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How do I prove that the space of real numbers, under the lower limit topology, is a normal space.

I could prove very easily that it is regular, by using an argument of basic sets, but I haven't been able to generalise that argument.

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HINT: Prove that it's Lindelof, because every regular Lindelof space is paracompact and thus normal. Alternatively, you can use the following approach. Let us denote the lower limit topology on $\mathbb{R}$ as $\mathbb{R}_\ell$. Suppose $A$ and $B$ are two disjoint closed sets in $\mathbb{R}_\ell$. Then, note that $\mathbb{R}\setminus A$ and $\mathbb{R}\setminus B$ are open and that $A\subset \mathbb{R}\setminus B$ and $B \subset \mathbb{R}\setminus A$). Given any $a \in A$, there is a basic open set $U_a :=[a,a+\rho_a)\subset \mathbb{R}\setminus B$ for some $\rho_a >0$. Similarly, for each $b \in B$, we can find a $\rho_b >0$ such that $V_b:=[b,b+\rho_b) \subset \mathbb{R}\setminus A$. Let $U=\bigcup_{A} U_a $ and $V=\bigcup_B V_b$. Clearly $A \subset U$ and $B \subset V$. Last, we show that $U$ and $V$ are disjoint. Suppose $U_a\cap V_b= [a,a+\rho_a)\cap[b,b+\rho_b) \not= \emptyset$, then $max\{a,b\}\in U_a\cap V_b$. W.l.o.g say $a = max\{a,b\}$. Then, $a \in A$ and $a \in V_b \subset \mathbb{R} \setminus A$, a contradiction.

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  • $\begingroup$ I do not know about Lindelof or paracompact spaces. I had this as a problem in my exam, so I believe there's another way of doing this. $\endgroup$
    – MathManiac
    Mar 9, 2016 at 20:04
  • $\begingroup$ I added a direct approach, hopefully you find this more concrete. $\endgroup$
    – sqtrat
    Mar 9, 2016 at 20:23

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