Study the relationship between two topologies

Question

Someone could help me solve the following statement:

Let $T'$ be the topology generated by $B = \{(a,b): a,b ∈ \Bbb R -\Bbb Q, a<b\}$, study the relationship between the topologies $T'$ and the usual topology $Tu$ in $\Bbb R$.

Can someone tell me if the next resolution is correct?

Since (a, b) ∈ Tu : a, b ∈ R-Q you have that T' ⊂ Tu and as (c, d) ∉ T': (c, d) ∈ R you have to Tu ≠ T '

Answer 1

I assume you define the usual topology by means of open balls: $U\subseteq \Bbb R$ is open iff for each $x_0\in U$, there exists $r>0$ such that $B_r(x_0):=\{\,x\in\Bbb R:|x-x_0<r\,\}\subseteq U$.

Each $(a,b)\in B$ is $T_u$-open because it is the open ball of radius $r:=\frac{b-a}2>0$ around $\frac{a+b}2$. Then also every $T'$-open set is $T_u$-open.

On the other hand, let $U$ be $T_u$-open. Given $x_0\in U$, there exists $r>0$ with $B_r(x_0)\subseteq U$. the non-empty open interval $(x_0-r,x_0)$ is not a subset of $\Bbb Q$, hence there exists $a\in\Bbb R-\Bbb Q$ with $x_0-r<a<x_0$. Similarly, we find $b\in\Bbb R-\Bbb Q$ with $x_0<a<x_0+r$. Then $(a,b)\in B$ and $x_0\in (a,b)\subseteq U$. We concldue that $U$ is the union of elements of $B$, hence is $T'$-open.

In summary, $T'=T_u$.

Answer 2

As all members of $B$ are open intervals, and so members of $\mathcal{T}_u$, $\mathcal{T}' \subseteq \mathcal{T}_u$.

On the other hand, if $(a,b)$ is any real open interval, find a sequence of irrationals $a_n$ such that $a_n \in (a,b)$ for all $n$ and $a_n \to a$. Likewise find $b_n \in (a,b)$, all irrational, such that $b_n \to b$. Then

$$(a,b) = \bigcup_n (a_n, b_n)$$ and so $(a,b) \in \mathcal{T}'$. So $\mathcal{T}_u \subseteq \mathcal{T}'$, showing the topologies to be the same.