Why Separate Points and Closed Sets?
Separation axioms describe how effectively the open sets of a topological space distinguish points and closed subsets. At one end, an axiom may require only that two different points be distinguishable by some open set. Stronger axioms require disjoint neighborhoods, or even disjoint neighborhoods of entire closed sets. These conditions matter because they determine which familiar arguments from metric spaces remain valid in more general topological spaces.
In the previous tutorial, paths and path connectedness were considered in arbitrary topological spaces. Separation axioms address a different structural question: how much can the topology distinguish? We will use the standard notation \(T_0,T_1,T_2,T_3,T_4\). There is a convention difference in the use of “regular” and “normal,” so the definitions below specify exactly what is meant here.
The Point-Separation Axioms
The \(T_0\) condition asks only for one-sided distinction: an open set must tell the points apart in at least one direction. The \(T_1\) condition requires both directions, though the two open sets need not be disjoint. Hausdorffness requires a stronger pair of neighborhoods: they must be disjoint.
Proof. Suppose \(X\) is \(T_1\), and fix \(x\in X\). For every \(y\ne x\), the \(T_1\) condition gives an open set \(U_y\) containing \(y\) but not \(x\). Thus every \(y\ne x\) belongs to \(X\setminus\{x\}\) and has an open neighborhood contained in \(X\setminus\{x\}\). Therefore \(X\setminus\{x\}\) is open, so \(\{x\}\) is closed.
Conversely, suppose every singleton is closed. If \(x\ne y\), then \(X\setminus\{y\}\) is an open set containing \(x\) but not \(y\), and \(X\setminus\{x\}\) is an open set containing \(y\) but not \(x\). Hence \(X\) is \(T_1\). \(\square\)
This characterization is often the quickest way to check \(T_1\). In contrast, \(T_0\) alone does not require singletons to be closed, and it does not require that each point be distinguishable from the other in both directions.
Worked Example: A \(T_0\) Space That Is Not \(T_1\)
Let \(X=\{a,b\}\) have the topology \(\{\varnothing,\{a\},X\}\). The open set \(\{a\}\) contains \(a\) but not \(b\), so the two points are distinguishable and \(X\) is \(T_0\).
However, \(\{a\}\) is not closed, since its complement \(\{b\}\) is not open. By the characterization of \(T_1\) spaces, \(X\) is not \(T_1\). Notice that \(\{b\}\) is closed: its complement \(\{a\}\) is open. The failure is specifically that \(\{a\}\) is not closed.
Separating Points from Closed Sets
The regularity condition separates one point from a closed set not containing it. A singleton is itself closed in a \(T_1\) space, so the definitions already suggest that \(T_3\) spaces should be Hausdorff. Normality strengthens the separation requirement from one point and a closed set to two closed sets.
The \(T_1\) clause in \(T_3\) and \(T_4\) is included explicitly. Some authors build it into the words “regular” and “normal,” while others do not. Keeping the convention visible prevents an implication from depending on an unstated assumption.
Proof. A \(T_4\) space is \(T_1\) and normal by definition. To see that it is regular, take a closed set \(F\) and \(x\notin F\). The singleton \(\{x\}\) is closed because the space is \(T_1\), and \(\{x\}\cap F=\varnothing\). Normality gives disjoint open neighborhoods of \(\{x\}\) and \(F\). Thus the space is regular, hence \(T_3\).
Now suppose \(X\) is \(T_3\), and take distinct \(x,y\in X\). Since \(X\) is \(T_1\), the set \(\{y\}\) is closed. Regularity separates \(x\) from \(\{y\}\) by disjoint open sets, so \(X\) is Hausdorff.
If \(X\) is Hausdorff and \(x\ne y\), disjoint open neighborhoods \(U\) of \(x\) and \(V\) of \(y\) in particular give an open set containing \(x\) but not \(y\), and one containing \(y\) but not \(x\). Thus \(X\) is \(T_1\). Finally, the two open sets supplied by \(T_1\) include at least one open set distinguishing each pair, so every \(T_1\) space is \(T_0\). This proves the chain. \(\square\)
Worked Example: A \(T_1\) Space That Is Not Hausdorff
Let \(X\) be an uncountable set with the cocountable topology: a set is open if it is empty or its complement in \(X\) is countable. For each \(x\in X\), the complement \(X\setminus\{x\}\) is open because its complement \(\{x\}\) is countable. Therefore every singleton is closed, and \(X\) is \(T_1\).
Any two nonempty open sets \(U,V\) have countable complements. Their intersection has complement \((X\setminus U)\cup(X\setminus V)\), which is countable because a union of two countable sets is countable. Since \(X\) is uncountable, this complement cannot be all of \(X\). Hence \(U\cap V\ne\varnothing\). There are no disjoint nonempty open sets, so distinct points cannot have disjoint open neighborhoods. The space is not Hausdorff.
This example shows why the \(T_1\) condition is weaker than \(T_2\): closed singletons do not by themselves guarantee disjoint neighborhoods.
Regularity and Closures of Neighborhoods
Regularity can be tested by asking whether an open neighborhood of a point can be replaced by a smaller open neighborhood whose closure still stays inside the original one. This formulation is useful in proofs because it turns a separation condition into a controlled choice of neighborhood.
Proof. Suppose \(X\) is regular, and let \(x\in G\) with \(G\) open. The set \(F=X\setminus G\) is closed and does not contain \(x\). By regularity, there are disjoint open sets \(V,W\) with \(x\in V\) and \(F\subseteq W\). Since \(W\) is open and disjoint from \(V\), no point of \(W\) lies in \(\overline V\): each point of \(W\) has the neighborhood \(W\), which misses \(V\). Thus \(\overline V\subseteq X\setminus W\). Since \(F\subseteq W\), we have \(X\setminus W\subseteq X\setminus F=G\). Therefore \(x\in V\subseteq\overline V\subseteq G\).
Conversely, suppose the stated neighborhood property holds. Let \(F\) be closed and \(x\notin F\). Apply the property to the open set \(G=X\setminus F\) to obtain an open \(V\) with \(x\in V\) and \(\overline V\subseteq G\). The set \(W=X\setminus\overline V\) is open because \(\overline V\) is closed. It contains \(F\), since \(\overline V\subseteq X\setminus F\), and it is disjoint from \(V\), since \(V\subseteq\overline V\). Hence \(X\) is regular. \(\square\)
The closure inclusion is essential. Merely finding an open \(V\) with \(x\in V\subseteq G\) says nothing beyond the original fact that \(G\) is a neighborhood of \(x\). Regularity gives room to choose \(V\) so that even its closure avoids the closed set outside \(G\).
Normality and Shrinking Open Sets
There is a parallel closure criterion for normality. Instead of starting with one point inside an open set, we start with a closed set \(A\) inside an open set \(G\). Normality lets us find an open neighborhood of all of \(A\) whose closure remains in \(G\).
Proof. Suppose \(X\) is normal, with \(A\subseteq G\), where \(A\) is closed and \(G\) is open. If \(A=\varnothing\), take \(V=\varnothing\). Otherwise, \(B=X\setminus G\) is closed and disjoint from \(A\). By normality, there are disjoint open sets \(V,W\) with \(A\subseteq V\) and \(B\subseteq W\). Since \(W\) is open and disjoint from \(V\), it is also disjoint from \(\overline V\). Hence \(\overline V\subseteq X\setminus W\subseteq X\setminus B=G\). This gives the required inclusions.
Conversely, suppose the shrinking property holds. Take disjoint closed sets \(A,B\), and apply the property with \(G=X\setminus B\). Then there is an open \(V\) with \(A\subseteq V\subseteq\overline V\subseteq X\setminus B\). Set \(W=X\setminus\overline V\). This set is open and contains \(B\), while \(V\cap W=\varnothing\). Thus \(A\) and \(B\) have disjoint open neighborhoods, so \(X\) is normal. \(\square\)
Worked Example: Discrete Spaces Are \(T_4\)
Let \(X\) have the discrete topology, so every subset is open. Every subset is also closed, since its complement is open. If \(A\) and \(B\) are disjoint closed subsets, then \(A\) and \(B\) themselves are disjoint open sets containing them. Thus \(X\) is normal. Its singletons are closed, so it is \(T_1\), and consequently \(X\) is \(T_4\).
This includes the cases where one of the closed sets is empty: the empty set is open and is a neighborhood of the empty set in the containment sense, while the other set can be used as its own open neighborhood. The example is a direct check of the definition and does not depend on \(X\) being finite.
Metric Spaces Meet the Stronger Axioms
Metric spaces provide an important source of normal spaces. The proof uses distances from points to closed sets. For a nonempty set \(A\) in a metric space \((X,d)\), define \(d(x,A)=\inf\{d(x,a):a\in A\}\). The distance-to-a-set function is continuous: for any \(x,y\in X\), the triangle inequality gives \(d(x,A)\leq d(x,y)+d(y,A)\), and interchanging \(x,y\) yields \(|d(x,A)-d(y,A)|\leq d(x,y)\).
Proof. Let \(A,B\) be disjoint closed subsets of a metric space \(X\). If either set is empty, the definition of normality is satisfied by taking the open neighborhood of the empty set to be empty and the other neighborhood to be \(X\). Now suppose both sets are nonempty. Define $$ U=\{x\in X:d(x,A)<d(x,B)\},\qquad V=\{x\in X:d(x,B)<d(x,A)\}. $$ The two distance functions are continuous by the estimate above. Their difference is continuous, so the strict-inequality sets \(U\) and \(V\) are open. They are disjoint, because no pair of real numbers can satisfy both \(r<s\) and \(s<r\).
Take \(a\in A\). Then \(d(a,A)=0\). Since \(B\) is closed and \(a\notin B\), some open ball about \(a\) misses \(B\). Thus there is an \(\varepsilon>0\) such that \(d(a,b)\geq\varepsilon\) for every \(b\in B\), so \(d(a,B)\geq\varepsilon>0\). Therefore \(a\in U\). The same reasoning, with the roles of \(A\) and \(B\) reversed, shows that every \(b\in B\) belongs to \(V\). Thus \(A\subseteq U\), \(B\subseteq V\), and \(U,V\) are disjoint open neighborhoods. This proves normality. The earlier theorem that every metric space is Hausdorff gives \(T_1\), so \(X\) is \(T_4\). \(\square\)
The closedness hypothesis is used when showing that the distance from a point outside a set to that set is positive. For an arbitrary nonclosed set, a point outside it can still have distance zero. The proof therefore depends on closedness, not merely on disjointness.
How to Use the Separation Hierarchy
The implication chain gives a dependable first test: a \(T_4\) space is \(T_3\), Hausdorff, \(T_1\), and \(T_0\), but the reverse implications do not follow just from the definitions. The cocountable example shows that \(T_1\) need not imply Hausdorffness, and the two-point example shows that \(T_0\) need not imply \(T_1\). The stronger axioms are not just labels; they provide particular tools, such as separating a point from a closed set or shrinking an open neighborhood around a closed subset.
A common pitfall is to confuse “closed singleton” with “disjoint neighborhoods.” The \(T_1\) characterization concerns whether complements of singletons are open; it does not produce disjoint neighborhoods of two points. Another is to use “regular” or “normal” without checking whether the author includes \(T_1\) in those terms. Stating the convention, as we have done, makes each implication precise.
Check Your Understanding
Use the definitions and proved characterizations to answer the following questions.
- What additional requirement does \(T_1\) impose beyond \(T_0\)?
- Why does the closed-singleton characterization not imply that every \(T_1\) space is Hausdorff?
- In the neighborhood characterization of regularity, where is the closedness of \(X\setminus G\) used?
- How does normality produce an open set \(V\) with \(A\subseteq V\subseteq\overline V\subseteq G\)?
- Why must the closedness of \(B\) be used to show \(d(a,B)>0\) when \(a\notin B\)?