(a) Apply lemma 13.2 to show that the countable collection $\mathcal{B}=\{(a,b)| a\lt b,$ $a,b\in \Bbb{Q}\}$ is a basis that generates the standard topology on $\Bbb{R}$.
(b) Show that the collection $\mathcal{C}=\{[a,b)| a\lt b$, $a,b\in \Bbb{Q}\}$ is a basis that generates a topology different from the lower limit topology on $\Bbb{R}$.
My attempt: (a) By theorem 2.13 of Baby Rudin, $\mathcal{B}$ is countable. First we show $\mathcal{B}$ is basis. $\forall x\in \Bbb{R}, \exists p,q\in \Bbb{Q}$ such that $p\lt x \lt q$. So $x\in (p,q)$. Let $x\in (a,b)\cap (c,d)$; $a,b,c,d\in \Bbb{Q}$, $a\lt b$, $c\lt d$. WLOG $a\lt d$. Since we have finite number of points(in this case four), we can make only finite number of cases. Each point(out of a,b,c,d) is related with the other by order relation. If $b\leq c$, then $(a,b)\cap (c,d)=\phi$. So $c\lt b$. Case (1) $c\in (a,b)$/$a\lt c \lt b$ and $d\gt b$. Case (2) $c\leq a$ and $d\gt b$. Case(3) $a\lt c \lt b$ and $a\lt d \leq b$. Case(4) $c\leq a$ and $a\lt d\leq b$. It is easy to check, in each case $(a,b)\cap (c,d)\in \mathcal{B}$. So $\mathcal{B}$ satisfy second condition for basis. Thus $\mathcal{B}$ is basis.
First condition of lemma 13.2, $\mathcal{B} \subseteq \mathcal{B}_{s}$(basis of standard topology)$\subseteq \mathcal{T}$. So $\mathcal{B}\subseteq \mathcal{T}$. Let $U\in \mathcal{T}$ and $x\in U$. $\exists(m,n)\in \mathcal{B}_{s}$ such that $x\in (a,b)\subseteq U$. $m,n \in \Bbb{R}$ and $n\lt m$. By theorem 1.20 of Baby Rudin, $\exists p,q \in \Bbb{Q}$ such that $a\lt p \lt x \lt q \lt b$. So $x\in (p,q)\subset (a,b)$. So, for $x\in U$, take $(p,q)\in \mathcal{B}$ such that $x\in (p,q)\subset U$. This completes the proof. Is this proof correct?
(b) It is easy check $\mathcal{C}$ is a basis. To show two topology are not equal. Try to show $\mathcal{T}\nsubseteq \mathcal{T}^\prime$(generated by $\mathcal{C}$). Show hypothesis of lemma 13.3 don’t hold. And that’s how you solve part (b).
Best Answer
For a) you just apply lemma 13.2 directly:
Let $O$ be open in $\Bbb R$ and $x \in O$. Munkres has defined the standard topology by the base $\{(a,b)\mid a,b \in \Bbb R\}$. Ergo, for some $a < b$ we have $x \in (a,b) \subseteq O$. So $a < x < b$ and we can indeed (by the so-called order-denseness of $\Bbb Q$ in $\Bbb R$) find rationals $q_1, q_2$ so that $a < q_1 < x$ and $x < q_2 < b$. It then follows that $(q_1,q_2) \in \mathcal B$ and $x \in (q_1, q_2) \subseteq (a,b) \subseteq O$. Lemma 13.2 now tells us directly that $\mathcal B$ is a base for $\Bbb R$ in its standard topology. No need for further checks: just note that the condition is satisfied and all sets in $\mathcal B$ are themselves already open in the standard topology.
summary: your proof was in essence OK, but had redundant elements and did more work than needed. I like my proofs neat and tidy. Matter of taste (and laziness). Also easier to check, which is important, IMO.
For b) the proof that $\mathcal C$ is a base for some topology is very easy: covering is trivial and $\mathcal C$ is closed under (non-empty) binary intersection, so the two base conditions are easy (see the part where Munkres checks the bases for the lower limit topology and and the standard topology in section 13)
Members of $\mathcal C$ are not standard open but are lower-limit open. But it is nevertheless not a base for it because we cannot find a member $C$ of $\mathcal C$ such that $\sqrt{2} \in C \subseteq [\sqrt{2},3)$ so the open set $[\sqrt{2},3)$ is not a union of members of $\mathcal C$.. So this base generates a topology strictly smaller then that of $\Bbb R_l$.