f from sorgenfrey line to R which is continuous Suppose $f:X \mapsto \mathbb{R}$, such that  $f(x)= \cases{ \frac{1}{x} & , x $\neq$ 0 \\
                       0 & , x=0 }$
where $X$ is the topological subspace $[0,1]$ of the Sorgenfrey line.
Show that $f$ is continuous.
I have to show that $f$ inverses intervals of the form $(a,b)$ which are basic open sets, to open sets in the Sorgenfrey line.
So I take cases $0<a<b$ and $f^{-1}((a,b))= (\frac{1}{b}, \frac{1}{a})=\cup_{n=1}^\infty (\frac{1}{b},\frac{1}{a}-\frac{1}{n}]$ which belongs to Sorgenfrey topology $a<b<0$ which gives the empty set and $a<0<b$  and $f^{-1}((a,b))=(\frac{1}{b},1] \cup \{0\}=(\frac{1}{b},1] \cup (\cup_{n=1}^{\infty} (-\frac{1}{n}, \frac{1}{n}])$ which also belongs to Sorgenfrey topology.
Is this approach correct?
 A: If you look at it locally: $f$ is continuous at $x=0$ because $\{0\} =[0,1]\cap (-1,0]$ is an open set in $[0,1]$ and so in the local continuity definition or continuity at $x=p$:

For all open neighbourhoods $O$ of $f(p)$ there exists some neighbourhood $U$ of $p$ such that $f[U] \subseteq O$.

applied to $p=0$ we can just take $U = \{0\}$ for any $O$.
Then for $p>0$ we have that $f$ is already continuous as a map from standard reals to standard reals, and we can use that: 
Suppose $O$ is an open neighbourhood of $f(p)$, then by standard continuity there is some relativised open interval $U=(p-\delta, p+\delta) \cap [0,1]$ such that $f[U] \subseteq O$ and this is already a valid neighbourhood for $p$ in $X$ as well. ($f$ is already continuous at $p$ in a smaller/coarser topology on the domain, or use $(p-\delta,p] \cap [0,1]$ if you want to use the basic open sets directly). So $f$ is continuous at all points (only $0$ is a special case) and so $f$ is continuous.
A: It is almost correct. The only problem is that the last equality is false. But$$\left(\frac1b,1\right]\cup\{0\}=\left(\left(\frac1b,1\right]\cup(-1,0]\right)\cap[0,1],$$which is an open set.
