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.

Show that the open unit disc is open in the product topology on $\mathbb{R^2}$ . Show further that it is not of form $U \times V$ for any open subsets $U$ and $V$ of $\mathbb{R}$.

 Consider the space $\mathbb{R}_l \times \mathbb{R}$ with product topology, where $\mathbb{R}_l$ denotes  the real line with lower limit topology. The basis for the product topology consists of $\{(x,y) \in \mathbb{R^2} : a\le x<b ,c<y<d \}.$​​​​​​​

Let $X_1$ denote the topological space $\mathbb{R}$ with discrete topology and let $X_2$ be $\mathbb{R}$ with usual topology. Then the product topology $\Omega$ on $\mathbb{R} \times \mathbb{R}$ is nothing but the dictionary order topology on $\mathbb{R}^2$ . Since the basis for the product topology on $\mathbb{R} \times \mathbb{R}$ is given by $\{\{x_1\} \times (a,b):x_1,a,b \in \mathbb{R} \}$, any open set in the dictionary order topology is union of open sets in the product topology. We also note that the product topology $\Omega$ is finer than the usual topology on $\mathbb{R}^2$ . In fact, any basis element $(a,b) \times (c,d)$ of the usual topology can be  expressed as the union $\cup_{a<x<b}\{x\} \times \left(c,d \right)$  of open sets $\{x\} \times \left( c,d \right)$ in the product toplogy   $\Omega$ .

Any interval $(x,y)$ in a simply ordered set is convex: If $a,b \in (x,y)$ then $x<a<b<y$ and hence $(a,b) \subseteq (x,y)$.

Consider the subset [0, 1]×[0, 1] of $\mathbb{R^2}$ with dictionary order. Then 0 × 0, 1 × 1 belong to [0, 1] × [0, 1], however, $\frac{1}{2} \times 2 \in (0 \times 0,1 \times1)$does not belong to [0, 1] × [0, 1]. The first quadrant is not a convex subset of $\mathbb{R^2}$ . However, the right half plane is convex.

If $X$ is Hausdorff then show that the complement of any finite set is open.

Consider the real line $\mathbb{R}$ with usual topology. Then the set $\mathbb{Q}$ of rationals is not closed. This follows since any neighbourhood of an irrational number contains rationals. If we enumerate $\mathbb{Q}$ as a sequence $\{r_n\}$ then$\mathbb{Q}=\cup_n\{r_n\}$, which shows that countable union of closed sets need not be closed.

Show that $X$ is Hausdorff iff the diagonal $\Delta :=\{(x,x) \in X \times X \}$ is closed in $X \times X$.

Show that every topological space with order topology is Hausdorff.