topological indistinguishability has all three properties, so it groups points into classes of points that the open sets cannot tell apart.
Proof. For reflexivity, take any \(x\in X\). For every open set \(U\), the statement \(x\in U\) has the same truth value as itself, so \(x\) is indistinguishable from \(x\). For symmetry, if \(x\in U\) if and only if \(y\in U\) for every open \(U\), then \(y\in U\) if and only if \(x\in U\) for every open \(U\). Thus \(y\) is indistinguishable from \(x\). For transitivity, suppose \(x\) is indistinguishable from \(y\) and \(y\) is indistinguishable from \(z\). For each open \(U\), \(x\in U\) if and only if \(y\in U\), and \(y\in U\) if and only if \(z\in U\). Therefore \(x\in U\) if and only if \(z\in U\) for every open \(U\). Hence \(x\) is indistinguishable from \(z\). \(\square\)
This relation is about the topology, not about whether two points are equal as elements of \(X\). A point is always indistinguishable from itself, but distinct points can also be indistinguishable in some spaces. The \(T_0\) condition rules out precisely those distinct pairs.
Proof. Suppose first that \(X\) is \(T_0\). If \(x\) and \(y\) are distinct, the definition of \(T_0\) gives an open set containing exactly one of them. They therefore do not have the same membership in every open set, so they are distinguishable. Thus no distinct points are indistinguishable.
Conversely, suppose no two distinct points are indistinguishable. Take any distinct \(x,y\in X\). Since they are not indistinguishable, the defining condition fails: there is an open set \(U\) for which it is not true that \(x\in U\) if and only if \(y\in U\). For two membership statements, failure of “if and only if” means exactly one statement is true. Thus \(U\) contains one of \(x,y\) and not the other. This holds for every distinct pair, so \(X\) is \(T_0\). If \(X\) has fewer than two points, there are no distinct pairs; both conditions hold vacuously. \(\square\)
Worked Tests in Three Topologies
Worked Example: The Indiscrete Topology
Let \(X=\{p,q,r\}\), with the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). Determine which points are indistinguishable and whether the space is \(T_0\).
There are only two open sets to test. The empty set contains none of \(p,q,r\), and \(X\) contains all three. Thus each of \(p,q,r\) has exactly the same membership in every open set. In particular, \(p\) and \(q\) are indistinguishable, as are \(p\) and \(r\), and \(q\) and \(r\). Since \(p\ne q\), the \(T_0\) condition fails.
Check: There is no other open set available to distinguish \(p\) from \(q\); the two listed open sets contain either neither point or both. This agrees with the \(T_0\) test.
Worked Example: A Three-Point Topology
On \(X=\{a,b,c\}\), let \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\). First check that this is a topology, then test every pair of distinct points.
The collection contains \(\varnothing\) and \(X\). The intersection of any two of its sets is one of \(\varnothing,\{a\},\{a,b\},X\), so it is closed under finite intersections. For arbitrary unions, a union of sets in the collection is again one of those four sets: the only possible nonempty sets are nested as \(\{a\}\subseteq\{a,b\}\subseteq X\). Thus \(\mathcal{T}\) is a topology.
The open set \(\{a\}\) contains \(a\) and not \(b\), so it distinguishes \(a\) from \(b\). It also contains \(a\) and not \(c\), so it distinguishes \(a\) from \(c\). The open set \(\{a,b\}\) contains \(b\) and not \(c\), so it distinguishes \(b\) from \(c\). Every distinct pair is distinguished, and therefore this space is \(T_0\).
Check: The three pairs have each been tested: \(\{a,b\}\), \(\{a,c\}\), and \(\{b,c\}\). Each has an explicit open-set witness, so there is no untested distinct pair that could violate \(T_0\).
Worked Example: Two Points That Open Sets Cannot Separate
On \(X=\{a,b,c\}\), let \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Determine all indistinguishable pairs and decide whether \(X\) is \(T_0\).
This is a topology: it contains \(\varnothing\) and \(X\); intersections of its sets belong to the collection; and any union of its sets is one of \(\varnothing,\{a\},X\). The open sets containing \(b\) are just \(X\), and the open sets containing \(c\) are also just \(X\). Hence \(b\) and \(c\) have the same membership in every open set and are indistinguishable.
The point \(a\) is distinguishable from \(b\), because \(\{a\}\) contains \(a\) and not \(b\). The same open set distinguishes \(a\) from \(c\). Thus the only indistinguishable pair of distinct points is \(b,c\), and the space is not \(T_0\).
Check: For \(\varnothing\), neither point is present; for \(\{a\}\), neither \(b\) nor \(c\) is present; and for \(X\), both are present. These are all the open sets, confirming the indistinguishability of \(b\) and \(c\).
Worked Example: The Standard Topology on the Real Line
Show that \(\mathbb{R}\) with its standard topology is \(T_0\). Take any distinct \(x,y\in\mathbb{R}\). If necessary, interchange their names so that \(x<y\). Set \(m=(x+y)/2\). Then \(x<m<y\): indeed, \(m-x=(y-x)/2>0\) and \(y-m=(y-x)/2>0\).
The ray \(U=(-\infty,m)\) is open in the standard topology. To check this from the interval basis, for each \(t\in U\), choose \(r=(m-t)/2>0\). Then \(t\in(t-r,t+r)\), and \(t+r=(t+m)/2<m\), so \((t-r,t+r)\subseteq U\). The ray contains \(x\) because \(x<m\), and it does not contain \(y\) because \(y>m\). Thus every distinct pair is distinguishable, proving that \(\mathbb{R}\) is \(T_0\).
Check: The midpoint satisfies \(x<(x+y)/2<y\), so the same open ray cannot contain both points or neither. It gives a valid witness for the chosen pair.
A Common Mistake: Requiring Disjoint Neighborhoods
A frequent error is to treat “distinguishable” as meaning “contained in disjoint open neighborhoods.” That is a stronger demand than the \(T_0\) condition. The finite topology in the previous example makes the difference concrete. For \(a\) and \(b\), the open set \(\{a\}\) contains \(a\) and not \(b\), so it distinguishes them. But every open neighborhood of \(b\) is \(X\). Since every open neighborhood of \(a\) is a nonempty set and \(X\) intersects every such set, no open neighborhoods of \(a\) and \(b\) are disjoint.
The incorrect calculation would be: “There are no disjoint neighborhoods of \(a\) and \(b\), so they are indistinguishable.” This conclusion does not follow. The correction is to test the definition directly: \(\{a\}\in\mathcal{T}\), \(a\in\{a\}\), and \(b\notin\{a\}\). Therefore \(a\) and \(b\) are distinguishable, even though disjoint neighborhoods do not exist. The \(T_0\) test concerns a single open set that contains one point but not the other.
Equivalence classes of indistinguishable points record exactly which points have identical open-set membership. In particular, every open set either contains all the points in such a class or contains none of them: if it contains one point, indistinguishability forces it to contain every point indistinguishable from that point. This is a useful way to read a topology: it can distinguish points only when some open set changes membership between them.
Check Your Understanding
For each question, use the definition in terms of membership in every open set.
- In the indiscrete topology on a set \(X\) with at least two points, which distinct pairs are indistinguishable, and is \(X\) \(T_0\)?
- On \(X=\{u,v,w\}\), consider \(\mathcal{T}=\{\varnothing,\{u\},\{u,v\},X\}\). Which open sets distinguish \(u\) from \(v\), \(u\) from \(w\), and \(v\) from \(w\)?
- Explain why topological indistinguishability is reflexive, symmetric, and transitive.
- Can two points be distinguishable without having disjoint open neighborhoods? Verify your answer using one of the finite spaces above.
- Prove directly from the definition that a space with exactly one point is \(T_0\).