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.


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 $x$. $\square$


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.)

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  • $\begingroup$ Excellent! Now I wonder if this can help to answer the configuration space question for $n>2$... $\endgroup$ – user31373 Jun 11 '12 at 0:20
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    $\begingroup$ I put a print screen of the theorem here math.stackexchange.com/a/157196/31682 $\endgroup$ – Sigur Jun 12 '12 at 1:06
  • $\begingroup$ Excellent answer! In retrospect I don't think it matters very much to the other question, but I am still very pleased to know this. $\endgroup$ – Niels J. Diepeveen Jun 19 '12 at 10:20
  • $\begingroup$ @Dejan: In your answer, is the second theorem strictly stronger than the first? That is, can the second theorem be proven using the first? $\endgroup$ – Herng Yi Aug 27 '12 at 14:44
  • $\begingroup$ @HerngYi: Well, my guess would be that it depends on what else you are willing to assume. As far as I understand things, they are both theorems of $\textrm{ZFC}$, thus both are true and equivalent over $\textrm{ZFC}$. But if you work over some other theory, I guess one or the other might be stronger ... But logic isn't exactly my strong point, so we should probably ask logicians about this. $\endgroup$ – Dejan Govc Aug 27 '12 at 16:44

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