From Separating Points to Separating a Compact Set
In a Hausdorff space, any two distinct points have disjoint open neighborhoods. But separating one point from every point of a set is not yet the same as separating it from the whole set: the neighborhoods chosen for different points may vary. Compactness supplies the key step. It lets us keep only finitely many of those neighborhoods, so that they can be combined into a single neighborhood of the point and a single open set containing the compact set.
Throughout, \((X,\mathcal{T})\) is a topological space. A space is Hausdorff if, for any distinct \(x,y\in X\), there are disjoint open sets containing \(x\) and \(y\), respectively. A subset \(K\subseteq X\) is compact if every cover of \(K\) by open sets of \(X\) has a finite subcover. We seek disjoint open sets containing a given point \(x\) and a compact set \(K\), under the assumption \(x\notin K\).
The theorem strengthens pointwise Hausdorff separation. For each \(y\in K\), Hausdorffness gives a pair of disjoint open neighborhoods, one of \(x\) and one of \(y\). The neighborhoods of the points \(y\) cover \(K\). Compactness says that finitely many suffice. We intersect the corresponding neighborhoods of \(x\), but take the union of the neighborhoods covering \(K\).
Proof. If \(K=\varnothing\), take \(U=X\) and \(V=\varnothing\); these sets are open, contain \(x\) and \(K\), respectively, and are disjoint. Now suppose \(K\ne\varnothing\). For each \(y\in K\), we have \(x\ne y\). By Hausdorffness, choose open sets \(U_y,V_y\) such that \(x\in U_y\), \(y\in V_y\), and \(U_y\cap V_y=\varnothing\). The family \(\{V_y:y\in K\}\) covers \(K\), since each \(y\) belongs to \(V_y\). By compactness, there are \(y_1,\dots,y_n\in K\), where \(n\ge 1\), such that \(K\subseteq\bigcup_{i=1}^{n}V_{y_i}\). Define \(U=\bigcap_{i=1}^{n}U_{y_i}\) and \(V=\bigcup_{i=1}^{n}V_{y_i}\). Finite intersections and finite unions of open sets are open, so \(U\) and \(V\) are open. Since every \(U_{y_i}\) contains \(x\), \(x\in U\); and the finite cover gives \(K\subseteq V\). If a point \(z\) belonged to \(U\cap V\), it would belong to \(V_{y_j}\) for some \(j\), and also to \(U\subseteq U_{y_j}\). This contradicts \(U_{y_j}\cap V_{y_j}=\varnothing\). Therefore \(U\cap V=\varnothing\). \(\square\)
Following the Finite-Subcover Construction
Worked Example: A Point and a Compact Interval
Take \(X=\mathbb{R}\) with its standard topology, \(K=[0,1]\), and \(x=2\). We construct finite families of point-separating neighborhoods and then combine them.
For \(y_0=0\), choose \(V_{y_0}=(-1,0.8)\) and \(U_{y_0}=(0.9,2.5)\). These are open, \(0\in V_{y_0}\), \(2\in U_{y_0}\), and they are disjoint because \(0.8<0.9\). For \(y_1=1\), choose \(V_{y_1}=(0.4,1.5)\) and \(U_{y_1}=(1.6,2.4)\). These are open, \(1\in V_{y_1}\), \(2\in U_{y_1}\), and they are disjoint because \(1.5<1.6\).
The two sets around the points of \(K\) cover it: if \(t\in[0,1]\) and \(t\le 0.4\), then \(t\in(-1,0.8)\); if \(0.4<t<0.8\), then \(t\in(-1,0.8)\) and \(t\in(0.4,1.5)\); if \(0.8\le t\le 1\), then \(t\in(0.4,1.5)\). Thus \(V_{y_0}\cup V_{y_1}=(-1,1.5)\) contains \([0,1]\).
The combined neighborhood of \(x\) is the intersection \(U_{y_0}\cap U_{y_1}=(0.9,2.5)\cap(1.6,2.4)=(1.6,2.4)\). It contains \(2\). The combined open set around \(K\) is \(V_{y_0}\cup V_{y_1}=(-1,1.5)\), and these combined sets are disjoint because every point of \((1.6,2.4)\) is greater than \(1.5\). This explicitly checks the theorem’s conclusion.
Check: the endpoints matter. The intervals \((-1,0.5)\) and \((0.5,1.5)\) would not cover \([0,1]\), because neither contains \(0.5\). In the construction above, the overlap between \((-1,0.8)\) and \((0.4,1.5)\) ensures that there is no such gap.
Worked Example: A Finite Discrete Space
Let \(X=\{a,b,c,d\}\) with the discrete topology, in which every subset is open. Put \(K=\{a,b\}\) and \(x=c\). The set \(K\) is compact: from any open cover of \(K\), select one member containing \(a\) and one member containing \(b\). Those at most two members cover \(K\).
For \(a\), take \(V_a=\{a\}\) and \(U_a=\{b,c,d\}\). For \(b\), take \(V_b=\{b\}\) and \(U_b=\{a,c,d\}\). All four sets are open. Each \(V\)-set contains its designated point, each \(U\)-set contains \(c\), and \(U_a\cap V_a=\varnothing\), \(U_b\cap V_b=\varnothing\). The two \(V\)-sets cover \(K\).
Combining them gives \(V=V_a\cup V_b=\{a,b\}\) and \(U=U_a\cap U_b=\{c,d\}\). Thus \(c\in U\), \(K\subseteq V\), and \(U\cap V=\varnothing\). This also checks the finite construction directly in a space where every set and every membership claim can be listed.
Why Both Hypotheses Matter
Hausdorffness provides the separate pairs of neighborhoods; compactness reduces the cover to finitely many pairs. Without either ingredient, the stated conclusion can fail. The next examples check each failure against the definitions.
Worked Example: Compactness Cannot Be Dropped
Use \(X=\mathbb{R}\) with its standard topology, \(K=(0,1)\), and \(x=0\). The set \(K\) is not compact. Indeed, the open sets \(V_n=(1/n,2)\), for \(n\in\mathbb{N}\), cover \((0,1)\): for any \(t\in(0,1)\), choose \(n\) large enough that \(1/n<t\), and then \(t\in V_n\). No finite subfamily covers \((0,1)\), since for a finite selection the largest index \(N\) gives the smallest left endpoint \(1/N>0\), leaving points of \((0,1/N]\) uncovered.
Suppose, contrary to the desired failure, that there were disjoint open sets \(U,V\subseteq\mathbb{R}\) with \(0\in U\) and \((0,1)\subseteq V\). Since \(U\) is open and contains \(0\), there is an \(\varepsilon>0\) such that \((-\varepsilon,\varepsilon)\subseteq U\). The number \(t=\min(\varepsilon,1)/2\) satisfies \(0<t<\varepsilon\) and \(t<1\), so \(t\in U\cap(0,1)\). Since \((0,1)\subseteq V\), this gives \(t\in U\cap V\), contradicting disjointness. Thus pointwise separation in this Hausdorff space does not give a neighborhood of \(0\) disjoint from all of this noncompact set.
Check: for each fixed \(y\in(0,1)\), the distinct points \(0\) and \(y\) do have disjoint open neighborhoods, for example \((-y/3,y/3)\) and \((2y/3,4y/3)\). The failure is in finding one neighborhood of \(0\) disjoint from a neighborhood containing the entire set.
Worked Example: Hausdorffness Cannot Be Dropped
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). This space is compact: given any open cover, at least one member contains \(a\); to cover \(b\), some member must contain \(b\), and the only such open set is \(X\). Therefore the cover has a one-set subcover. Set \(K=\{a\}\) and \(x=b\). The compact set \(K\) does not contain \(x\).
The open neighborhoods of \(b\) consist only of \(X\), because \(X\) is the only open set containing \(b\). The open neighborhoods of \(a\) are \(\{a\}\) and \(X\). Any open set containing \(b\) is therefore \(X\), which intersects every open set containing \(a\). Hence there cannot be disjoint open neighborhoods of \(b\) and \(K\). In particular, this space is not Hausdorff: the two distinct points \(a,b\) have no disjoint open neighborhoods.
A Consequence: Compact Subsets Are Closed
Proof. Let \(K\) be compact in the Hausdorff space \(X\). If \(K=\varnothing\), it is closed because its complement \(X\) is open. Otherwise, take any \(x\in X\setminus K\). By the finite-neighborhood separation theorem, there is an open neighborhood \(U_x\) of \(x\) disjoint from an open set containing \(K\). In particular, \(U_x\subseteq X\setminus K\). Thus every point of \(X\setminus K\) has an open neighborhood contained in \(X\setminus K\), so \(X\setminus K\) is open. By the definition of closedness, \(K\) is closed. \(\square\)
This consequence also clarifies the role of the theorem: it does more than show that individual points outside \(K\) have neighborhoods missing \(K\). It constructs a neighborhood of the point together with an open neighborhood of the whole compact set, with the two neighborhoods disjoint.
A Common Construction Pitfall
Common mistake: combine the neighborhoods of \(x\) by taking their union. The proof requires their intersection. In the interval example, \(U_{y_0}=(0.9,2.5)\) and \(U_{y_1}=(1.6,2.4)\). Their union is \((0.9,2.5)\), which intersects \(V=(-1,1.5)\), for instance at \(1\). So that union does not separate \(x\) from \(K\). Their intersection is \((1.6,2.4)\), which is disjoint from \(V=(-1,1.5)\). Conversely, for the neighborhoods covering \(K\), we take the union: an intersection of the \(V\)-sets need not contain all of \(K\). The direction of each operation follows its purpose: retain points common to all neighborhoods of \(x\), and include points from every selected piece of the cover of \(K\).
The compactness step is specifically what makes these operations finite. Arbitrary intersections of open sets need not be open, so one cannot replace the finite subcover by an unrestricted family and assume its intersection is an open neighborhood. The theorem uses Hausdorffness to start with pointwise separations and compactness to ensure that only finitely many are needed.
Check Your Understanding
For each question, identify which hypothesis or set operation is being used.
- In the finite-neighborhood separation theorem, why is the family \(\{V_y:y\in K\}\) a cover of \(K\)?
- Why do we intersect the finitely many neighborhoods of \(x\), rather than take their union?
- In the interval example, verify that \(0.8\in V_{y_1}\) and that \(1.5\notin V_{y_1}\). Does the union of the two \(V\)-sets still contain all of \([0,1]\)?
- Which part of the proof fails if \(K\) is not compact: choosing disjoint neighborhoods for each pair of points, or obtaining a finite subcover?
- In the two-point example, list the open neighborhoods of \(b\). Why does this rule out disjoint neighborhoods of \(b\) and \(\{a\}\)?