What Does It Mean to Cut a Space?
A connected space can become disconnected when a point is removed, but this does not happen for every point. The answer depends both on the space and on the point: removing an interior point of an interval separates its two sides, while removing a point from a circle leaves one connected arc. We will make these claims precise and verify them using the definitions.
A subset \(A\) of a space \(X\) is given the subspace topology: its open sets are intersections \(A\cap U\), where \(U\) is open in \(X\). A separation of a space \(Y\) is a pair of disjoint, nonempty, relatively open subsets whose union is \(Y\). A space is connected if it has no separation. In particular, the empty space and every one-point space are connected, since neither has two nonempty disjoint subsets.
Thus it is not enough to observe that deleting \(p\) produces more than one visible piece: one must show that the remaining space has a separation. To identify its components, one must also show that the pieces are connected and cannot be enlarged to a larger connected subset.
Two Connectedness Facts We Will Use
Proof. Suppose an interval \(J\subseteq\mathbb{R}\) has a separation \(U,V\). Choose \(u\in U\) and \(v\in V\). After exchanging the names of \(U,V\) if necessary, assume \(u<v\). The set \(E=U\cap[u,v]\) is nonempty because \(u\in E\), and it is bounded above by \(v\). Let \(s=\sup E\). Since \(J\) is an interval and \(u\le s\le v\), we have \(s\in J\), so \(s\in U\) or \(s\in V\). If \(s\in U\), relative openness of \(U\) gives points of \(U\cap[u,v]\) strictly greater than \(s\), since \(s<v\); this contradicts that \(s\) is an upper bound for \(E\). If \(s\in V\), then \(s\ne u\), and relative openness of \(V\) gives an interval immediately to the left of \(s\) lying in \(V\). The definition of supremum gives a point of \(E\) in that interval, contradicting \(U\cap V=\varnothing\). Both possibilities are impossible, so \(J\) has no separation. \(\square\)
Proof. Let \(f:X\to Y\) be continuous, with \(X\) connected. Suppose \(f(X)\) has a separation \(C,D\). Then \(f^{-1}(C)\) and \(f^{-1}(D)\) are open in \(X\), because \(C,D\) are open in the subspace \(f(X)\) and \(f\) is continuous. They are disjoint and cover \(X\). They are both nonempty: every point of \(C\) and every point of \(D\) is in \(f(X)\), so each has a preimage. This would be a separation of \(X\), contrary to connectedness. Therefore \(f(X)\) is connected. \(\square\)
We will also use the following consequence. A space is path-connected if every two of its points can be joined by a continuous map from \([0,1]\) into the space. A path-connected space is connected: if a separation existed, a path from a point in one side to a point in the other would have preimages forming a separation of \([0,1]\). Those preimages are open in \([0,1]\), nonempty, disjoint, and cover it, contradicting Proposition 1.
Worked Example: Deleting Points from a Closed Interval
Give \([0,1]\) its usual subspace topology from \(\mathbb{R}\). Determine whether each \(c\in[0,1]\) is a cut point, and find the components after deletion.
Interior point. Suppose \(0<c<1\). Then \( [0,1]\setminus\{c\}=[0,c)\cup(c,1]\). Both pieces are nonempty. They are relatively open in the punctured space: \([0,c)=(-1,c)\cap[0,1]\), and \((c,1]=(c,2)\cap[0,1]\). They are disjoint and cover the punctured space, so they form a separation. Each piece is an interval, hence connected by Proposition 1. Any connected subset meeting both pieces would be separated by its intersections with these two relatively open pieces. Therefore the two pieces are exactly the connected components, and \(c\) is a cut point.
Left endpoint. For \(c=0\), deletion gives \([0,1]\setminus\{0\}=(0,1]\). This is an interval, so it is connected. Thus \(0\) is not a cut point.
Right endpoint. For \(c=1\), deletion gives \([0,1]\setminus\{1\}=[0,1)\), also an interval and therefore connected. So \(1\) is not a cut point either.
Conclusion and check. The cut points of \([0,1]\) are exactly the points \(c\) satisfying \(0<c<1\). For such \(c\), the two components contain \(0\) and \(1\), respectively; the strict inequalities in the formulas leave \(c\) out of both pieces, as required.
Deleting a Point from the Circle
Let \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\), with the subspace topology inherited from the Euclidean topology on \(\mathbb{R}^2\). The usual parametrization \(t\mapsto(\cos t,\sin t)\) is continuous and maps \([0,2\pi]\) onto \(S^1\). Proposition 1 and Proposition 2 therefore show that \(S^1\) is connected, so it makes sense to ask which of its points are cut points.
Worked Example: Removing an Arbitrary Point from the Circle
Fix \(p\in S^1\). Choose a real angle \(\alpha\) with \(p=(\cos\alpha,\sin\alpha)\), and define \(f:(0,2\pi)\to S^1\) by \(f(t)=(\cos(\alpha+t),\sin(\alpha+t))\). The map is continuous. It never takes the value \(p\), because \(0<t<2\pi\) is not a full turn. Conversely, every point of \(S^1\) other than \(p\) has an angle differing from \(\alpha\), modulo \(2\pi\), by a unique number \(t\in(0,2\pi)\). Hence \(f((0,2\pi))=S^1\setminus\{p\}\).
The interval \((0,2\pi)\) is connected by Proposition 1, so its continuous image \(S^1\setminus\{p\}\) is connected by Proposition 2. This holds for every choice of \(p\), so the circle has no cut points. The result is a single connected component after deletion, not two components: the parametrization travels along the circle through all points other than the deleted one.
Check. The argument applies to an arbitrary \(p\), not only to \((1,0)\). Changing \(\alpha\) changes the starting angle, while the parameter interval still covers exactly one turn without including the deleted point.
A Finite Graph: The Y-Shaped Example
For a concrete finite graph, take the three vectors \(u_1=(1,0)\), \(u_2=(-1/2,\sqrt{3}/2)\), and \(u_3=(-1/2,-\sqrt{3}/2)\) in \(\mathbb{R}^2\), and let \(X=\{tu_i:0\le t\le1,\ i\in\{1,2,3\}\}\), with the subspace topology from \(\mathbb{R}^2\). This is a Y-shaped graph: its three edges meet at the vertex \(o=(0,0)\).
The space \(X\) is connected. Indeed, any point \(tu_i\) can be joined to \(o\) by the path \(r\mapsto(1-r)tu_i\), for \(0\le r\le1\). Joining two such paths gives a path between any two points of \(X\), so \(X\) is path-connected and hence connected.
Worked Example: Deleting the Centre or an Endpoint of the Y
Delete the central vertex. The remaining space is the disjoint union of the three arms \(A_i=\{tu_i:0<t\le1\}\), for \(i=1,2,3\). Each arm is connected: it is the continuous image of the interval \((0,1]\) under \(t\mapsto tu_i\).
Each \(A_i\) is relatively open in \(X\setminus\{o\}\). To check this, take \(x=tu_i\in A_i\), where \(0<t\le1\). The distance from \(x\) to \(o\) is \(t\). For \(j\ne i\) and \(0<s\le1\), the angle between \(u_i\) and \(u_j\) is \(2\pi/3\), so \(\|tu_i-su_j\|^2=t^2+s^2+ts\ge t^2\). Thus the open ball of radius \(t/2\) about \(x\) meets no other arm and does not contain \(o\). It follows that \(A_i\) is relatively open. Its complement in the punctured space is the union of the other two relatively open arms, so \(A_i\) is also relatively closed there.
A connected subset of \(X\setminus\{o\}\) cannot meet both \(A_i\) and its complement, because those intersections would separate that subset. Since each \(A_i\) is itself connected, the three arms are exactly the components. They are nonempty, so their union is a separation; consequently \(o\) is a cut point.
Delete an endpoint instead. Let \(q=u_1\), the endpoint at \(t=1\) on the first arm. Then \(X\setminus\{q\}\) is the union of \(\{tu_1:0\le t<1\}\), \(\{tu_2:0\le t\le1\}\), and \(\{tu_3:0\le t\le1\}\). Each set is an interval mapped continuously into \(X\), hence connected, and all three contain \(o\). Their union is connected: if a separation existed, the side containing \(o\) would have to contain each whole connected set, leaving the other side empty. Thus \(q\) is not a cut point.
Check. Deleting \(o\) leaves three nonempty components; deleting \(q\) leaves one connected space. The distinction is verified by the actual subspaces after deletion, rather than by the appearance of the drawing.
A Common Endpoint Error
A common mistake is to treat a deleted endpoint as though it were an interior point, or to silently drop an endpoint that has not been deleted. For example, the incorrect calculation \([0,1]\setminus\{0\}=(0,1)\) omits \(1\), which is still in the set. The correct calculation is \([0,1]\setminus\{0\}=(0,1]\). This is one interval and is connected; therefore \(0\) is not a cut point. By contrast, when \(0<c<1\), deleting \(c\) really does leave the two intervals \([0,c)\) and \((c,1]\). Checking endpoint membership before testing connectedness prevents the error.
The method used throughout is direct: write down the remaining subspace, exhibit a separation if it is disconnected, and verify connectedness of the proposed components. A picture can suggest the answer, but the separation and connectedness checks establish it.
Check Your Understanding
For each question, specify the topology and check carefully whether the endpoints belong to the set after deletion.
- For \(c\in(0,1)\), what are the connected components of \([0,1]\setminus\{c\}\)?
- Is \(1/2\) a cut point of the space \((0,1)\) with its usual subspace topology? Give the components after deletion.
- Why does the continuous parametrization of \(S^1\setminus\{p\}\) use the open interval \((0,2\pi)\), rather than \([0,2\pi]\)?
- In the Y-shaped graph, why does each arm after deleting the central vertex qualify as a component rather than merely a connected subset?
- For the Y-shaped graph, what is the space left after deleting an endpoint, and why is that space connected?