Show that the usual topology and co-countable topology on $\mathbb{R}$ are not comparable.

The collection $\{(a,b) \subseteq \mathbb{R}:a,b \in \mathbb{Q} \}$ is a basis for a topology on $\mathbb{R}$.

Show that collection of balls (with rational radii) in a metric space forms a basis.

(Arithmetic Progression Basis) Let $X$ be the set of positive integers and consider the collection $\mathbb{B}$ of all arithmetic progressions of positive integers. Then $\mathbb{B}$ is a basis. If $m \in X$ then $B:=\{m+(n-1)p\}$ contains m. Next consider two arithmetic progressions $B_1=\{a_1+(n-1)p_1\}$and $B_2=\{a_2+(n-1)p_2\}$containing an integer $m$. Then$B:=\{m+(n-1)(p)\}$ does the job for $p:=lcm\{p_1,p_2\}$.

Show that the topology $\Omega{_B}$ generated by the basis $\mathbb{B}:=\{(a,b) \subseteq \mathbb{R}:a,b \in \mathbb{Q}\}$ is the usual topology on $\mathbb{R}$.

The collection $\{[a,b) \subseteq \mathbb{R}:a,b \in\mathbb{R}\}$is a basis for a topology on $\mathbb{R}$. The topology generated by it is known as lower limit topology on $\mathbb{R}$.

Note that $\mathbb{B}:=\{p\} \cup \{\{p,q\}: q\in X,q\neq p \}$ is a basis. We check that the topology $\Omega_{B}$ generated by $\mathbb{B}$ is the VIP topology on $X$. Let $U$ be a subset of $X$ containing $p$. If $x \in U$ then choose $B=\{p\}$if $x=p$, and $B=\{p,x\}$ otherwise. Note further that if $p\notin U$then there is no $B \in \mathbb{B}$ such that $B \subseteq U$. This shows that $\Omega_{\mathbb{B}}$ is precisely the VIP topology on $X$.

Show that the topology generated by the basis $\mathbb{B}:={X}\cup\{\{q\}:q\in X,q\neq p \}$ is the outcast topology.

Show that the topological space $\mathbb{N}$ of positive numbers with topology generated by arithmetic progression basis is Hausdorff.

Let $X:=\mathbb{R}$. Consider the pairs of bases:

(1)$\mathbb{B_1}:=\{(a,b) \subseteq \mathbb{R} :a,b \in \mathbb{R} \}$ and $\mathbb{B_2}:=\{(a,b) \subseteq \mathbb{R} :a,b \in \mathbb{Q} \}$.

(2) $\mathbb{B_1}:=\{[a,b) \subseteq \mathbb{R} :a,b \in \mathbb{R} \}$ and $\mathbb{B_2}:=\{[a,b) \subseteq \mathbb{R} :a,b \in \mathbb{Q} \}$.

(3) $\mathbb{B_1}:=\{[a,b) \subseteq \mathbb{R} :a,b \in \mathbb{R} \}$ and $\mathbb{B_2}:=\{(a,b) \subseteq \mathbb{R} :a,b \in \mathbb{R} \}$.

Do they generate comparable topologies ? If so then do they generate the same topology ?

([1, H. F¨urstenberg]) Consider $\mathbb{N}$ with the arithmetic progression topology. Verify the following:

(1) For a prime number p, the basis element {np : n ≥ 1} is closed.

(2) There are infinitely prime numbers.