4.2Separation Axioms

Updated 20 Aug 2026Chapter (PDF)

[4.2.1]Definition(Topologically Distinguishability, Separation)#

Let $x, y$ be two points in a topological space $X$ . Then $x, y$ are topologically indistinguishable if the set of all open neighborhoods about $x$ coincides with the set of all open neighborhoods about $y$ . Thus, $x, y$ are topologically distinguishable if one point has an open neighborhood that excludes the other. Moreover, $x, y$ are separated if there exist open neighborhoods about $x$ and $y$ that are disjoint. Separation generalizes easily to a point and a set or two sets.

Note that when dealing with connected spaces, we also use separations, but only separations whose union is the entire space. In dealing with separation axioms, we still deal with separations (i.e., disjoint, open sets), but we do not care if their union is the entire space.

[4.2.2]Definition(Regular, Normal Spaces)#

Suppose a space $X$ is $T_1$ (i.e., all singletons are closed). We say that $X$ is regular if, for any point $x$ and a closed set $B \subseteq X$ disjoint from $x$ , we may separate $x$ and $B$ . We say that $X$ is normal if any pair of disjoint, closed sets can be separated.

Note that we specify $B$ to be closed in the previous definition since, if $B \subseteq X$ is not closed, then we would be unable to separate a point $x \in \bar{B} \setminus B$ from $B$ , making most spaces $X$ not regular. The same is true for the definition of normality (consider a set $A$ and a singleton set $B$ that contains a point from $\bar{A} \setminus A$ ).

Obviously, a regular space is Hausdorff (consider a point $x$ and the closed space $\{y\}$ ), and analogously, normal spaces are regular.

[4.2.3]Theorem(Characterizations of Regularity and Normality)#

Suppose $X$ is $T_1$ .

$X$ is regular if and only if, for any $x \in X$ with any open neighborhood $U$ of $x$ , there is some open neighborhood $V$ of $x$ such that $\bar{V} \subseteq U$ .
Shrinking Lemma. $X$ is normal if and only if, for any closed set $A \subseteq X$ contained in any open neighborhood $U$ , there is some open set $V$ that contains $A$ such that $\bar{V} \subseteq U$ .

Proof.

$(\Longrightarrow)$ : Suppose $X$ is regular. Consider a point $x \in X$ and some open neighborhood $U$ of $x$ . Set $B = X \setminus U$ , a closed set; by regularity of $X$ , there exist disjoint, open neighborhoods $V, W$ of $x$ and $B$ . If we show that $\bar{V} \cap B = \emptyset$ , then we have shown that $\bar{V} \subseteq U$ . Indeed, if $y \in B$ , then there is an open neighborhood $W$ of $y$ that does not intersect $V$ , and so $y \notin \bar{V}$ . Thus $\bar{V} \subseteq U$ .
$(\Longleftarrow)$ : Let $x \in X$ be a point disjoint from some closed set $B \subseteq X$ . Then note $U = X \setminus B$ is an open neighborhood of $x$ . By hypothesis, there is an open neighborhood $V$ of $x$ such that $\bar{V} \subseteq U$ . Then $U, X \setminus \bar{V}$ are disjoint open neighborhoods separating $x$ and $B$ respectively.
The proof is analogous to $(1)$ .

It is quick to see that the subspace $V$ of a Hausdorff space $X$ is Hausdorff. Indeed, if $x, y \in V$ , then there exist disjoint open neighborhoods $X \cap V, Y \cap V$ of $x, y$ respectively, where $X, Y$ are disjoint open neighborhoods of $x, y$ in the space $X$ .

It is also quick to see that the product $X$ of Hausdorff spaces $\{X_\lambda\}_{\lambda \in \Lambda}$ is Hausdorff. Indeed, for distinct $\vb{x} = (x_\lambda)_{\lambda \in \Lambda}$ and $\vb{y} = (y_\lambda)_{\lambda \in \Lambda}$ , there is some $j \in \Lambda$ for which $x_j \ne y_j$ . Since $X_j$ is Hausdorff, there exist disjoint, open neighborhoods $O_x, O_y \subseteq X_j$ of $x_j, y_j$ . Then the sets

O_X = O_x \times \left( \prod_{\lambda \in \Lambda \setminus \{j\}} X_\lambda \right) \quad O_Y = O_y \times \left( \prod_{\lambda \in \Lambda \setminus \{j\}} X_\lambda \right),

are disjoint, open neighborhoods of $\vb{x}$ , $\vb{y}$ in the product topology.

[4.2.4]Theorem(Subspace and Product of Regular Spaces is Regular)#

A subspace of regular spaces is regular. The product of a collection of regular spaces is regular.

Proof.

Suppose $X$ is a regular space and $Y \subseteq X$ is a subspace. Note $Y$ is Hausdorff and thus $T_1$ . Let $x$ be a point in $Y$ and $B$ a closed subset of $Y$ . Note that $\Cl_X(B) \cap Y = \Cl_Y(B) = B$ , with the last equality by $B$ closed. Considering $x \in X$ and $\bar{B} \subseteq X$ a closed set, by regularity of $X$ we have that there exist disjoint, open neighborhoods $U, V$ of $x$ and $\bar{B}$ . Then $U \cap Y, V \cap Y$ are disjoint, open neighborhoods of $x$ and $B$ , proving $Y$ is regular.

Now suppose $\{X_\lambda\}_{\lambda \in \Lambda}$ is a collection of regular spaces. We show $X = \prod_{\lambda \in \Lambda} X_\lambda$ is regular by the alternative characterization. Note that $X$ is Hausdorff and thus $T_1$ , so it satisfies that part. Let $\vb{x} = (x_\lambda)_{\lambda \in \Lambda} \in X$ , and let $U$ be any open neighborhood of $\vb{x}$ . Since $U$ is open, there is a basic, open set $\prod_{\lambda} U_\lambda$ such that $x \in \prod_{\lambda} U_\lambda \subseteq U$ . From this basis, we can easily construct an open neighborhood $V$ of $x$ such that $\bar{V} \subseteq U$ . We will define $V = \prod_{\lambda \in \Lambda} V_\lambda$ . Indeed, for each $\lambda \in \Lambda$ , consider $U_\lambda$ . If $U_\lambda = X_\lambda$ , then let $V_\lambda = X_\lambda$ . Otherwise, let $V_\lambda$ be some open set such that $\bar{V_\lambda} \subseteq U_\lambda$ . Then $V$ is an open neighborhood for $x$ , and since $\bar{V} = \prod_{\lambda \in \Lambda} \bar{V_\lambda}$ , we have that $\bar{V} \subseteq \prod_{\lambda \in \Lambda} U_\lambda \subseteq U$ , such that $X$ is regular.

[4.2.5]Example(Hausdorff But Not Regular)#

Consider $\bR_K$ , the reals endowed with the topology given by basis elements of the form $(a, b) \setminus K$ , where $K = \{1/n\}_{n \in \bZ_+}$ . It is quick to see $\bR_K$ is Hausdorff: for any $x \ne y \in \bR_K$ , let $O_x, O_y$ be disjoint, open neighborhoods of $x, y$ in the standard topology for $\bR$ , then note $O_x \setminus K, O_y \setminus K$ are disjoint, open neighborhoods of $x, y$ in the $K$ -topology. However, it is not regular by a straightforward proof.

Note that normality of a space is not preserved under subspace nor product. For products, consider the Sorgenfrey line $\bR_\ell$ , which is normal, and the Sorgenfrey plane $\bR_\ell^2$ , which is not normal.

[4.2.6]Recap#

Two points $x, y$ are topologically indistinguishable if the collection of open neighborhoods about $x$ is the exact same as the collection of open neighborhoods about $y$ (otherwise, the points are topologically distinguishable). We say two points $x, y$ are separated if there exist disjoint, open neighborhoods of $x$ and $y$ . The same idea applies to separating a point from a subset or a subset from a subset.
Suppose $X$ is a $T_1$ space. Then $X$ is regular if any point $x$ disjoint from a closed set $B$ can be separated. $X$ is normal if any pair of disjoint, closed sets $A, B$ can be separated.
The requirement that the sets are closed are important. If we drop that restriction, then there is limit point disjoint from a non-closed set, but separating them is impossible, so regularity would become far too restrictive. The same idea applies for normal spaces.
We may also characterize regularity and normality in a different way. A space $X$ is regular if and only if, for any open neighborhood $U$ of $x$ , there is some open neighborhood $V$ of $x$ such that $\bar{V} \subseteq X$ . Analogously, a space $X$ is normal if and only if, for any open neighborhood $U$ of a closed set $A$ , there is an open neighborhood $V$ of $A$ such that $\bar{V} \subseteq A$ .
Note that the equivalent definition of regularity closely resembles local compactness. Recall that a space $X$ is locally compact Hausdorff if, for any open neighborhood $U$ of $x$ , there is some open neighborhood $V$ of $x$ such that $\bar{V}$ is compact and $\bar{V} \subseteq X$ . At once, we have that locally compact Hausdorff spaces are regular.
Subspace and product operations on regular spaces produce regular spaces. This is not true for normal spaces.

4.2.1Exercises#

[4.2.7]Problem#

Show that if $X$ is regular, then prove that any two distinct points in $X$ have open neighborhoods whose closures are disjoint.

Proof.

Let $x \ne y \in X$ . By Hausdorffness of $X$ , we have disjoint, open neighborhoods $U_x, U_y$ about $x, y$ respectively. By regularity of $X$ , there are neighborhoods $V_x, V_y$ of $x, y$ such that $\bar{V_x} \subseteq U_x$ and $\bar{V_y} \subseteq U_y$ , and so $\bar{V_x} \cap \bar{V_y} = \emptyset$ , completing the proof.

[4.2.8]Problem#

Let $f, g: X \to Y$ be continuous maps into the Hausdorff space $Y$ . Show the equalizer $\Eq(f, g) = \{x: f(x) = g(x)\}$ is closed in $X$ .

Proof.

We show $C = \Eq(f, g)$ contains all its limit points. For the sake of contradiction, suppose $x \in \bar{C} \setminus C$ . By Hausdorffness of $Y$ , there exist disjoint, open neighborhoods $Y_f, Y_g$ of $f(x), g(x)$ respectively. Again by continuity of $f, g$ , there exist open neighborhoods $X_f, X_g$ of $x$ for which $f(X_f) \subseteq Y_f$ and $g(X_g) \subseteq Y_g$ . Then consider the single open neighborhood $U = X_f \cap X_g$ of $x$ , which also satisfies $f(U) \subseteq Y_f$ and $g(U) \subseteq Y_g$ . Since $x \in \bar{C}$ , $U$ intersects $C$ at a point whose image under both $f$ and g is $y$ (since the point lies in the equalizer). Thus $y \in Y_f \cap Y_g$ , a contradiction, and so the construction of $x$ is invalid. Thus $C$ contains all its limit points, completing the proof.