Separating One Point from an Entire Closed Set
Hausdorff separation asks for disjoint open neighborhoods of two distinct points. A related question replaces one of those points by a whole closed set: can we find one open neighborhood of the point and another open set containing every point of the closed set, with the two open sets disjoint? The requirement that a single open set contain the whole closed set is important. It is stronger than merely finding a neighborhood of the point that avoids the closed set.
Throughout, \((X,\mathcal{T})\) is a topological space. The closure \(\overline{A}\) of \(A\subseteq X\) is the intersection of all closed subsets of \(X\) containing \(A\). In particular, \(\overline{A}\) is closed and contains \(A\). A set \(U\) is an open neighborhood of \(x\) if \(U\in\mathcal{T}\) and \(x\in U\).
If \(F=\varnothing\), this condition holds: take \(U=X\) and \(V=\varnothing\). The interesting case is when \(F\) is nonempty. We will use the following equivalent neighborhood condition: there is an open neighborhood \(U\) of \(x\) whose closure misses \(F\). The closure matters because an open neighborhood can avoid \(F\) while its closure still meets \(F\).
The Closure-Neighborhood Test
- There are disjoint open sets \(U,V\) with \(x\in U\) and \(F\subseteq V\).
- There is an open neighborhood \(U\) of \(x\) such that \(\overline{U}\cap F=\varnothing\).
Proof. Suppose first that \(U,V\) are disjoint open sets, with \(x\in U\) and \(F\subseteq V\). Since \(U\cap V=\varnothing\), we have \(U\subseteq X\setminus V\). The set \(X\setminus V\) is closed because \(V\) is open. It contains \(U\), so the definition of closure gives \(\overline{U}\subseteq X\setminus V\). Since \(F\subseteq V\), it follows that \(\overline{U}\cap F=\varnothing\).
Conversely, suppose \(U\) is open, \(x\in U\), and \(\overline{U}\cap F=\varnothing\). The closure \(\overline{U}\) is closed, so \(V=X\setminus\overline{U}\) is open. The disjointness condition implies \(F\subseteq V\). Also \(U\subseteq\overline{U}\), so \(U\cap V=\varnothing\). Thus \(U,V\) are the required disjoint open sets. Both implications apply when \(F=\varnothing\) as well: in the second implication \(V=X\setminus\overline{U}\) contains the empty set. \(\square\)
This test recasts a condition involving two open sets as a condition involving one open neighborhood and its closure. It also explains why the direct fact \(X\setminus F\) is an open neighborhood of \(x\) is not enough: it ensures only that the neighborhood itself avoids \(F\), not that its closure does. We now check the condition in spaces where the relevant neighborhoods and closed sets can be described explicitly.
Worked Examples
Worked Example: A Point and an Interval in the Real Line
Use the standard topology on \(\mathbb{R}\). Let \(x=0\) and \(F=[2,3]\). The set \(F\) is closed, and \(0\notin F\). We will separate them by checking the closure-neighborhood condition.
Take \(U=(-1,1)\). This is open and contains \(0\). Its closure in the standard topology is \(\overline{U}=[-1,1]\): each point strictly between \(-1\) and \(1\) is already in \(U\); every open interval around either endpoint meets \(U\); and if \(y<-1\), then the open interval \((y, -1)\) around no point is not needed—more directly, \(B(y,(-1-y)/2)\) lies to the left of \(-1\) and misses \(U\). If \(y>1\), \(B(y,(y-1)/2)\) lies to the right of \(1\) and misses \(U\). Thus no point outside \([-1,1]\) is in the closure.
Now compare the intervals: every point of \([-1,1]\) is at most \(1\), and every point of \([2,3]\) is at least \(2\). Hence \([-1,1]\cap[2,3]=\varnothing\). The test gives separation. Indeed, the open sets \(U=(-1,1)\) and \(V=(1,4)\) are disjoint, \(0\in U\), and \([2,3]\subseteq V\).
Check: \(\overline{U}\cap F=[-1,1]\cap[2,3]=\varnothing\), as required. The explicit set \(V\) confirms that the entire closed interval, not merely one of its points, is covered.
Worked Example: A Finite Discrete Space
Let \(X=\{a,b,c,d\}\) with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), so every subset is open. Set \(x=a\) and \(F=\{c,d\}\). Since every set is also closed in this topology, \(F\) is closed.
Choose \(U=\{a\}\). It is open and contains \(a\). It is also closed, so its closure is itself: \(\overline{U}=\{a\}\). Since \(a\notin\{c,d\}\), \(\overline{U}\cap F=\{a\}\cap\{c,d\}=\varnothing\). Thus the closure-neighborhood condition holds.
The corresponding open set on the other side can be taken to be \(V=\{c,d\}\). Both sets are open, \(a\in U\), \(F\subseteq V\), and \(U\cap V=\varnothing\). Here the calculation is especially direct because every subset is open and closed.
Check: the closure test and the direct separator agree: \(\overline{\{a\}}\cap\{c,d\}=\varnothing\), and \(\{a\}\cap\{c,d\}=\varnothing\).
Worked Example: Failure in the Cofinite Topology
Let \(X=\mathbb{Z}\) with the cofinite topology: the open sets are \(\varnothing\) and all subsets whose complements in \(\mathbb{Z}\) are finite. The closed sets are \(\mathbb{Z}\) and all finite subsets. Take \(x=0\) and \(F=\{1\}\); the set \(F\) is closed and \(0\notin F\).
Suppose \(U\) is any open neighborhood of \(0\). It is nonempty, so its complement \(\mathbb{Z}\setminus U\) is finite. In particular \(U\) is infinite. Any closed set containing \(U\) must therefore be \(\mathbb{Z}\), since every proper closed set is finite. By the definition of closure, \(\overline{U}=\mathbb{Z}\). Consequently \(\overline{U}\cap F=\mathbb{Z}\cap\{1\}=\{1\}\ne\varnothing\), for every open neighborhood \(U\) of \(0\). The closure-neighborhood condition fails, so the point and the closed set cannot be separated by disjoint open sets.
We can also verify the failure directly. Every nonempty open subset of \(\mathbb{Z}\) has finite complement. If \(U,V\) are nonempty and open, then \(\mathbb{Z}\setminus(U\cap V)=(\mathbb{Z}\setminus U)\cup(\mathbb{Z}\setminus V)\), a finite union of finite sets. Thus \(U\cap V\) is nonempty. Any proposed separator must have \(0\in U\) and \(1\in V\), so both sets are nonempty and cannot be disjoint.
Check: \(X\setminus F=\mathbb{Z}\setminus\{1\}\) is open and contains \(0\), and it avoids \(F\). But that set alone does not separate \(0\) from \(F\): an open set containing \(F\) would also have to be nonempty, and any two nonempty open sets in this topology intersect.
Worked Example: Failure in the Sierpinski Space
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). The closed sets, obtained by taking complements of the open sets, are \(\varnothing,\{b\},X\). Thus \(F=\{b\}\) is closed. Take \(x=a\), which is not in \(F\).
The open neighborhoods of \(a\) are \(\{a\}\) and \(X\). The only closed set containing \(\{a\}\) is \(X\), so \(\overline{\{a\}}=X\). Also \(\overline{X}=X\). Therefore every open neighborhood \(U\) of \(a\) has \(\overline{U}=X\), and \(\overline{U}\cap F=X\cap\{b\}=\{b\}\ne\varnothing\). The closure-neighborhood condition fails.
For a direct check, the only open set containing \(b\) is \(X\). Any proposed open set \(V\) containing \(F\) must therefore equal \(X\), which intersects every open neighborhood of \(a\). There are no disjoint open sets placing \(a\) and \(F\) on opposite sides.
Check: although \(\{a\}\) is an open neighborhood of \(a\) and is disjoint from \(F\), its closure is \(X\), which meets \(F\). This is a concrete instance of why merely avoiding the closed set is weaker than the closure-neighborhood condition.
What the Condition Does—and Does Not—Say
The examples show that the condition depends on the topology, not just on the underlying set or on whether \(F\) is closed. In the discrete example, the point has a small open neighborhood whose closure remains small. In the Sierpinski example, an open neighborhood of the point may avoid \(F\), but its closure reaches the closed set. In the cofinite example, every open neighborhood of the point has the whole space as its closure.
A common mistake is to treat \(X\setminus F\) as a complete separator. Since \(F\) is closed and \(x\notin F\), \(X\setminus F\) is indeed open, contains \(x\), and misses \(F\). But the definition asks for a second open set containing every point of \(F\), disjoint from the first. The cofinite example demonstrates the gap: \(X\setminus\{1\}\) is an open neighborhood of \(0\), yet no disjoint open set can contain \(1\).
Another incorrect shortcut is to replace the closure condition by \(U\cap F=\varnothing\). In the Sierpinski example, \(U=\{a\}\) satisfies \(U\cap F=\varnothing\), but \(\overline{U}\cap F=\{b\}\). The correct test uses \(\overline{U}\), not just \(U\). The theorem proves exactly why: a disjoint open set containing \(F\) forces the closure of \(U\) to stay outside that open set, and conversely a closure that misses \(F\) provides the open set \(X\setminus\overline{U}\) containing \(F\).
Thus, for a specified point \(x\) and closed set \(F\), the useful question is whether \(x\) has an open neighborhood whose closure avoids all of \(F\). This local formulation is often easier to verify than searching independently for two open sets, and it records the full separation requirement without confusing it with the weaker condition of simply avoiding \(F\).
Check Your Understanding
For each question, keep track of whether the condition concerns \(U\) or its closure \(\overline{U}\).
- In the standard topology on \(\mathbb{R}\), separate \(x=0\) from \(F=[4,5]\) by giving an open neighborhood \(U\) and checking that \(\overline{U}\cap F=\varnothing\).
- In a discrete space \(X\), show directly that any point \(x\notin F\) can be separated from any subset \(F\) by disjoint open sets.
- In the cofinite topology on an infinite set \(X\), explain why every nonempty open set has closure \(X\).
- In the Sierpinski space above, list all open neighborhoods of \(a\) and compute the closure of each.
- Explain why \(U\cap F=\varnothing\) does not, by itself, imply \(\overline{U}\cap F=\varnothing\), using one of the examples in this tutorial.