Why Closed Discrete Sets Are Useful
Normality asks whether every pair of disjoint closed sets can be placed in disjoint open neighborhoods. A closed discrete set supplies many closed sets at once: every way of dividing it into two parts gives a pair of disjoint closed sets in the ambient space. This makes such sets useful for testing proposed neighborhood separations. If one partition cannot be separated, the space is not normal.
Throughout, an open neighborhood of a point is an open set containing it. A space is normal if every two disjoint closed subsets have disjoint open neighborhoods containing them, respectively. In the convention that \(T_4\) includes \(T_1\), a \(T_4\) space must also be \(T_1\). The separation test below concerns the normality condition itself; a failure of that condition also rules out \(T_4\).
Discreteness concerns neighborhoods relative to the points of \(D\). It does not say that \(D\) is open in \(X\), nor does it say that \(D\) is closed. The closedness assumption is essential for the result that follows.
Every Subset Is Closed
Proof. Let \(A\subseteq D\). For each \(d\in D\), discreteness gives an open set \(U_d\) with \(U_d\cap D=\{d\}\). We show that \(X\setminus A=(X\setminus D)\cup\bigcup_{d\in D\setminus A}U_d\). A point outside \(D\) belongs to \(X\setminus D\). A point of \(X\setminus A\) that lies in \(D\) is some \(d\in D\setminus A\), so it belongs to \(U_d\). This proves that the left side is contained in the right side. Conversely, \(X\setminus D\) misses \(A\), and each \(U_d\) in the union meets \(D\) only at \(d\notin A\), so the right side misses \(A\). The equality follows. Since \(D\) is closed, \(X\setminus D\) is open; every \(U_d\) is open; and an arbitrary union of open sets is open. Thus \(X\setminus A\) is open, so \(A\) is closed. \(\square\)
The proof identifies the two hypotheses separately. The isolating neighborhoods \(U_d\) let us cover points of \(D\setminus A\) without picking up points of \(A\). Closedness of \(D\) ensures that points outside \(D\) also lie in an open part of \(X\setminus A\).
Worked Example: A Subset of the Integers
Take \(X=\mathbb{R}\) with its standard topology and \(D=\mathbb{Z}\). First verify that \(D\) is closed discrete. For each \(n\in\mathbb{Z}\), the interval \((n-\tfrac12,n+\tfrac12)\) meets \(\mathbb{Z}\) only in \(\{n\}\), so \(D\) is discrete. If \(x\notin\mathbb{Z}\), write \(x=n+r\), where \(n\in\mathbb{Z}\) and \(0<r<1\). Set \(\delta=\min(r,1-r)/2\). Then \(\delta>0\), and \((x-\delta,x+\delta)\) contains no integer: its points lie strictly between \(n\) and \(n+1\). Thus \(\mathbb{R}\setminus\mathbb{Z}\) is open and \(D\) is closed.
Now let \(A=\{2k:k\in\mathbb{Z}\}\), the even integers. Since \(A\subseteq D\), the theorem gives that \(A\) is closed in \(\mathbb{R}\). This also checks directly through the theorem’s complement description: every odd integer has an isolating interval \((n-\tfrac12,n+\tfrac12)\) disjoint from \(A\), and every point outside \(\mathbb{Z}\) has an open interval disjoint from \(A\). Hence every point of \(\mathbb{R}\setminus A\) has an open neighborhood contained in \(\mathbb{R}\setminus A\), as required.
Partitions as Separation Tests
Proof. The theorem shows that \(A\) and \(B\) are closed because both are subsets of \(D\). They are disjoint by the assumption \(A\cap B=\varnothing\). If \(X\) is normal, its defining separation condition applied to these two closed sets gives disjoint open neighborhoods \(U,V\) containing \(A,B\), respectively. \(\square\)
This proposition gives a practical direction of implication: a normal space must pass every such partition test. To show a space is not normal, it is enough to find one closed discrete \(D\), a partition \(D=A\cup B\) into disjoint nonempty parts, and a proof that every open neighborhood of \(A\) intersects every open neighborhood of \(B\). The proposition does not say that passing the tests for one particular \(D\) proves normality. Normality concerns all pairs of disjoint closed sets, not just subsets of that \(D\).
Worked Example: Separating the Even and Odd Integers
In \(\mathbb{R}\) with its standard topology, partition \(D=\mathbb{Z}\) into the even integers \(A\) and the odd integers \(B\). Define \(U=\bigcup_{a\in A}(a-\tfrac13,a+\tfrac13)\) and \(V=\bigcup_{b\in B}(b-\tfrac13,b+\tfrac13)\). Each is open because it is a union of open intervals, and \(A\subseteq U\), \(B\subseteq V\).
To check disjointness, suppose an interval centered at \(a\in A\) and one centered at \(b\in B\) intersect. A point in their intersection would imply, by the triangle inequality, \(|a-b|<\tfrac13+\tfrac13=\tfrac23\). But \(a\) and \(b\) are distinct integers, so \(|a-b|\geq 1\), contradicting \(|a-b|<\tfrac23\). Thus \(U\cap V=\varnothing\). The explicit neighborhoods verify the separation that normality requires for this partition.
Check: the sum of the two radii is \(1/3+1/3=2/3\), strictly less than the minimum distance \(1\) between distinct integers. The strict inequality is what rules out an intersection; merely knowing the radii are each less than \(1\) would not suffice.
A Partition That Cannot Be Separated
The next example shows how a closed discrete set can detect failure of normality. The key is not just that its two parts are closed: every open neighborhood of either part is forced to be large, so the two neighborhoods must overlap.
Worked Example: A Countable Discrete Set in the Cocountable Topology
Let \(X\) be an uncountable set with the cocountable topology: its open sets are \(\varnothing\) and all subsets whose complements in \(X\) are countable. This is a topology. An arbitrary union of nonempty open sets contains one such set, so its complement is a subset of a countable set and is countable. The intersection of two nonempty open sets also has countable complement, since that complement is the union of two countable sets. The empty set and \(X\) are open as well. Each singleton is closed because its complement is open, so \(X\) is \(T_1\).
Choose a countably infinite subset \(D=\{d_n:n\in\mathbb{N}\}\) of \(X\), with distinct \(d_n\). It is closed, since \(X\setminus D\) is open. It is discrete: for each \(d_n\), the set \(X\setminus(D\setminus\{d_n\})\) is open, because \(D\setminus\{d_n\}\) is countable, and its intersection with \(D\) is exactly \(\{d_n\}\). So \(D\) is closed discrete.
Partition \(D\) into \(A=\{d_{2k}:k\in\mathbb{N}\}\) and \(B=\{d_{2k-1}:k\in\mathbb{N}\}\). Both are nonempty, disjoint, and closed by the theorem. Suppose \(U,V\) are open sets with \(A\subseteq U\) and \(B\subseteq V\). Because \(A\) and \(B\) are nonempty, \(U\) and \(V\) are nonempty, hence each has countable complement in \(X\). Therefore \(X\setminus(U\cap V)=(X\setminus U)\cup(X\setminus V)\) is countable. Since \(X\) is uncountable, \(U\cap V\ne\varnothing\). No such neighborhoods can be disjoint, so \(X\) is not normal.
Check: this argument applies to every pair of nonempty open sets, not only to neighborhoods containing \(A\) and \(B\). It therefore rules out the required separation for this particular pair of disjoint closed sets. The space is also \(T_1\), so the failure is not explained by a failure of the \(T_1\) condition.
A Common Pitfall: Discrete Does Not Mean Closed
It is incorrect to drop “closed” from “closed discrete.” In \(\mathbb{R}\), let \(D=\{1/n:n\in\mathbb{N}\}\). This set is discrete as a subspace: each \(1/n\) has an interval around it that contains no other point of \(D\). For instance, for \(n\geq 2\), the positive gaps to the adjacent terms are \(1/(n-1)-1/n=1/(n(n-1))\) and \(1/n-1/(n+1)=1/(n(n+1))\); choosing a radius smaller than half the smaller gap isolates \(1/n\). For \(n=1\), the interval \((3/4,5/4)\) isolates \(1\).
Nevertheless, \(D\) is not closed: every open interval around \(0\) contains some \(1/n\), since for any \(\varepsilon>0\) one can choose \(n>1/\varepsilon\), giving \(0<1/n<\varepsilon\). Thus \(0\in\overline{D}\) but \(0\notin D\). The false calculation would be “\(D\) is discrete, so \(D\) is closed.” The correction is that discreteness only supplies isolating neighborhoods at points of \(D\); it does not control points outside \(D\). The theorem applies only after both discreteness and closedness have been established.
When using this diagnostic, keep the logical roles distinct: first verify that \(D\) is closed discrete in the ambient space; then use the theorem to show its parts are closed; finally examine whether disjoint open neighborhoods can exist. A failure at the last step proves nonnormality, while a successful separation of one partition proves only that this particular test has not found an obstruction.
Check Your Understanding
For each question, identify which hypothesis or separation condition is being used.
- Explain why every subset of a finite closed discrete subset \(D\subseteq X\) is closed in \(X\).
- In the standard topology on \(\mathbb{R}\), show that \(3\mathbb{Z}\) is closed discrete and conclude that \(\{6k:k\in\mathbb{Z}\}\) is closed.
- In the cocountable topology on an uncountable set, why must any two nonempty open sets intersect?
- Does a successful separation of one partition of a closed discrete set prove that \(X\) is normal? Explain.
- Give the point in \(\overline{\{1/n:n\in\mathbb{N}\}}\setminus\{1/n:n\in\mathbb{N}\}\) that shows this discrete subset of \(\mathbb{R}\) is not closed.