# 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 iff the points $0$ and $y$ lies in different components of $X\setminus \{x\}$.

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

Proof. Suppose $x. 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”

1. Dejan Govc says:

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 $X$.

2. Dejan Govc says:

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

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