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 problementer image description here

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

thanks in advance


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:

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


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

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.

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

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