Cut point

A line (closed interval) has infinitely many cut-points between two end points. A circle has no cut-point. Since they have different numbers of cut-points, lines are not homeomorphic to circles

A cut-point of a connected T₁ topological space X, is a point p in X such that X - {p} is not connected. A point which is not a cut-point is called a non-cut point.

A non-empty connected topological space X is a cut-point space if every point in X is a cut point of X.

• A closed interval [a,b] has infinitely many cut-points. All points except for its end points are cut-points and the end-points {a,b} are non-cut points.
• An open interval (a,b) also has infinitely many cut-points like closed intervals. Since open intervals don't have end-points, it has no non-cut points.
• A circle has no cut-points and it follows that every point of a circle is a non-cut point.

• A cutting of X is a set {p,U,V} where p is a cut-point of X, U and V form a separation of X-{p}.
• Also can be written as X\{p}=U|V.

• Cut-points are not necessarily preserved under continuous functions. For example: f: [0, 2π] → R², given by f(x) = (cos x, sin x). Every point of the interval (except the two endpoints) is a cut-point, but f(x) forms a circle which has no cut-points.
• Cut-points are preserved under homeomorphisms. Therefore, cut-point is a topological invariant.

• Every continuum (compact connected Hausdorff space) with more than one point, has at least two non-cut points. Specifically, each open set which forms a separation of resulting space contains at least one non-cut point.
• Every continuum with exactly two noncut-points is homeomorphic to the unit interval.
• If K is a continuum with points a,b and K-{a,b} isn't connected, K is homeomorphic to the unit circle.

• Let X be a connected space and x be a cut point in X such that X\{x}=A|B. Then {x} is either open or closed. if {x} is open, A and B are closed. If {x} is closed, A and B are open.
• Let X be a cut-point space. The set of closed points of X is infinite.

A cut-point space is irreducible if no proper subset of it is a cut-point space.

The Khalimsky line: Let ${\displaystyle \mathbb {Z} }$ be the set of the integers and ${\displaystyle B=\{\{2i-1,2i,2i+1\}:i\in \mathbb {Z} \}\cup \{\{2i+1\}:i\in \mathbb {Z} \}}$ where ${\displaystyle B}$ is a basis for a topology on ${\displaystyle \mathbb {Z} }$. The Khalimsky line is the set ${\displaystyle \mathbb {Z} }$ endowed with this topology. It's a cut-point space. Moreover, it's irreducible.

• A topological space ${\displaystyle X}$ is an irreducible cut-point space if and only if X is homeomorphic to the Khalimsky line.