Prove that $f$ is $\tau_1 - \tau$ continuous on $\mathbb{R}$ We equip $\mathbb{R}$ with the standard topology $\tau$ and $\tau_1$ is topology generating by the basis $\mathcal{B}_1=\left\lbrace [a,b): a<b \right\rbrace$. The map $f:\mathbb{R} \rightarrow {\mathbb{R}}$ is given by $f(x)= \begin{cases} x, &x<0\\ x+1,& x \ge 0 \end{cases}$
Question 1: Prove that $f$ is $\tau_1 - \tau$ continuous on $\mathbb{R}$.
Question 2: Prove that $f$ is not $\tau-\tau_1$ continuous on $\mathbb{R}$.
With question 1, I have proved that for $x<0$, $f$ is identity map and $\tau_1 \supset \tau$, therefore $f$ is continuos, but how about the case $x \ge 0$?
With question 2, what is the solution of it ?
 A: In the topology $\tau_1$ (the Sorgenfrey line, or lower limit topology), both sets $A=\{x\in \Bbb R: x < 0\}$ and $B=\{x\in \Bbb R: x \ge 0\}$ are open, for $A$ because it's also $\tau$-open and $\tau \subseteq \tau_1$ and $B = \cup_{n=0}^\infty [n,n+1)$ is a union of basic open sets. 
Now $f\restriction_A$ is continuous (for the reason you stated) and $f\restriction_B$ is also continuous (either because it's the sum of two continuous functions, the identity and constant map, or because the inverse image of an open interval clearly is an open interval, etc.) 
And as $A$ and $B$ form a partition of open sets of $(X,\tau_1)$ and $f$ is continuous on both parts, $f$ is globally continuous. (A form of the pasting lemma.)
As to the same map under as a map from $(X,\tau) \to (X,\tau_1)$ it suffices to find a single open set $O$ in $\tau_1$ such that $f^{-1}[O]$ is not open. For this $O=[1,2)$, with $f^{-1}[O]= [0,1)$ and $[0,1) \notin \tau$ as $0$ is not an interior point.
An alternative: $x_n = -\frac1n \to 0$ in $\tau$ but $f(x_n)= -\frac{1}{n} \not \to f(0) =1$. As continuous functions must preserve convergence of sequences, this also shows the non-continuity.
