When Does Deleting Several Points Disconnect a Space?
A cut point is a single point whose deletion disconnects a space. With several deleted points, the answer can depend on how they are placed and on the space itself. Every nonempty finite subset disconnects the real line, but deleting one point from a circle leaves a connected space. In a graph, deleting a central vertex can disconnect several branches, while deleting additional vertices can leave just one connected branch.
Throughout, connectedness refers to the stated topology on each space. A space is disconnected if it is the union of two disjoint, nonempty open sets; otherwise it is connected. The empty space is connected, since it has no decomposition into two nonempty sets.
A useful distinction is that being a separating set is not generally preserved when more points are deleted. Removing points from a disconnected space can remove all but one of its connected pieces. We will see this directly in a graph.
The Real Line and a Closed Interval
We use the standard connectedness theorem for the real line: every interval in \(\mathbb{R}\), with its standard topology, is connected. In particular, \(\mathbb{R}\), open intervals, closed intervals, and half-open intervals are connected. This is the interval theorem from the earlier material on connectedness.
Proof. First consider \(\mathbb{R}\). The space \(\mathbb{R}\) is connected, so the empty set is not separating. Now let \(S\) be a nonempty finite subset of \(\mathbb{R}\), and write its elements in increasing order as \(s_1<s_2<\cdots<s_k\), where \(k\geq1\). The sets \(U=(-\infty,s_1)\) and \(V=(s_1,\infty)\setminus S\) are disjoint, nonempty, and open in \(\mathbb{R}\setminus S\). They are open there because they are intersections of \(\mathbb{R}\setminus S\) with open subsets of \(\mathbb{R}\); they are nonempty because \(s_1-1\in U\) and \(s_k+1\in V\). Their union is \(\mathbb{R}\setminus S\). Thus this deletion disconnects \(\mathbb{R}\).
Now let \(I=[a,b]\). If \(S\cap(a,b)=\varnothing\), the only possible deleted points are the endpoints \(a\) and \(b\). The remaining space is one of \([a,b]\), \((a,b]\), \([a,b)\), or \((a,b)\), according to which endpoints were removed. Each is an interval, hence connected. If \(S\cap(a,b)\ne\varnothing\), choose \(x\in S\cap(a,b)\). In \(I\setminus S\), the sets \(U=(I\setminus S)\cap(-\infty,x)\) and \(V=(I\setminus S)\cap(x,\infty)\) are disjoint and relatively open, and their union is \(I\setminus S\). They are nonempty: since \(S\) is finite, some point of \((a,x)\) and some point of \((x,b)\) is not in \(S\). Therefore they disconnect \(I\setminus S\). This proves both directions. \(\square\)
Worked Example: Finite Deletions from the Real Line
Let \(S=\{-2,1,4\}\subseteq\mathbb{R}\), with the standard topology. The points are already in increasing order: \(s_1=-2\), \(s_2=1\), and \(s_3=4\). The deletion leaves the four intervals \((-\infty,-2)\), \((-2,1)\), \((1,4)\), and \((4,\infty)\).
For a direct disconnection test, take \(U=(-\infty,-2)\) and \(V=(-2,\infty)\setminus\{1,4\}\). Both are open in \(\mathbb{R}\setminus S\), they are disjoint, and their union is \(\mathbb{R}\setminus S\). They are nonempty: \(-3\in U\) and \(0\in V\). Hence \(S\) separates \(\mathbb{R}\).
Check: the test uses the strict inequalities \(x<-2\) and \(x>-2\); the deleted point \(-2\) belongs to neither part. The point \(0\) lies in \(V\) because \(0>-2\), \(0\ne1\), and \(0\ne4\).
Worked Example: Endpoint and Interior Deletions
Work in \(I=[0,5]\) with the subspace topology inherited from \(\mathbb{R}\). First take \(S=\{0,5\}\). Then \(I\setminus S=(0,5)\), an interval, so it is connected and \(S\) is not separating. In particular, deleting both endpoints does not disconnect this interval.
Next take \(T=\{0,2,5\}\). The point \(2\) lies strictly between \(0\) and \(5\). The sets \(U=[0,5]\setminus T\;\cap\;(-\infty,2)\) and \(V=[0,5]\setminus T\;\cap\;(2,\infty)\) are disjoint, relatively open, and cover \([0,5]\setminus T\). They are nonempty since \(1\in U\) and \(3\in V\). Thus \(T\) separates \(I\).
Check: \(1\) satisfies \(0\leq1\leq5\), \(1<2\), and \(1\notin\{0,2,5\}\), while \(3\) satisfies \(0\leq3\leq5\), \(3>2\), and \(3\notin\{0,2,5\}\). The cut is at the interior point \(2\), not at either endpoint.
Deleting Points from a Circle
Take \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) with the subspace topology from the Euclidean plane. The circle is connected: any two points can be joined by an arc, so it is path-connected; a path-connected space is connected because a separation would pull back along a path joining points in its two parts to a separation of the interval.
Proof. If \(S=\varnothing\), the remaining circle is connected. If \(S=\{p\}\), a rotation of the circle can move \(p\) to \((0,1)\). On the circle with \((0,1)\) removed, the map \(f(x,y)=x/(1-y)\) is a homeomorphism to \(\mathbb{R}\): its inverse is \(t\mapsto(2t/(1+t^2),(t^2-1)/(1+t^2))\). Both maps are continuous on their stated domains, and substitution in either direction gives the identity. Therefore \(S^1\setminus\{p\}\) is connected because it is homeomorphic to \(\mathbb{R}\).
Suppose instead that \(S\) has at least two distinct points. List its points in their order around the circle. There are at least two open arcs between successive points of this finite list; each arc is nonempty and contains no point of \(S\). Choose one such arc \(A\). In \(S^1\setminus S\), both \(A\) and its complement are nonempty and relatively open: the complement is the union of the other open arcs. They are disjoint and cover \(S^1\setminus S\). Hence that space is disconnected. \(\square\)
Worked Example: One Point Versus Two Points on the Circle
Delete first \(p=(0,1)\) from \(S^1\). The formula in the proof gives \(f(x,y)=x/(1-y)\). For example, \((-1,0)\) maps to \(-1\), and \((1,0)\) maps to \(1\). The inverse formula sends \(t=0\) to \((0,-1)\). Since the punctured circle is homeomorphic to the connected space \(\mathbb{R}\), this single deletion does not disconnect it.
Now delete \(p=(1,0)\) and \(q=(-1,0)\). The remaining points lie in the two open semicircles \(A=\{(x,y)\in S^1:y>0\}\) and \(B=\{(x,y)\in S^1:y<0\}\). These sets are disjoint, nonempty, relatively open, and cover \(S^1\setminus\{p,q\}\). Thus the two-point deletion disconnects the circle.
Check: \((0,1)\in A\) because its second coordinate is \(1>0\), and \((0,-1)\in B\) because its second coordinate is \(-1<0\). The deleted points have second coordinate \(0\), so neither belongs to either part.
A Graph Example and a Common Pitfall
For a concrete graph, use the geometric realization of the star \(K_{1,3}\): three copies of \([0,1]\) are joined by identifying their \(0\)-endpoints as one central vertex \(c\). Give the realization the quotient topology from these intervals. Its three outer endpoints are vertices \(v_1,v_2,v_3\); each segment from \(c\) to an outer vertex is an edge. A vertex deletion means removing that point from the realization.
Worked Example: Vertex Deletions in a Three-Arm Star
Deleting the central vertex \(c\) leaves three disjoint arms, each homeomorphic to \((0,1]\). Each arm is connected because it is an interval. Each is also open in the remaining space: its complement is the union of the other two arms, which is closed there. Consequently, one arm and the union of the other two form a separation, so \(\{c\}\) is a cut point.
Now delete \(c,v_1,v_2\). Only the third arm remains, with its outer endpoint \(v_3\) and without \(c\); it is homeomorphic to \((0,1]\), hence connected. Thus \(\{c,v_1,v_2\}\) is not a separating set, even though it contains the cut point \(c\).
Finally, delete \(v_1,v_2\) but keep \(c\). The remaining space consists of the two arms ending at \(v_2\) and \(v_3\), joined at \(c\). It is connected: each arm is connected, and the two arms have the common point \(c\). The union of two connected sets with a common point is connected; a separation of their union would have to put each connected arm entirely in one side, and their shared point forces them into the same side.
Check: after deleting \(c,v_1,v_2\), exactly one arm remains, so there are not two nonempty parts to separate. After deleting \(v_1,v_2\), the central point remains and joins the two surviving arms.
Common pitfall. A tempting but false rule is: “If \(S\) is separating, every finite set containing \(S\) is separating.” In the star, \(S=\{c\}\) separates, but \(T=\{c,v_1,v_2\}\) does not: its complement is the connected single arm described above. Deletion can remove whole pieces of a disconnected space, leaving a connected remainder. Therefore, for a proposed multi-point deletion, test the space that remains; knowing that one of the deleted points is a cut point is not enough.
What the Examples Show
| Space and topology | Finite deletion separates exactly when |
|---|---|
| \(\mathbb{R}\), standard topology | The deleted set is nonempty. |
| \([a,b]\), subspace topology, \(a<b\) | The deleted set contains an interior point. |
| \(S^1\), subspace topology from \(\mathbb{R}^2\) | The deleted set contains at least two distinct points. |
| Realized star \(K_{1,3}\), vertex deletions | For the cases shown, deleting the center separates if at least two arms remain; deleting leaves alone does not. |
The criterion depends on the space and on the location of the deleted points, not just on their number. On the line, any one-point deletion separates; on the circle, one deletion does not. On an interval, endpoint deletions alone do not separate, but any interior deletion does. In a graph, the arrangement of branches supplies still more possibilities. These calculations are useful whenever connectedness is used to distinguish spaces or to analyze what remains after removing specified points.
Check Your Understanding
For each question, use the definition of separating set: the remaining subspace must be disconnected.
- In \(\mathbb{R}\) with its standard topology, does \(S=\{-1,0,3\}\) separate? Give a separation of the complement.
- In \([0,4]\), which of \(S=\{0,4\}\) and \(T=\{1,4\}\) separates?
- Does deleting exactly one point from \(S^1\) disconnect it? Does deleting two distinct points?
- In the star \(K_{1,3}\), what remains after deleting the center and exactly two outer vertices? Is the deletion set separating?
- Give an example from this tutorial showing that a finite set containing a cut point need not be separating.