Connected Components for $(\Bbb R, \mathcal T_{ lower limit})$

Question

$(\Bbb R, \mathcal T_{{ lower }{ limit}})$ 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 $(\Bbb R, \mathcal T_{ lower limit})$?

You need to prove the singletons and the empty set are the only connected subsets. See this http://math.stackexchange.com/questions/172964/is-r-with-j-d-topology-totally-disconnected — JonSK, May 04 '16 at 01:53
@JonSK so the connected components are ${x}$ for all $x\in \Bbb R$? — Belive, May 04 '16 at 02:25
@Belive basis of clopen sets. Do you see how to use that to show there can be no connected subset with more than one point? — Forever Mozart, May 04 '16 at 02:27
@ I haven't heard basis of clopen sets so far. Or are basis of clopen sets and bais of topology synonymous? — Belive, May 04 '16 at 03:41
I'll add a fun exercise: characterize all continous functions $f : \mathbb{R} \to \mathbb{R}^\ell$. — Guillermo Mosse, May 16 '19 at 13:34

score 14 · Accepted Answer · answered May 04 '16 at 07:38

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 $a=b$. $\blacksquare$

Connected Components for $(\Bbb R, \mathcal T_{ lower limit})$

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