choose the correct option about comparison of three topology?

let $$T_u$$and $$T_d$$ denote the usual topology and the discrete topology on $$\mathbb{R}$$ consider the following three topology

$$T_1$$ = usual topology on $$\mathbb{R}^2= \mathbb{R} \times \mathbb{R}$$

$$T_2$$ =Topology generated by the basis $$\{U \times V : U \in T_d, V\in T_u\}$$ on $$\mathbb{R} \times \mathbb{R}$$

$$T_3$$= Dictionary order topology on $$\mathbb{R} \times \mathbb{R}$$

then , choose the correct option

a) $$T_3 \subsetneq T_1 \subseteq T_2$$

b) $$T_1 \subsetneq T_2 \subsetneq T_3$$

c)$$T_3 \subseteq T_2 \subsetneq T_1$$

d)$$T_1 \subsetneq T_2 =T_3$$

This is the orginal problem

My attempts : i know that discrete topology is the finest(strongest) topology on universe ,so option a) is correct

honestly im not able to tackle this question so, i just put my logics that is option a) will be correct

any Hints/solution will be appreciated

It is best to consider bases for all three topologies. Notice that a sub-basis for $$\mathcal{T}_u$$ has a basis of the form $$\mathcal{S}_u:= \{(-\infty,a)\}_{a\in \mathbb{R}}\cup\{ (b,\infty) \}_{b\in \mathbb{R}}$$, which means that $$\mathcal{B}_u=\Big\{ \underset{i=1}{\overset{n}{\cap}} S_i: n\in \mathbb{N}, S_i\in \mathcal{S}_u \; \text{for all} \; i \Big\}$$ is a basis of $$\mathcal{T}_u$$. Since all $$U\in \mathcal{T}_u$$ can be written as a union of sets in the basis, we can see that:

$$\mathcal{S}_2:=\{(-\infty,a)\times\{c\}\}_{a,c\in \mathbb{R}}\cup\{ (b,\infty) \times\{d\} \}_{b,d\in \mathbb{R}}$$ is a sub-basis of $$\mathcal{T}_2$$.

The usual topology on $$\mathbb{R}_2$$ is generated by the basis $$\mathcal{B}_1= \{ (a,b)\times (c,d) \}_{a,b,c,d\in \mathbb{R}}$$ and thus by the sub-basis:

$$\mathcal{S}_1:=\{(-\infty,a)\times(-\infty,c)\}_{a,c\in \mathbb{R}}\cup\{ (b,\infty) \times(-\infty,c) \}_{b,c\in \mathbb{R}}\cup \{(b,\infty)\times(d,\infty)\}_{b,d\in \mathbb{R}}\cup \{(-\infty,a)\times(d,\infty)\}_{a,c\in \mathbb{R}}$$

Finally $$\mathcal{T}_3$$ is generated by the sub-basis:

$$\mathcal{S}_3:= \Big\{ (c,d) :(c,d)< (a,b) \Big\}_{a,b\in \mathbb{R}} \cup \Big\{ (c,d) :(c,d)> (a,b) \Big\}_{a,b\in \mathbb{R}}$$

You can notice that:

$$\Big\{ (c,d) :(c,d)< (a,b) \Big\}_{a,b\in \mathbb{R}}=\Big\{ (c,d) :cd \Big\}=\Big\{ \{a\}\times (-\infty, b) \Big\}_{a,b\in \mathbb{R}}\cup \Big\{ (-\infty,a)\times \mathbb{R} \Big\}_{a\in \mathbb{R}}$$

Similarly:

$$\Big\{ (c,d) :(c,d)> (a,b) \Big\}_{a,b\in \mathbb{R}}=\Big\{ \{a\}\times (b,\infty) \Big\}_{a,b\in \mathbb{R}}\cup \Big\{ (a,\infty)\times \mathbb{R} \Big\}_{a\in \mathbb{R}}$$

Which shows that $$\mathcal{S}_3 \supseteq \mathcal{S}_2 \supsetneq \mathcal{S}_1$$ and thus $$\mathcal{T}_3 \supseteq \mathcal{T}_2 \supsetneq \mathcal{T}_1$$. Since for all $$a\in \mathbb{R}$$, we know that $$(a,\infty)\times \mathbb{R},(-\infty,a)\times \mathbb{R}\in \mathcal{T}_1$$, you can conclude that $$\mathcal{S}_3 =\mathcal{S}_2$$. And (D) is the correct answer.

• thanks u @Keen -ameteur – jasmine Oct 4 '18 at 17:59