Homeomorphism between lower limit topology and another topology Given a basis $B$  for a topology  $T$  on R with $B=\left\{[a,b): a,b\in R-\left\{0\right\} \cup \left\{(-x,x): x>0\right\}\right\}$. Show that $(R,T)$ is homemorphic to the lower limit topology $R_l$.
My Progress: Since the basis of $R_l$ consists of open sets of the form $[a,b)$, I am trying to show that there exists a function $f: R\rightarrow R_l$ such that for any basis open sets $[a,b)$,  $f^{-1} ([a,b))$ is open in $B$. I haven't been able to find such a function $f$ to work with, so can anyone give some hints on this issue?
 A: HINT: For $n\in\Bbb N$ let $$I_n=\left[\frac1{2^{n+1}},\frac1{2^n}\right)\;.$$(Note: $0\in\Bbb N$.) Let $L=(\leftarrow,0)$ and $R=[1,\to)$. The sets $L,R$, and $I_n$ for $n\in\Bbb N$ are all clopen in $\Bbb R_\ell$. Now rearrange them, together with the point $0$, as shown here:
$$L\quad I_1\quad I_3\quad I_5\;\ldots\; 0\;\ldots\; I_4\quad I_2\quad I_0\quad R$$
Added: For $n\in\Bbb N$ let
$$J_n=\left[-\frac1{2^n},-\frac1{2^{n+1}}\right)\;.$$
For each $n\in\Bbb N$ there are order-isomorphisms $f_n:I_{2n+1}\to J_n$ and $g_n:I_{2n}\to I_n$. Define
$$h:\Bbb R\to\Bbb R:x\mapsto\begin{cases}
x,&\text{if }x\in R\cup\{0\}\\
g_n(x),&\text{if }x\in I_{2n}\\
f_n(x),&\text{if }x\in I_{2n+1}\\
x-1,&\text{if }x\in L\;.
\end{cases}$$
Show that if the domain has the lower-limit topology, and the range has the topology $T$, then $h$ is a homeomorphism. 
$$\begin{array}{rcc}
L&I_1&I_3&I_5&\ldots&0&\ldots&I_4&I_2&I_0&R\\
&&&&&\;\;\;\downarrow h\\
(\leftarrow,-1)&J_0&J_1&J_2&\ldots&0&\ldots&I_2&I_1&I_0&R
\end{array}$$
Added2: Let $x\in\Bbb R\setminus\{0\}$; then a basic open nbhd of $h(x)$ has the form $[h(x),y)$ for some $y>h(x)$. If $x\in R$, then $[x,y)$ is an open nbhd of $x$ that maps onto $[h(x),y)$. If $x\in I_{2n}$ for some $n\in\Bbb N$, choose $z\in I_{2n}$ such that $x<z$ and $h(z)=g_n(z)\le y$; this is always possible, and $h$ maps $[x,z)$ into $[h(x),y)$. If $x\in I_{2n+1}$ for some $n\in\Bbb N$, choose $z\in I_{2n+1}$ such that $x<z$ and $h(z)=f_n(z)\le y$; this is always possible, and $h$ maps $[x,z)$ into $[h(x),y)$. Finally, if $x\in L$, let $z=y+1$; then $h$ maps $[x,z)$ onto $[h(x),y)$.
Now let $(-x,x)$ be a basic open nbhd of $h(0)=0$. There is an $n\in\Bbb N$ such that $2^{-n}<x$; clearly $I_k\subseteq(0,x)$ for all $k\ge n$. Let $B=\{0\}\cup\bigcup_{k\ge n}I_{2k}$; then $B$ is an open nbhd of $0$ in the domain of $h$, and $h[B]=\{0\}\cup\bigcup_{k\ge n}I_k\subseteq(-x,x)$.
We’ve now shown that $h$ is continuous at every point of its domain and therefore is continuous.
