Let
$B = \{ \mathbb{R} \} \cup \{ (a,b) \cap\mathbb {Q} \ ,\ a\lt b \ ,\ a,b \in\mathbb{Q}\}$
Thus, a set $V \in B$ if it is either equal to $\mathbb{R}$ or if it is in the intersection of $\mathbb{Q}$ with an open interval with rational endpoints.
a) Show that $B$ forms a base for a topology $T$ on $\mathbb{R}$.
b) Show that this topology does not contain nor be contained in the usual topology of $\mathbb{R}$.
c) Is $(\mathbb{R}, T)$ compact? Is it separable? Is it connected?
I think have solved a)
All irrational numbers are contained in the set $\mathbb{R}$ and all the other numbers are contained in some intersection of an open set with $\mathbb{Q}$. So each point $x$, is contained in some set $v \in B$. Also if a point $x$ belongs to the intersection of two base sets then I can find some set containing $x$ that is contained in that intersection.
Best Answer
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.