Separating Two Points by Open Sets
A topological space may have many open neighborhoods around each point, but the neighborhoods of two different points need not be disjoint. The Hausdorff condition asks for exactly this stronger form of separation: given two distinct points, can we find open neighborhoods that do not meet? Metric spaces provide a systematic way to do so, while the discrete topology provides an especially direct construction.
Throughout, \((X,\mathcal{T})\) denotes a topological space. A neighborhood used below is an open set containing the point in question. Two sets \(U\) and \(V\) are disjoint when \(U\cap V=\varnothing\).
The quantifiers matter: the neighborhoods may depend on the pair \(x,y\), and the condition is required for every distinct pair. If \(X\) has fewer than two points, there are no distinct pairs to test, so the condition holds vacuously. Merely finding different neighborhoods is not enough; they must have empty intersection.
Metric Balls Give Separation
A metric on \(X\) is a function \(d:X\times X\to[0,\infty)\) satisfying, for all \(x,y,z\in X\), the four axioms \(d(x,y)\geq 0\), \(d(x,y)=0\) if and only if \(x=y\), \(d(x,y)=d(y,x)\), and \(d(x,z)\leq d(x,y)+d(y,z)\). The topology induced by \(d\) consists of those sets \(U\subseteq X\) for which every \(x\in U\) has some \(r>0\) with \(B_d(x,r)\subseteq U\), where \(B_d(x,r)=\{z\in X:d(x,z)<r\}\). In this topology, each open ball is open and contains its center.
Proof. Let \(x,y\in X\) be distinct. By the metric axioms, \(d(x,y)>0\). Set \(r=d(x,y)/3\), and let \(U=B_d(x,r)\) and \(V=B_d(y,r)\). These are open neighborhoods of \(x\) and \(y\), respectively. Suppose that \(z\in U\cap V\). Then \(d(x,z)<d(x,y)/3\) and \(d(z,y)=d(y,z)<d(x,y)/3\), where symmetry gives the equality. The triangle inequality would imply \(d(x,y)\leq d(x,z)+d(z,y)<2d(x,y)/3\). Since \(d(x,y)>0\), this is impossible. Thus \(U\cap V=\varnothing\), and the neighborhoods separate the arbitrary distinct points \(x,y\). If \(X\) has no distinct points, the defining condition is vacuous. Hence every metric space is Hausdorff. \(\square\)
The proof relies on positive distance between distinct points and the triangle inequality. The choice \(d(x,y)/3\) leaves room for a strict contradiction if a point were in both balls. More generally, any positive radius less than \(d(x,y)/2\) works by the same argument.
Worked Example: Separating Two Points in the Real Line
Give disjoint open neighborhoods of \(0\) and \(3\) in \(\mathbb{R}\) with its standard topology. The standard metric is \(d(s,t)=|s-t|\), so \(d(0,3)=|0-3|=3\). The theorem's choice of radius is \(3/3=1\). The corresponding open balls are \(B(0,1)=\{t:|t|<1\}=(-1,1)\) and \(B(3,1)=\{t:|t-3|<1\}=(2,4)\). The first contains \(0\), the second contains \(3\), and the intervals do not overlap because every point of the first is less than \(1\), while every point of the second is greater than \(2\). Thus they are disjoint open neighborhoods. This explicit construction verifies the Hausdorff separation for this pair.
Euclidean Spaces
For \(n\geq 1\), give \(\mathbb{R}^n\) its standard topology, the topology induced by the Euclidean metric \(d(x,y)=\|x-y\|\). The Euclidean norm satisfies \(\|v\|\geq0\), \(\|v\|=0\) exactly when \(v=0\), and \(\|v\|=\|-v\|\). Its triangle inequality follows from the Cauchy–Schwarz inequality: for vectors \(u,v\), \(\|u+v\|^2=\|u\|^2+2\langle u,v\rangle+\|v\|^2\leq(\|u\|+\|v\|)^2\), so taking nonnegative square roots gives \(\|u+v\|\leq\|u\|+\|v\|\). These norm properties give all four metric axioms for \(d\). The metric-space theorem therefore proves that \(\mathbb{R}^n\) is Hausdorff. The next example carries out the construction for particular points in the plane.
Worked Example: Two Points in the Euclidean Plane
In \(\mathbb{R}^2\) with its standard Euclidean topology, separate \(p=(1,2)\) and \(q=(4,6)\). Their distance is \(\|p-q\|=\|(1-4,2-6)\|=\|(-3,-4)\|=\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=5\). Choose radius \(5/3\). The open balls \(B(p,5/3)\) and \(B(q,5/3)\) contain their respective centers, since each center is at distance \(0<5/3\) from itself. To check disjointness, if a point \(z\) belonged to both, the triangle inequality would give \(5=\|p-q\|\leq\|p-z\|+\|z-q\|<5/3+5/3=10/3\), contradicting \(5>10/3\). Thus these are disjoint open neighborhoods. The calculation uses the strict inequality in the definition of an open ball.
The same argument works for every pair of distinct vectors in \(\mathbb{R}^n\): their Euclidean distance is positive, and balls of one-third that distance are disjoint by the triangle inequality. When \(n=0\), \(\mathbb{R}^0\) is a one-point space; it is Hausdorff because there are no distinct points to separate.
Discrete Spaces
The discrete topology on a set \(X\) is the topology \(\mathcal{P}(X)\), in which every subset of \(X\) is open. No metric or coordinates are needed to separate points.
Proof. Take any distinct \(x,y\in X\). The singleton sets \(U=\{x\}\) and \(V=\{y\}\) are open because every subset is open in the discrete topology. They contain \(x\) and \(y\), respectively, and \(U\cap V=\varnothing\) because \(x\ne y\). Thus every distinct pair has disjoint open neighborhoods. If \(X\) has no distinct pair, the condition holds vacuously. Therefore the discrete space is Hausdorff. \(\square\)
Worked Example: A Finite Discrete Space
Let \(X=\{a,b,c\}\) have the discrete topology \(\mathcal{P}(X)\). To separate \(a\) and \(c\), choose \(U=\{a\}\) and \(V=\{c\}\). Both belong to \(\mathcal{P}(X)\), so both are open; \(a\in U\) and \(c\in V\); and \(U\cap V=\varnothing\). For the pair \(a,b\), the same calculation gives \(\{a\}\cap\{b\}=\varnothing\), and for \(b,c\), it gives \(\{b\}\cap\{c\}=\varnothing\). These are all three distinct pairs, so the example verifies the condition for every pair in this space. The proof for an arbitrary discrete space uses exactly the same singleton neighborhoods.
Why the Topology Matters
The same underlying set can carry different topologies, and Hausdorffness depends on the topology, not just on the points. In particular, the discrete construction does not work unless singletons are open. An infinite set with the cofinite topology shows that even a topology in which many sets are open may fail to separate points.
Worked Example: An Infinite Cofinite Space Is Not Hausdorff
Let \(X=\mathbb{Z}\) with the cofinite topology: the open sets are \(\varnothing\) and all subsets whose complements in \(\mathbb{Z}\) are finite. Choose distinct integers \(x\) and \(y\), and let \(U,V\) be any open neighborhoods of \(x,y\). Neither is empty. Therefore \(\mathbb{Z}\setminus U\) and \(\mathbb{Z}\setminus V\) are finite. Their union is finite, and \(\mathbb{Z}\setminus(U\cap V)=(\mathbb{Z}\setminus U)\cup(\mathbb{Z}\setminus V)\). If \(U\cap V\) were empty, this equality would make \(\mathbb{Z}\) finite, which it is not. Hence \(U\cap V\ne\varnothing\) for every such pair of neighborhoods. In particular, \(x\) and \(y\) cannot have disjoint open neighborhoods, so this space is not Hausdorff.
This example is a useful contrast with metric and discrete spaces. In a metric space, distance lets us shrink balls until they separate two points. In a discrete space, singletons already separate them. In the cofinite topology on an infinite set, every two nonempty open sets must intersect, so no pair of distinct points can be separated in the required way.
A Common Error About Neighborhoods
Common pitfall. A radius must be small enough for the two balls to be disjoint. For example, in \(\mathbb{R}\), the distance between \(0\) and \(2\) is \(2\). Choosing radius \(2\) gives \(B(0,2)=(-2,2)\) and \(B(2,2)=(0,4)\). Their intersection is \((0,2)\), not the empty set, so these balls do not prove Hausdorffness. With the theorem's radius \(2/3\), the balls are \(B(0,2/3)=(-2/3,2/3)\) and \(B(2,2/3)=(4/3,8/3)\); the first ends below \(2/3\) and the second begins above \(4/3\), so they are disjoint. The corrected neighborhoods contain their respective centers and satisfy the required empty-intersection condition.
For a metric-space proof, the reliable procedure is to compute the positive distance \(d(x,y)\), choose a radius strictly less than half that distance, and use the triangle inequality to rule out a point in both balls. For a discrete-space proof, use the two singleton sets and verify both openness and disjointness.
Check Your Understanding
For each question, use the definition of Hausdorffness and verify the required openness and disjointness.
- In \(\mathbb{R}\) with its standard topology, construct disjoint open neighborhoods of \(-2\) and \(1\) using balls of radius one-third their distance.
- Explain why every metric space with its metric topology is Hausdorff, identifying where positivity of \(d(x,y)\) and the triangle inequality enter.
- In a discrete space \(X\), what neighborhoods separate two distinct points \(x,y\)? Why does the argument also cover finite spaces?
- Show directly that any two nonempty open sets in an infinite cofinite space intersect.
- For points at distance \(d>0\) in a metric space, why does choosing radius \(d\) fail to guarantee disjoint balls, while choosing radius \(d/3\) does guarantee them?