Separation order

I follow up on the previous post to report that the question stated there was answered by Dejan Govc. I’ll try to put together a self-contained version of his proof.

Let $X$ be a connected topological space. Fix a base point $0$ and define the separation order on $X$ as follows: $x<y$ iff the points $0$ and $y$ lies in different components of $X\setminus \{x\}$ .

Theorem: at most one of $x<y$ and $y<x$ holds.

Proof. Suppose $x<y$ . Let $Y$ be the connected component of $y$ in $X\setminus \{x\}$ . Note that $0,x\in X\setminus Y\subset X\setminus \{y\}$ . If we can show that $X\setminus Y$ is connected, then $0$ and $x$ are in the same component of $X\setminus \{y\}$ , which is to be proved. Suppose to the contrary that $X\setminus Y$ has a nonempty proper closed-open subset $A$ . We may assume $x\notin A$ . Note two things about $A$ :

$\overline{A}\subset A\cup Y$ , because $A$ is closed in $X\setminus Y$
$A$ is open in $X$ . Indeed, $A$ open in $X\setminus Y$ . The closure of $Y$ is contained in $Y\cup \{x\}$ and therefore does not meet $A$ . Hence $A$ is open in $X\setminus \overline{Y}$ , and therefore in $X$ .

Since $Y\subsetneq A\cup Y\subset X\setminus \{x\}$ , the set $A\cup Y$ is not connected. So it has a nonempty proper closed-open subset $Z$ . Then either $Y\subset Z$ or $Y\cap Z=\varnothing$ ; by replacing $Z$ with its complement in $A\cup Y$ we may assume the latter holds: that is, $Z\subset A$ . We claim that $Z$ is both closed and open in $X$ , contradicting the connectedness of $X$ . Indeed,

$Z$ is closed in $A\cup Y$ and $Z\subset \overline{A}\subset A\cup Y$ . Hence $Z$ is closed in $\overline{A}$ , and in $X$ .
$Z$ is open in $A\cup Y$ and $Z\subset A$ , hence $Z$ is open in $A$ . Since $A$ is open in $X$ , it follows that $Z$ is open in $X$ .

2 thoughts on “Separation order”

Nice exposition! The $<$ notation gives is very suggestive. I think there's a slight problem, though: I'm not sure $Y$ is necessarily closed in $X$ .

I think it is quite easy to get rid of this problem though: $Z$ is closed in $\overline A$ which is closed in $X$ ; and since $x\notin A$ , we have that $A$ is open in $X\setminus\overline Y$ . I hope this helps. Best regards, Dejan

Separation order

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2 thoughts on “Separation order”

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