Showing that the topologies are incomparable (?) Consider the following topologies on $X = \Bbb R$

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*The topology generated by the basis $B_1 = \left \{(a, b) | a, b ∈ \Bbb Q \text{ with } a < b \right \}$

*The topology generated by the basis $B_2 = \left \{(−a, a) | a ∈ \Bbb R \text{ with } a > 0 \right \}$.

I think these two topologies should be incomparable since there is no intersection between the two bases whatsoever. Am I missing something?
 A: For any $a \in \Bbb R$ with $a \gt 0$, let $\{q_n \mid n \in \Bbb N \}$ be an increasing sequence of positive rationals such that $\lim_{n \to \infty} q_n = a$.  Then $\bigcup (-q_n, q_n) = (-a, a)$ so every element of $B_2$ is open in the topology generated by $B_1$ and the two topologies are comparable.
Note also that every element of $B_2$ contains $0$ so every non-empty open set in the topology generated by $B_2$ contains $0$.  Consequently, there are sets in $B_1$ that are not in the topology generated by $B_2$.  Therefore, the topology generated by $B_2$ is coarser than the topology generated by $B_1$.
A: It is easy to show since $\Bbb Q $ is dense in $\Bbb R$ that I can approximate every real number with a rational number taking the truncated decimal development. Let show $\tau_e$ the euclidian topology is the same as $\tau_1$. Let $\tau_e=${$(a,b)|a,b\in \Bbb R,a<b$} since $\tau_e\ge\tau_1$ is trivial, just take an $A=(a,b)$ open set for $\tau_e$.
Since $A= \bigcup_{a<c,d<b}(c,d)$ with $c,d\in\Bbb Q $ so $A\in\tau_1$,since is an arbitrary union of open set of $\tau_1$. This shows that $\tau_e\le\tau_1$
This prove that $\tau_1=\tau_e\ge\tau_2$. Since for example the open set $(\sqrt 3,\sqrt2)$ is open for $ \tau_1 \land\tau_e $ and not for $\tau_2$
A: Actually  $B_2\cup\{\Bbb{R}, \emptyset\}$ is a topology since it is closed under finite intersections & arbitrary unions. It is thus the topology $\mathscr{T}_2$ generated by $B_2$.
On the other hand —as already noted by Anne Bauval— the topology $\mathscr{T}_1$ generated by $B_1$ is the usual one, thus $\mathscr{T}_1\supseteq\mathscr{T_2}$.
Since we have (for example) $]0,1\mathclose[\in\mathscr{T}_1\smallsetminus\mathscr{T}_2$, the former inclusion is strict: $\mathscr{T}_1\supsetneq\mathscr{T_2}$.
