"Antisymmetry" among cut points We might define a ternary relation between points of a topological space $X$ by writing $x|yz$ whenever $y$ is not in the same quasi-component of $X\setminus \{x\}$ as $z$. It is not hard to prove that if $X$ is connected, at most one of $x|yz$, $y|xz$ and $z|xy$ is true for distinct $x, y, z$. Informally, at most one of the points can be "between" the other two.
What I can't figure out is whether this still holds if quasi-components are replaced by real components. If it did, I think it would simplify the answer to the question Is the configuration space of a connected space connected?
Any help would be appreciated.

Addendum:
For comparison, here is the proof I had in mind for the
case of quasi-components. Clearly $x|yz$ is symmetric w.r.t. $y$ and $z$,
so it suffices to prove that 

if $x|yz$ and $y|xz$ for distinct $x, y, z \in X$,
  then $X$ is disconnected.

If $x|yz$ then there is a clopen neighbourhood $U$ of $z$ in
$X \setminus \{x\}$ that does not include $y$.
This means that one of $U$ and $U \cup \{x\}$ must be open in
$X$ and one must be closed. Similarly, if $y|xz$ there must be a set $V$ that
includes $z$ but not $x$, such that one of $V$ and $V \cup \{y\}$ is open
and one is closed.
Since $y \notin U \cup \{x\}$ and $x \notin V \cup \{y\}$ we have
$$
  (U \cup \{x\}) \cap (V \cup \{y\}) = (U \cup \{x\}) \cap V =
  U \cap (V \cup \{y\}) = U \cap V
$$
Thus $U \cap V$ is always the intersection of two open sets and the
intersection of two
closed sets, therefore it is clopen, and since it contains $z$ but not
$x$ it shows that $X$ is disconnected. $\square$
The analogous statement about path-components in a path-connected space
is also not hard to prove. This statement is equivalent to

For distinct points $x, y, z$ in a path-connected space $X$, there is
  a path from $z$ to $x$ that does not pass through $y$, or a path from
  $z$ to $y$ that does not pass through $x$.

Let $f: [0,1] \to X$ be any path from $z$ to $x$. Consider the path
$$
  g(t) = \cases{
    y, & if $y \in f([0, t])$  \cr
    f(t), & otherwise.  \cr
  }
$$
If $g(a) = x$ for some $a \in (0,1]$. then
$h: [0,1] \to X: t \mapsto g(t/a)$ is a path from $z$ to $x$ that doesn't
pass through $y$. If not then $g$ is a path from $z$ to $y$ that
doesn't pass through $x$. $\square$
 A: I have found the following theorem in Kuratowski - Topology (vol. 2, page 140), which is helpful here:
Theorem. Let $X$ be a connected topological space, $A\subseteq X$ a connected subset and $C$ a component of $X\setminus A$. Then $X\setminus C$ is connected.
For distinct points $x,y,z$, define $x|yz$ to mean that $y$ and $z$ lie in different connected components of $X\setminus\{x\}$. We shall now show how the theorem helps us prove what we want:
Proposition. Suppose $X$ is connected and $x,y,z\in X$. Then at most one of $x|yz$, $y|xz$, $z|xy$ holds.
Proof. Suppose $x|yz$ holds, i.e. $y$ and $z$ lie in different connected components of $X\setminus\{x\}$. Let $M$ be the component of $X\setminus\{x\}$ that contains $y$. By the theorem above, $X\setminus M$ is connected. By definition, $y\notin X\setminus M$, so $X\setminus M$ is a connected subset of $X\setminus\{y\}$. Since $y$ and $z$ lie in different components, we further have $z\in X\setminus M$ and since $M\subseteq X\setminus\{x\}$ we also have $x\in X\setminus M$. Therefore $x$ and $z$ lie in the same connected subset of $X\setminus\{y\}$, so $y|xz$ cannot hold. This completes the proof. $\square$
Since I cannot find a useful link to the proof of the theorem, I shall just briefly describe the ideas Kuratowski uses to prove it. First, he establishes the following theorem of decomposition:
Theorem. Let $C$ be a connected subset of a connected topological space $X$. If $M$ and $N$ are separated sets (i.e. $(M\cap\overline{N})\cup(\overline{M}\cap N)=\emptyset$) such that $X\setminus C=M\cup N$, then the sets $C\cup N$ and $C\cup M$ are connected. (Furthermore, if $C$ is closed, $C\cup N$ and $C\cup M$ are also closed.)
The idea of proof is pretty straightforward: first we suppose that $C\cup M=A\cup B$, where $A$ and $B$ are separated. Without loss of generality we can assume that $A\cap C=\emptyset$ and then we show that in $X=C\cup M\cup N=A\cup B\cup N$ the sets $A$ and $B\cup N$ are separated, so at least one of them must be empty.
The proof of the first theorem uses a similar idea. Suppose $X\setminus C= M\cup N$, where $M$ and $N$ are separated. Without loss of generality, $A\cap M=\emptyset$. Next we show that $C\subseteq C\cup M\subseteq X\setminus A$. Since by the second theorem, $C\cup M$ is connected, $C=C\cup M$ follows, proving that $M$ is empty.
(I suppose it shouldn't be too hard for the reader to supply the missing details here. In any case, Kuratowski's exposition of connectedness is really good, so I warmly recommend the book to anyone interested in this subject.)
