# Cut points in Bowditch Boundary of Relatively hyperbolic groups 2

Understanding Swenson (large excerpt.. some diagrams .. some remarks). Please be cautious. Potentially wrong remarks lie ahead.

### Continuum

A continuum is a compact connected Hausdorff space.

### Cut Point

In a continuum Z, $$c \in Z$$ is a cut point if $$Z = A \cup B$$ where A and B are non-singleton continua and $$A \cap B = \{c\}$$. If in addition $$D \subset A – \{c\}$$ and $$E \subset B – \{ c \}$$, we say that c separates D from E.

For the remainder of this section, Z will be a metric continuum, G will be a group (possibly trivial) of homeomorphisms of Z, and $$C \subset Z$$ will be a G-equivariant (GC = C) set of cut points of Z.

### Is a member of (interval)

For $$a, b \in Z$$ and $$c \in C$$, we define $$c \in (a, b)$$ if there exist non-singleton continua A containing a and B containing b with $$A \cup B = Z$$ and $$A \cap B = \{ c \}$$.

Any point in an interval is a cut point by definition of interval.

We define the closed and half open intervals in the obvious way i.e., $$[a, b] = \{a, b \} \cup (a, b)$$, and $$[a, b) = \{a\} \cup (a, b)$$ for $$a \neq b$$ ( $$[a, a) = \phi$$ )

Notice that if $$c \in (a, b)$$ then for any subcontinuum $$Y \subset Z$$ from a to b ($$a, b \in Y$$ ), $$c \in Y$$. Why? Suppose Y is a subcontinuum containing a and b. $$c \in (a, b)$$ implies there are continua A and B such that $$a \in A, b \in B, A \cup B = Z, A \cap B = \{ c \}$$.

$$A \cap Y$$ and $$B \cap Y$$ compact and Hausdorff (hence are subcontinua) containing a and b respectively. Their union is Y (as all y in Y are either in A or B as A union B is Z and Y is contained in Z). Their intersection is either $$\phi$$ or {c}.

confusion

### Equivalence of points

For $$a, b \in Z – C$$ we say that a is equivalent to b, a ~ b, if $$(a, b) = \phi$$. That is there is no cut point between a and b.For $$c \in C$$, c is equivalent only to itself. This is clearly an equivalence relation, so let P bet the set of equivalence classes of Z. We will abuse notation and say $$C \subset P$$ since each element of C is its own equivalence class.

Observe that for $$a, b, d\in Z$$ if a ~ b, then (a, d) = (b, d). We can therefore translate the interval relation on Z to P and we also enlarge it as follows.

### Membership in P (interval in P)

For $$x, y, z \in P$$, we say $$y \in (x, z)$$ if either

1. $$y \in C$$ where $$y \in (a, b)$$ for some $$a, b \in Z$$ with $$a \in x$$ and $$b \in z$$ (note that x, and z are equivalence classes ) or
2. $$y \notin C and if \( a, b, d \in Z$$ with $$a\in x , b \in Y$$ and $$d \in z$$, then $$[a, b) \cup (b, d] = \phi$$

Since C was chosen to be G invariant, the action of G on Z gives an action of G on P which preserves the interval structure ( we have not given P a topology so it doesn’t make sense to ask if the action is by homeomorphism).