Hausdorff Spaces

Separating Points in a Topological Space

In the previous tutorial, Hausdorffness was a hypothesis that helped turn compactness into closedness and make certain continuous bijections into homeomorphisms. We now examine this separation condition on its own. A metric space is Hausdorff, but general topological spaces need not come from metrics. The Hausdorff condition expresses, using open sets alone, the idea that distinct points can be kept apart.

The essential requirement is not that every pair of open sets be disjoint. Rather, for each pair of distinct points, we must be able to choose open neighborhoods of those points that are disjoint. This pointwise condition has useful consequences: it passes to subspaces and products, it can be recognized by looking at the diagonal in a product, and it guarantees that sequence limits are unique.

Definition: A topological space $X$ is Hausdorff if, whenever $x,y\in X$ and $x\ne y$, there are disjoint open sets $U,V\subseteq X$ such that $x\in U$ and $y\in V$. Such a space is also called a $T_2$ space.

The empty space and every one-point space are Hausdorff: in either case, there are no two distinct points for which the condition needs to be checked. More generally, any space with at most one point is Hausdorff. By the result established earlier in this course, every metric space is Hausdorff. The definition above applies even when no metric is available.

Worked Example: Separating Two Real Numbers

Let $a,b\in\mathbb R$ with $a\ne b$, and set $r=|a-b|/3$. The open intervals $U=(a-r,a+r)$ and $V=(b-r,b+r)$ contain $a$ and $b$, respectively. They are disjoint. Indeed, if some $z$ belonged to both, then $$ |a-b|\leq |a-z|+|z-b|<r+r=\frac{2|a-b|}{3}, $$ which is impossible because $|a-b|>0$. Thus the usual topology on $\mathbb R$ separates every pair of distinct points, as the general theorem for metric spaces also guarantees.

Hausdorffness Passes to Subspaces

A subspace uses the open sets inherited from its ambient space: an open set in $A\subseteq X$ has the form $A\cap O$, where $O$ is open in $X$. If two points of $A$ can be separated in $X$, intersecting their separating open sets with $A$ gives neighborhoods in the subspace. The intersections remain disjoint.

Theorem: Every subspace of a Hausdorff space is Hausdorff.

Proof. Let $X$ be Hausdorff, let $A\subseteq X$, and take distinct points $a,b\in A$. By Hausdorffness of $X$, there are disjoint open sets $U,V\subseteq X$ with $a\in U$ and $b\in V$. The sets $A\cap U$ and $A\cap V$ are open in the subspace $A$, contain $a$ and $b$, respectively, and are disjoint, since $U\cap V=\varnothing$. Thus every pair of distinct points of $A$ has disjoint open neighborhoods in $A$, so $A$ is Hausdorff. If $A$ has at most one point, the conclusion holds directly from the definition as well. $\square$

Worked Example: The Irrational Numbers as a Hausdorff Subspace

Let $H=\mathbb R\setminus\mathbb Q$, with the subspace topology inherited from $\mathbb R$. Consider the distinct points $a=\sqrt{2}$ and $b=\pi$, both in $H$. Set $r=(\pi-\sqrt{2})/3$, which is positive. The sets $$ H\cap(\sqrt{2}-r,\sqrt{2}+r) \qquad\text{and}\qquad H\cap(\pi-r,\pi+r) $$ are open in $H$, contain $\sqrt{2}$ and $\pi$, respectively, and are disjoint. The disjointness follows from the same interval calculation as in the preceding example. The subspace theorem gives the stronger conclusion that every pair of distinct irrational numbers can be separated in $H$, not only this particular pair.

This result is useful when working with subsets of familiar Hausdorff spaces. For example, a circle or an interval regarded as a subspace of a Euclidean space is Hausdorff without requiring a new separation argument for each pair of its points. The inherited topology matters: the theorem concerns the subspace topology, not an arbitrary topology one might separately place on the same set.

Products of Hausdorff Spaces

The product topology is designed so that conditions on finitely many coordinates define basic open sets. In particular, if $X=\prod_{i\in I}X_i$, then for each coordinate $i$, the coordinate projection $p_i:X\to X_i$ is continuous. To separate two distinct product points, it is enough to find a coordinate where they differ and use separation in that one factor.

Theorem: A product of Hausdorff spaces, equipped with the product topology, is Hausdorff.

Proof. Let $X=\prod_{i\in I}X_i$, where each $X_i$ is Hausdorff. If $X$ is empty, it is Hausdorff because it has no distinct points. If $I$ is empty, the product is a one-point space and is Hausdorff. In the remaining case, take distinct points $x=(x_i)_{i\in I}$ and $y=(y_i)_{i\in I}$ in $X$. Since they are not equal, there is an index $i\in I$ for which $x_i\ne y_i$. Because $X_i$ is Hausdorff, there are disjoint open sets $U,V\subseteq X_i$ with $x_i\in U$ and $y_i\in V$. Then $p_i^{-1}(U)$ and $p_i^{-1}(V)$ are open in $X$, contain $x$ and $y$, respectively, and are disjoint. Indeed, a point in both inverse images would have its $i$-th coordinate in $U\cap V$, which is empty. Therefore $X$ is Hausdorff. $\square$

The proof also explains why there is no need to assume that every factor has at least two points. A factor with at most one point is Hausdorff by definition. If two distinct points of the product exist, however, they must differ in some coordinate; that particular factor then has two distinct points to separate. If no two distinct product points exist, the product is Hausdorff immediately.

Worked Example: Separating Points in a Product

Give $\mathbb R\times\{0,1\}$ the product topology, where $\mathbb R$ has its usual topology and $\{0,1\}$ has the discrete topology. The discrete topology makes every subset open, so $\{0\}$ and $\{1\}$ are disjoint open sets. Take distinct points $(a,i)$ and $(b,j)$.

If $a\ne b$, let $r=|a-b|/3$. The open rectangles $$ (a-r,a+r)\times\{0,1\} \qquad\text{and}\qquad (b-r,b+r)\times\{0,1\} $$ contain the two points and are disjoint by the interval calculation above. If $a=b$, distinctness of the product points forces $i\ne j$. In that case, $\mathbb R\times\{i\}$ and $\mathbb R\times\{j\}$ are disjoint open neighborhoods of the two points. Both cases give the required separation.

The Diagonal Detects Hausdorffness

For a space $X$, its diagonal is the subset of $X\times X$ consisting of pairs whose coordinates agree. The criterion below recasts point separation as a closed-set property in the product. It is useful because closed sets and product topologies can sometimes be easier to analyze than neighborhoods for every pair of points.

Theorem: A topological space $X$ is Hausdorff if and only if its diagonal $\Delta=\{(x,x):x\in X\}$ is closed in $X\times X$.

Proof. First suppose that $X$ is Hausdorff. Take any $(x,y)\in(X\times X)\setminus\Delta$, so $x\ne y$. Choose disjoint open sets $U,V\subseteq X$ with $x\in U$ and $y\in V$. The set $U\times V$ is open in the product topology and contains $(x,y)$. It is disjoint from $\Delta$: if $(z,z)\in U\times V$, then $z\in U\cap V$, contradicting $U\cap V=\varnothing$. Thus every point outside $\Delta$ has an open neighborhood outside $\Delta$, so $(X\times X)\setminus\Delta$ is open and $\Delta$ is closed.

Conversely, suppose $\Delta$ is closed, and take distinct $x,y\in X$. Then $(x,y)\notin\Delta$, so the complement of $\Delta$ is an open neighborhood of $(x,y)$ in $X\times X$. By the basis property for the product topology, there are open sets $U,V\subseteq X$ such that $$ (x,y)\in U\times V\subseteq(X\times X)\setminus\Delta. $$ In particular, $x\in U$ and $y\in V$. If a point $z$ belonged to $U\cap V$, then $(z,z)\in U\times V$, which would put a point of $\Delta$ inside its complement. Hence $U\cap V=\varnothing$. We have found disjoint open neighborhoods of $x$ and $y$, so $X$ is Hausdorff. If $X$ has no distinct points, the separation condition is vacuous and the same equivalence remains valid. $\square$

Sequence Limits Are Unique

In a Hausdorff space, a sequence cannot converge to two different points. Here convergence has its neighborhood meaning: a sequence $(x_n)$ converges to $x$ if every open neighborhood of $x$ contains all terms $x_n$ for sufficiently large $n$. The separation condition makes two distinct proposed limits incompatible.

Theorem: In a Hausdorff space, every convergent sequence has at most one limit.

Proof. Suppose $(x_n)$ converges to both $x$ and $y$. If $x\ne y$, Hausdorffness gives disjoint open neighborhoods $U$ of $x$ and $V$ of $y$. Convergence to $x$ gives an index $N_1$ such that $x_n\in U$ whenever $n\geq N_1$. Convergence to $y$ gives an index $N_2$ such that $x_n\in V$ whenever $n\geq N_2$. For $n\geq\max\{N_1,N_2\}$, the term $x_n$ must lie in $U\cap V$, contradicting $U\cap V=\varnothing$. Therefore $x=y$, and the limit is unique. $\square$

Worked Example: A Sequence Cannot Also Converge to Another Real Number

In $\mathbb R$, the sequence $x_n=1/n$ converges to $0$. To check this, let $\varepsilon>0$. Choose a positive integer $N$ with $N>1/\varepsilon$. For every $n\geq N$, $$ |x_n-0|=\frac{1}{n}\leq\frac{1}{N}<\varepsilon. $$ It cannot also converge to $1$. For instance, the open intervals $U=(-1/3,1/3)$ and $V=(2/3,4/3)$ are disjoint neighborhoods of $0$ and $1$. If the sequence converged to $0$, its terms would eventually lie in $U$; if it converged to $1$, its terms would eventually lie in $V$. A term sufficiently far along would have to lie in both, which is impossible. The uniqueness theorem gives this conclusion for any two proposed limits in $\mathbb R$, not just $0$ and $1$.

What the Hausdorff Condition Does—and Does Not—Say

Hausdorffness is a condition on pairs of distinct points, not a requirement that the space have many open sets or that every subset be closed. It does imply that individual points are closed: if $x\in X$ and $y\ne x$, choose disjoint open neighborhoods $U_y$ of $y$ and $V_y$ of $x$. Then $U_y\subseteq X\setminus\{x\}$. Taking the union of these open neighborhoods over all $y\ne x$ shows that $X\setminus\{x\}$ is open, so $\{x\}$ is closed.

The converse is not built into the definition: knowing that every one-point set is closed does not itself provide disjoint open neighborhoods for each pair. Keep the distinction between separating points by open sets and merely having closed points. The Infinite Cofinite Topology theorem from earlier in this course provides a useful warning: an infinite set with the cofinite topology is compact but not Hausdorff. Compactness alone does not guarantee separation.

A practical way to apply these results is to choose the form best suited to the problem. For a subspace, intersect separating neighborhoods with the subset. For a product, find a coordinate in which the points differ and separate them in that factor. For a question involving all pairs at once, the closed-diagonal characterization may be more convenient. In each case, the core requirement is the same: distinct points must have disjoint open neighborhoods.

Check Your Understanding

Use the definition and results in this tutorial to answer the following questions.

Why is every space with at most one point Hausdorff?
How do disjoint open sets in a Hausdorff space produce disjoint open neighborhoods in one of its subspaces?
In the product theorem, why must two distinct product points differ in at least one coordinate?
Explain why a closed diagonal implies that distinct points have disjoint open neighborhoods.
Where does the contradiction arise if a sequence is assumed to converge to two distinct points in a Hausdorff space?

What You'll Learn