Cut-Point Profiles of Lines, Intervals, and Circles

What Does One Deleted Point Do?

A cut point is a point whose deletion disconnects a connected space. The effect of deleting a point depends on the structure of the space: removing an interior point from an interval separates its left and right portions, while removing a point from a circle leaves a connected arc. Endpoints require separate attention, because deleting an endpoint of a closed interval leaves one interval rather than two.

Unless stated otherwise, \(\mathbb{R}\) has its standard topology, intervals inherit the subspace topology from \(\mathbb{R}\), and \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) has the subspace topology from the Euclidean plane.

When we describe two sets as open in \(X\setminus\{x\}\), we mean open in its subspace topology. For example, if \(X\subseteq\mathbb{R}\), then \(X\cap(a,b)\) is open in \(X\). This relative-topology detail is what makes a proposed split a separation of the deleted space.

Why Intervals Are Connected

Proof. Suppose, to obtain a contradiction, that \(J=U\cup V\) is a separation. Choose one point from each set. After interchanging \(U\) and \(V\) if needed, there are \(u\in U\) and \(v\in V\) with \(u<v\). The interval property implies \([u,v]\subseteq J\). The set \(E=U\cap[u,v]\) is nonempty because \(u\in E\), and it is bounded above by \(v\). Let \(c=\sup E\), which exists by completeness of \(\mathbb{R}\). Since \(u\in E\), \(c\ge u\). Also \(c\ne u\): because \(U\) is open relative to \(J\) and \(u<v\), there is a point of \(U\cap[u,v]\) strictly greater than \(u\). Thus \(c>u\).

The point \(c\) belongs to \(J\), so it lies in exactly one of \(U,V\). If \(c\in U\) and \(c<v\), relative openness gives points of \(U\cap[u,v]\) strictly greater than \(c\), contradicting that \(c\) is an upper bound of \(E\). If \(c\in U\) and \(c=v\), then \(v\in U\cap V\), contradicting disjointness. If \(c\in V\), relative openness gives a neighborhood of \(c\) in \(J\) contained in \(V\). Since \(c>u\) and \(c=\sup E\), there is a point of \(E\) sufficiently close to \(c\) from below; that point would belong to both \(U\) and \(V\). Every possibility is a contradiction. Therefore \(J\) has no separation and is connected. The empty and singleton cases are connected directly from the definition. \(\square\)

We will also use two elementary facts about connectedness. A continuous image of a connected space is connected: if the image had a separation, its preimages would be disjoint, nonempty, relatively open sets covering the domain. A homeomorphism therefore preserves connectedness, since it and its inverse are continuous. These facts let us use explicit parametrizations to study the circle.

The Cut-Point Profile of the Line and Intervals

Proof. Let \(c\in\mathbb{R}\). The deleted space is the disjoint union \((-\infty,c)\cup(c,\infty)\). Both sets are nonempty and open in \(\mathbb{R}\setminus\{c\}\), and together they cover that space, so they form a separation. Thus every \(c\in\mathbb{R}\) is a cut point.

Now let \(a<c<b\). In \((a,b)\setminus\{c\}\), the sets \((a,c)\) and \((c,b)\) are disjoint, nonempty, relatively open, and cover the deleted space. In \([a,b]\setminus\{c\}\), the sets \([a,c)\) and \((c,b]\) have the same properties. So every interior point is a cut point in both intervals.

At the left endpoint of the closed interval, \([a,b]\setminus\{a\}=(a,b]\), which is an interval and hence connected by the preceding theorem. At the right endpoint, \([a,b]\setminus\{b\}=[a,b)\), also an interval and connected. Therefore neither endpoint is a cut point. Finally, if \(a=b\), the space \([a,a]=\{a\}\) is connected and its deletion is empty, which is connected by definition. It has no cut points. \(\square\)

Deleting a Point from the Circle

The circle behaves differently from a line: deleting one point does not divide it into two components. First, \(S^1\) is connected. The continuous map \(t\mapsto(\cos t,\sin t)\) from the interval \([0,2\pi]\) onto \(S^1\) is surjective, and \([0,2\pi]\) is connected by the interval theorem. The continuous-image fact above implies that \(S^1\) is connected.

Proof. We first show that \(S^1\setminus\{(1,0)\}\) is homeomorphic to \(\mathbb{R}\). Define \(F:\mathbb{R}\to S^1\setminus\{(1,0)\}\) by \(F(t)=\left(\frac{t^2-1}{t^2+1},\frac{2t}{t^2+1}\right)\). For every real \(t\), \(t^2+1>0\), and \((t^2-1)^2+(2t)^2=t^4+2t^2+1=(t^2+1)^2\), so \(F(t)\in S^1\). Its first coordinate is not \(1\), since \(t^2-1=t^2+1\) would imply \(-1=1\). Thus its image avoids \((1,0)\).

For any \((x,y)\in S^1\setminus\{(1,0)\}\), define \(G(x,y)=y/(1-x)\). The denominator is nonzero because \(x\ne1\). Substituting the coordinates of \(F(t)\) gives \(1-x=2/(t^2+1)\) and \(y=2t/(t^2+1)\), so \(G(F(t))=t\). Conversely, using \(x^2+y^2=1\) and \(x\ne1\), substitution of \(t=y/(1-x)\) into \(F\) gives first coordinate \(x\) and second coordinate \(y\): indeed \(t^2=(1+x)/(1-x)\), so \((t^2-1)/(t^2+1)=x\) and \(2t/(t^2+1)=y\). Hence \(F\) and \(G\) are inverse bijections. Their coordinate formulas are continuous wherever their denominators are nonzero, so they are continuous. Thus they are a homeomorphism. Since \(\mathbb{R}\) is connected, \(S^1\setminus\{(1,0)\}\) is connected.

For an arbitrary point \(p=(u,v)\in S^1\), where \(u^2+v^2=1\), the rotation \(R(x,y)=(ux+vy,-vx+uy)\) maps \(p\) to \((1,0)\). Its inverse is the rotation \((x,y)\mapsto(ux-vy,vx+uy)\), so it is a homeomorphism of \(S^1\). It maps \(S^1\setminus\{p\}\) onto \(S^1\setminus\{(1,0)\}\). Therefore \(S^1\setminus\{p\}\) is connected for every \(p\in S^1\). Since \(S^1\) itself is connected, none of its points is a cut point. \(\square\)

Profiles and a Common Pitfall

Common pitfall. A tempting but incorrect claim is that deleting the left endpoint of \([a,b]\) disconnects it into a left piece and a right piece. For \([-2,4]\), this would require a nonempty part of the interval to the left of \(-2\), but no such point belongs to \([-2,4]\). The actual remainder is \((-2,4]\), a single interval, and is connected. The two-piece separation applies only when the deleted point is strictly between the endpoints.

The distinction is useful because cut-point information is a quick way to recognize how a space is organized. A line and an open interval have a cut point at every point; a nondegenerate closed interval omits its endpoints from that profile; and the circle has no cut points at all. The conclusion comes from the topology of the deleted space, not from a drawing alone.