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In the usual subspace topology from the real line, a subset $A\subseteq \mathbb{R}$ is connected exactly when it has the defining property of a real interval:Every real number between two members must be present. Thus open intervals, closed intervals, intervals with only one endpoint included, and unbounded real intervals are connected. Singleton sets are connected too. With the no-separation of a topological space convention used here, the empty set satisfies the same property; some texts reserve the word connected for nonempty topological spaces.^[ref]

Removing an interior point changes the answer:These two real intervals are exactly its connected components and its path components. Removing an endpoint instead leaves a real interval, for example $[-2,2]\setminus \{2\}=[-2,2)$, which stays connected. This is why an interior point of a closed interval is a cut point, whereas an endpoint is not. A nonempty finite set of points in the real line is connected exactly when it is a singleton set.

Proof outline: if $a,b\in A$ but $c\in (a,b)$ is missing, the subsetsgive a separation of a topological space, since the two open intervals make both pieces open in $A$. Conversely, a nonempty real interval contains the continuous path $t\mapsto (1-t)a+tb$ between any two of its points, so it is path-connected and hence connected. Together with continuous image of a connected space, this characterization gives the intermediate value theorem: a continuous function on a real interval cannot jump over a value lying between two attained values.

Let $t_0$ be the finite cut time and put $q=\gamma (t_0)$. Continuity of Riemannian distance shows that the segment to $q$ still minimizes and $d(p,q)=t_0$. Choose $t_j\downarrow t_0$ with $t_j>t_0$. By the Hopf-Rinow theorem, there is a minimizing geodesic from $p$ to $q_j=\gamma (t_j)$ with length ${\ell }_j=d(p,q_j)<t_j$ and initial unit vector $v_j$. Continuity gives ${\ell }_j\to t_0$.

The unit tangent sphere is compact, so after a subsequence $v_j\to v$. Smooth dependence of the Riemannian exponential map gives ${\exp }_p(t_{0}v)=q$. The geodesic with initial vector $v$ has length $t_0=d(p,q)$ up to $q$, hence minimizes. If $v\ne \dot{\gamma }(0)$, this is the required second minimizing geodesic.

Suppose instead $v=\dot{\gamma }(0)$ and $q$ is not a conjugate point. Then the differential of ${\exp }_p$ at $t_0\dot{\gamma }(0)$ is nonsingular. The inverse function theorem makes ${\exp }_p$ injective on a neighborhood of this vector. For large $j$, both ${\ell }_j{v}_j$ and $t_j\dot{\gamma }(0)$ belong to that neighborhood and exponentiate to $q_j$. They must be equal, implying ${\ell }_j=t_j$, a contradiction. Thus either a distinct minimizing geodesic exists or $q$ is conjugate. In the latter case it is the first conjugate point, since part (a) excludes any earlier one on a minimizing segment. We have proved the cut-point dichotomy for complete Riemannian manifolds.

Here the relevant concept is a Riemannian cut point, not the topological cut point obtained by disconnecting a space through removal of a point.

No. TakeIt contains more than one point and has the fixed-point property of a closed interval, but it is not homeomorphic to the closed unit disc. Indeed, deleting any point $(t,0)$ with $0<t<1$ disconnects $B$, so such a point is a cut point. Deleting one point from a two-dimensional closed disc leaves a path-connected space. Since being a cut point is preserved by homeomorphisms, the two spaces cannot be homeomorphic. This gives the required counterexample.

Removing $1/2$ from the closed interval $I=[0,1]$ leaves the separation of a topological spaceThus $1/2$ is a cut point of $I$.

In contrast, removing any point $p$ from the square $Q=[0,1{]}^2$ leaves a path-connected space. To prove this, take distinct $u,v\in Q\setminus \{p\}$. Choose an interior point $w$ outside the two lines through $(u,p)$ and $(v,p)$. Such a choice exists: a small circle in the interior meets each of those lines in at most two points. Since $Q$ is a convex set, the line segments $[u,w]$ and $[w,v]$ stay in $Q$; our choice of $w$ ensures that both avoid $p$. Their concatenation is a continuous path from $u$ to $v$ in $Q\setminus \{p\}$.

The product topology on $I\times I$ agrees with its usual subspace topology in the Euclidean plane. If $h:Q\to I$ were a homeomorphism, its restriction would give a homeomorphismThe domain is connected, since path connectedness implies connectedness, while the target is not connected. This contradicts the preservation of connectedness by a homeomorphism.

For a unit-speed geodesic $\gamma (t)={\exp }_p(tv)$, a finite cut time $c(v)$ is the last time to which its segment from $p$ minimizes Riemannian distance. Its endpoint is the Riemannian cut point along that geodesic. It lies in the cut locus; it need not be a cut point in the topological sense. The segment up to the cut time still minimizes by continuity of distance.