Prove that the Sorgenfrey line is totally disconnected

Question

Problem:

Let $ \mathbb{R}_l $ denote the topological space whose underlying set is the real line $ \mathbb{R} $ and the topology is generated by the half closed intervals $ [a,b) $. Prove that the topological space $ \mathbb{R}_l $ is totally disconnected.

My proof:

A space is totally disconnected if its only connected components are one-point sets. Given any set $ I\in \mathbb{R}_l $, $ I=[a,b) $ for some $ a\le b $ in $ R $. If $ a=b $, $ I $ is a one-point set and clearly a connected component (there are no $ A,B\in I $ both non-empty and proper). If $ a<b $, $ I $ is not a one-point set and there exists a $ c $ with $ a<c<b $. Then $ A=[a,c) $ and $ B=[c,b) $ are both non-empty, proper open subsets of $ I $ and they constitute a separation of $ I $ because $ A\cap B=\emptyset $ and $ A\cup B=I $. It is therefore clear that the only connected components in $ \mathbb{R}_l $ are the one-point sets, and hence the topological space $ \mathbb{R}_l $ is totally disconnected.

Question: Is my proof correct?

Answer 1

Your proof is almost complete, and what you have shown is done very nicely.

However:

What about sets that are not half-closed intervals?

Your proof only shows that all intervals are not connected components. It should be easy to complete it by showing that

• closed intervals are not connected
• open intervals are not connected
• if a set $A\subseteq \mathbb R$ is not an interval, it also cannot be connected.

Hint:

$X\subseteq$ is not an interval if and only if there exist some $a<b<c$ such that $a\in X$, $b\notin X$ and $c\in X$

Answer 2

Let $F\subset \mathbb R_l$ be non empty, finite with atleast two elements. Take any $a$ in $F$. Set $d= \min_{x\in F-\{a\}} (|x-a|)$. $\{a\}= F\cap [a,a+d/2)$. So $\{a\}$ is open in subspace topology. $F$ is a discrete space so $F$ can't be connected.

If $S\subset \mathbb R_l$ is infinite, then take any $s$ in $S$ such that $s> \inf S$. $S= ((-\infty, s)\cap S)\cup ([s,\infty)\cap S)$ is a separation so not connected.

So the only connected subsets of $\mathbb R_l$ are the singletons hence $\mathbb R_l$ is totally disconnected.

Answer 3

Let $S$ be a subspace of $\mathbb R_l$ with $a,b \in S$ and $a<b.$

$A=(-\infty,b)=\cup_{x<b}[x,b)$ is open in $\mathbb R_l,$ and $B=[b,\infty)=\cup_{y>b}[b,y)$ is open in $\mathbb R_l$. So $A\cap S$ is not empty (because it contains $a$) and is not equal to $S$ (because it does not contain $b$), but it is open in the subspace $S$, and it is closed in $S$ (because its complement in $S,$ which is $B\cap S,$ is open in $S.$).

So any subspace $S$ of $\mathbb R_l$ with more than one point is a disconnected space.