The topology generated by open intervals of rational numbers

Let $B = mathcal{I} cup mathcal{J}$ where $mathcal{I} = left{ mathbb{R} right}$, $mathcal{J} = left{ (a,b) cap mathbb{Q} | a<b, a,b in mathbb{Q} right}$. Suppose that $U,V in B$. If $U,V in mathcal{I}$, then clearly $U cap V in mathcal{I}$. If one of $U,V$ is in $mathcal{I}$ whereas the other is in $mathcal{J}$, then clearly $U cap V in mathcal{J}$. If both $U$ and $V$ are in $mathcal{J}$, then $U cap V in mathcal{J}$ from the basic properties of open intervals. Hence $B$ forms a basis.

Let $mathcal{T}_B$ be the topology generated by $B$ and $mathcal{T}$ be the usual topology. Since the interval $(0,1) in mathcal{T}$ contains irrational numbers $(0,1) not in mathcal{T}_B$. Therefore $mathcal{T}_B$ does not contains $mathcal{T}$. Since the interval $(0,1) cap mathbb{Q} in mathcal{T}_B$ is not open in the usual topology, we have $mathcal{T}$ does not contains $mathcal{T}_B$.

The space $X = (mathbb{R}, mathcal{T}_B)$ is compact. To see this let $mathcal{O}$ be an open cover of $X$. Since $sqrt{2} in U in mathcal{O}$ implies $U = mathbb{R}$, we have a finite subcover $I subseteq mathcal{O}$.

The space $X$ is separable. To see this, consider $D = left{ 0 right}$. For every irrational $r$, there is unique neighborhood $mathbb{R}$ showing that the closure of $D$ is just $X$. Hence $X$ is separable.

The space $X$ is connected. To see this, suppose that $U$ and $V$ are proper non empty open subsets of $X$. Then they cannot cover $X$ simply because they do not have any irrational numbers.