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.

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.