From Disjoint Neighborhoods to Closed Singletons
The Hausdorff condition is a way to distinguish any two different points by their neighborhoods: each point has an open neighborhood that does not meet a chosen neighborhood of the other. The \(T_1\) condition asks for a related but weaker conclusion: each singleton is closed. The implication from Hausdorff to \(T_1\) follows by collecting, for a fixed point, neighborhoods that exclude it. This tutorial gives that argument in full and then checks it in metric spaces.
Let \((X,\mathcal{T})\) be a topological space. A neighborhood in this tutorial means an open set containing the point in question. A set \(C\subseteq X\) is closed if \(X\setminus C\) is open.
The two conditions use different kinds of separation. Hausdorffness gives a pair of disjoint open neighborhoods for each pair of distinct points. The \(T_1\) condition concerns whether the complement of each point is open. We will use the Hausdorff neighborhoods to show that every point other than a fixed \(x\) lies in an open set that omits \(x\).
The Main Implication
Proof. Fix \(x\in X\). Consider the union of all open sets that omit \(x\): $$ W=\bigcup\{V\in\mathcal{T}:x\notin V\}. $$ A union of open sets is open, so \(W\) is open. Also \(x\notin W\), since none of the sets in the union contains \(x\). Now let \(y\in X\setminus\{x\}\). By the Hausdorff condition, there are disjoint open sets \(U,V\) with \(x\in U\) and \(y\in V\). Disjointness and \(x\in U\) imply \(x\notin V\). Therefore \(V\) is one of the open sets in the union defining \(W\), and \(y\in V\subseteq W\). We have shown that every \(y\ne x\) belongs to \(W\), while \(x\notin W\). Thus \(W=X\setminus\{x\}\), so \(X\setminus\{x\}\) is open and \(\{x\}\) is closed.
This argument also covers the degenerate cases. If \(X=\varnothing\), there are no singletons to check, so the \(T_1\) condition holds vacuously. If \(X=\{x\}\), then \(X\setminus\{x\}=\varnothing\), which is open. In the proof, there are no points \(y\ne x\) to include in \(W\), and the union of the open sets omitting \(x\) is indeed empty. \(\square\)
The union formulation is useful because it avoids choosing one neighborhood separately for every \(y\ne x\). Instead, it takes the union of the entire collection of open sets that omit \(x\). The Hausdorff condition guarantees that each point other than \(x\) lies in at least one set from this collection.
Worked Example: A Singleton in the Real Line
Take \(\mathbb{R}\) with its standard topology and the point \(2\). We verify directly that \(\{2\}\) is closed. Its complement is $$ \mathbb{R}\setminus\{2\}=(-\infty,2)\cup(2,\infty). $$ Both rays are open in the standard topology, so their union is open and \(\{2\}\) is closed.
The neighborhoods that arise from the Hausdorff argument can also be written explicitly. Fix \(y\ne2\) and put \(r=|y-2|/3\). Since \(y\ne2\), \(r>0\), so \(V=(y-r,y+r)\) is an open neighborhood of \(y\). If \(y<2\), then \(2-y=|y-2|=3r\), so \(y+r=y+ r<y+3r=2\). Thus every point of \(V\) is less than \(2\). If \(y>2\), then \(y-2=|y-2|=3r\), so \(y-r=y-3r<y-r\) would be an invalid comparison; the correct endpoint calculation is \(y-r=y-3r+2r=2+2r>2\), using \(y-2=3r\). Hence every point of \(V\) is greater than \(2\). In either case, \(V\) omits \(2\).
Check: In the case \(y>2\), the identity \(y=2+3r\) gives \(y-r=2+2r>2\). In the case \(y<2\), the identity \(y=2-3r\) gives \(y+r=2-2r<2\). Thus the interval around \(y\) really does avoid the point \(2\), as required.
Worked Example: Separating Two Points in the Euclidean Plane
Let \(\mathbb{R}^2\) have its Euclidean topology, induced by the metric \(\|(a,b)-(c,d)\|=\sqrt{(a-c)^2+(b-d)^2}\). Consider \(x=(4,-2)\) and \(y=(1,-2)\). Their distance is $$ \|x-y\|=\sqrt{(4-1)^2+((-2)-(-2))^2} =\sqrt{3^2+0^2} =3. $$ The open balls \(B(x,1)\) and \(B(y,1)\) are disjoint. Indeed, if a point \(w\) belonged to both, the triangle inequality would give $$ 3=\|x-y\|\le \|x-w\|+\|w-y\|<1+1=2, $$ which is impossible. The weak inequality in the middle is the triangle inequality; the final strict inequality follows from membership in the two open balls.
For example, \(B(x,1)\) and \(B(y,1)\) are open neighborhoods of the stated centers and are disjoint, so they verify the Hausdorff separation for this pair. The general metric argument below shows that this separation is available for every pair of distinct points of \(\mathbb{R}^2\), and the theorem then gives that every singleton in \(\mathbb{R}^2\) is closed.
Check: The coordinate difference is \((4-1, -2-(-2))=(3,0)\), whose Euclidean norm is \(\sqrt{9}=3\). The contradiction is \(3<2\), which cannot hold.
Why Metric Spaces Are Hausdorff
The plane calculation is a special case of a general fact. In a metric space, distinct points have positive distance, and balls with sufficiently small radii around them cannot intersect.
Proof. Let \(x,y\in X\) be distinct, and set \(s=d(x,y)\). The metric axioms give \(s>0\). The sets \(B_d(x,s/3)\) and \(B_d(y,s/3)\) are open neighborhoods of \(x\) and \(y\), respectively. Suppose \(z\) belonged to both. Then \(d(x,z)<s/3\) and \(d(z,y)<s/3\), while the triangle inequality gives $$ s=d(x,y)\le d(x,z)+d(z,y)<\frac{s}{3}+\frac{s}{3}=\frac{2s}{3}. $$ Since \(s>0\), we have \(2s/3<s\), contradicting \(s<2s/3\). Thus the two balls are disjoint. This holds for every pair of distinct points, so the metric space is Hausdorff. \(\square\)
The radius \(s/3\) is a convenient choice, not the only possible one. What matters is that the sum of the two radii is no greater than \(s\); for open balls, the distances from a point in each ball are strictly less than the radii, which supplies the contradiction.
Worked Example: A Singleton in a Discrete Metric Space
Let \(X=\{a,b,c\}\) and define \(d(u,v)=0\) when \(u=v\), and \(d(u,v)=1\) when \(u\ne v\). This is a metric: it is nonnegative and symmetric, and \(d(u,v)=0\) exactly when \(u=v\). For the triangle inequality, if \(u=v\), its left side is \(0\); if \(u\ne v\), then for any \(w\), at least one of \(u\ne w\) or \(w\ne v\) holds, so \(d(u,w)+d(w,v)\ge1=d(u,v)\).
For each \(u\in X\), the open ball of radius \(1/2\) is $$ B_d(u,1/2)=\{v\in X:d(u,v)<1/2\}=\{u\}, $$ because \(d(u,u)=0<1/2\), while \(d(u,v)=1\not<1/2\) for each \(v\ne u\). In particular, \(\{a\}\) is open, and its complement \(\{b,c\}=B_d(b,1/2)\cup B_d(c,1/2)\) is open. Therefore \(\{a\}\) is also closed.
Check: The strict membership test \(d(u,v)<1/2\) includes \(v=u\), since \(0<1/2\), and excludes both other points, since \(1<1/2\) is false. The displayed ball is exactly \(\{u\}\).
What the Hypothesis Contributes
The theorem uses Hausdorffness to obtain neighborhoods that omit a chosen point. Without that hypothesis, the conclusion need not hold. For an explicit check, let \(X=\{a,b\}\) with the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). The only open neighborhood of either point is \(X\), so \(a\) and \(b\) have no disjoint open neighborhoods. Also \(X\setminus\{a\}=\{b\}\) is not open, and hence \(\{a\}\) is not closed. This example shows why the separation assumption cannot simply be dropped from the theorem.
Common pitfall. When proving two metric balls are disjoint, one might write $$ d(x,y)<d(x,z)+d(z,y). $$ That strict inequality is not the triangle inequality: the triangle inequality only guarantees \(d(x,y)\le d(x,z)+d(z,y)\). For the points \(x=(4,-2)\), \(y=(1,-2)\), and a hypothetical \(z\in B(x,1)\cap B(y,1)\), the valid calculation is $$ 3=d(x,y)\le d(x,z)+d(z,y)<1+1=2. $$ The middle comparison is weak, and the final one is strict because both ball memberships are strict. Together they give the contradiction \(3<2\). The correction matters: the proof must use exactly what the metric axioms provide, then obtain strictness from the open-ball conditions.
The central argument has a useful shape: fix a point \(x\), use separation to find open sets omitting \(x\) around every other point, and take their union. The result is the entire complement of \(\{x\}\), so that complement is open. Metric spaces supply the required disjoint neighborhoods by a direct distance calculation; the final step from Hausdorffness to closed singletons is topological and does not depend on a metric.
Check Your Understanding
For each question, keep track of which sets are open neighborhoods and which set is being shown open or closed.
- In the main proof, why does \(y\in V\) and \(U\cap V=\varnothing\), with \(x\in U\), imply \(x\notin V\)?
- For distinct points at distance \(s>0\) in a metric space, verify that balls of radius \(s/4\) are disjoint using the triangle inequality.
- In the Euclidean plane example, replace the radius \(1\) by \(3/2\). Does the same triangle-inequality argument still prove the two balls disjoint? Explain.
- What is the complement of a singleton in a one-point space, and why does this handle the \(T_1\) condition there?
- In the indiscrete topology on \(\{a,b\}\), list all open sets and use them to check both the failure of Hausdorffness and the failure of \(\{a\}\) to be closed.