TITLE: Connected Components for $(\Bbb R, \mathcal T_{ lower limit})$
QUESTION [6 upvotes]: $(\Bbb R, \mathcal T_{{ lower }{ limit}})$ is a topological space $\Bbb R$ with Lower limit topology.
As I know,  $(\Bbb R, \mathcal T_{{ lower }{ limit}})$ is disconnected. What 
are the Connected Components of $(\Bbb R, \mathcal T_{ lower limit})$ or how can we describe the Connected Components of $(\Bbb R, \mathcal T_{ lower limit})$?
I know that $[a,b)$ , $(-\infty,a)$, $[a,\infty)$ are clopen for any $a,b\in \Bbb R$. And $\Bbb R = (-\infty,\infty)=(-\infty,0)\cup [0,\infty)$, so $(-\infty,0)$,$[0,\infty)$ are Connected Components of $(\Bbb R, \mathcal T_{ lower limit})$?

REPLY [12 votes]: To see that the connected components are the singletons, it is enough to show the following:
Claim: If $A\subseteq\mathbb R$ is not empty, then $A$ is connected if and only if $A$ is a singleton.
Proof: Sufficiency is trivial. As for necessity, suppose that $A\neq \varnothing$ and take an arbitrary $a\in A$. Then, $A$ is the disjoint union of the two (relatively) open sets $(-\infty,a)\cap A$ and $[a,\infty)\cap A$. Since $A$ is connected, at most one of these open sets can be non-empty, and since $a\in[a,\infty)\cap A\neq\varnothing$, it follows that $(-\infty,a)\cap A=\varnothing$. Hence, if $b\in A$, then $b\notin(-\infty,a)$, so that $b\geq a$. Since $a$ and $b$ have been arbitrarily chosen, their roles can be interchanged, implying that $a\geq b$. Conclusion: if $a,b\in A$, then $a=b$. $\blacksquare$