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

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  • $\begingroup$ You need to prove the singletons and the empty set are the only connected subsets. See this math.stackexchange.com/questions/172964/… $\endgroup$ – JonSK May 4 '16 at 1:53
  • $\begingroup$ Isn't this space zero dimensional? $\endgroup$ – Forever Mozart May 4 '16 at 2:04
  • $\begingroup$ @JonSK so the connected components are $\{x\}$ for all $x\in \Bbb R$? $\endgroup$ – Belive May 4 '16 at 2:25
  • $\begingroup$ @ForeverMozart What is zero dimensional? $\endgroup$ – Belive May 4 '16 at 2:26
  • $\begingroup$ @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? $\endgroup$ – Forever Mozart May 4 '16 at 2:27
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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 $a=b$. $\blacksquare$

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