$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\T}{\mathcal{T}}$

This is textbook problem in Munkres' Topology. I know there are plenty of solutions online, but none of them are exactly the same as mine. I would really appreciate if anyone can review my solution to this problem and point out any mistakes and stuff that can be improved.

Problem: Let$I=[0,1]$. Compare the product topology on $I\times I$, the dictionary order topology on $I\times I$, and the topology $I\times I$ inherits as a subspace of $\R\times \R$ in the dictionary order topology.

Attempt: Let $\T_{1}$ denote the product topology on $I\times I$, $\T_{2}$ the dictionary topology on $I\times I$, $\T_{3}$ the topology inherits as a subspace of $\R\times\R$ in the dictionary topology and let $\mathcal{B}_{1},\mathcal{B}_{2}$ and $\mathcal{B}_{3}$ denote the corresponding set of basis. Note that for an arbitrary element $B\in\mathcal{B}_{2}$, we have \begin{align} B & =\left\langle \left(a,b\right),\left(c,d\right)\right\rangle \\ & =\left\{ a\right\} \times(b,1]\cup\left(a,c\right)\times[0,1]\cup c\times[0,d)\\ & =\left(\left\{ a\right\} \times\left(b,2\right)\cap I\times I\right)\cup\left(\left(a,c\right)\times\left(-1,2\right)\cap I\times I\right)\cup\left(\left\{ c\right\} \times\left(-1,d\right)\cap I\times I\right)\\ & =\left\{ \left[\left\{ a\right\} \times\left(b,2\right)\right]\cup\left[\left(a,c\right)\cup\left(-1,2\right)\right]\cup\left[\left\{ c\right\} \times\left(-1,d\right)\right]\right\} \cap\left(I\times I\right) \end{align} and notice that $\left[\left\{ a\right\} \times\left(b,2\right)\right]\cup\left[\left(a,c\right)\cup\left(-1,2\right)\right]\cup\left[\left\{ c\right\} \times\left(-1,d\right)\right]$ is open in the dictionary order topology of $\R\times\R$. Therefore, we have $B\in\mathcal{B}_{3}$, which implies $\mathcal{T}_{2}\subset\mathcal{T}_{3}.$ Furthermore, we claim that $\mathcal{T}_{3}$ in fact strictly finer than $\T_{2}$. Let $A=\left\{ 1/2\right\} \times(1/2,1].$ Since we can rewrite $A=\left[\left\{ 1/2\right\} \times\left(1/2,3/2\right)\right]\cap\left(I\times I\right)$, $A\in\T_{3}.$ However, suppose $A\in\mathcal{T}_{2};$ then $\exists\left\{ B_{i}\right\} _{i\in I}\in\mathcal{B}_{2}$ such that $A=\cup_{i\in I}B_{i}$. Since $\left(1/2,1\right)\in A$, $\exists B_{i}=\left\langle \left(a_{i},b_{i}\right),\left(c_{i},d_{i}\right)\right\rangle $ such that $\left(1/2,1\right)\in B_{i}$. Note that we have $a_{i}=c_{i}=1/2$ since otherwise we would have $\cup_{i}B_{i}\neq A$. But $\left(1/2,1\right)\in B_{i}$ implies $\left(1/2,a_{i}\right)\prec\left(1/2,1\right)\prec\left(1/2,d_{i}\right)$, which in turn implies $1<d\leq1$, a contradiction. Therefore, $A\notin\T_{2}.$

Next, we claim that $\mathcal{T}_{1}\subset\T_{3}.$ Indeed, by Problem 17 (which says the $\R_d\times\R$ is equal to $\R\times\R$ in dictionary order), we have $\T_{3}=\left(\R_{d}\times\R\right)\cap I\times I\supset\left(\R\times\R\right)\cap I\times I=\T_{1}.$ We also have that $\T_{1}$ and $\T_{2}$ are incomparable. To see this, we first note that $\left\{ 1/2\right\} \times\left(1/3,2/3\right)=\left\langle \left(1/2,1/3\right),\left(1/2,2/3\right)\right\rangle \in\T_{3}$. However, any open set in $\T_{2}$ containing $\left\{ 1/2,1/2\right\} $ will contain $\left\{ 1/2+\varepsilon,1/2\right\} $ for some $\varepsilon>0$. Thus $\T_{2}\nsubseteq\T_{1}$. For this other direction, we have $I\times(0,1]=\left(-1\times2\right)\times(0,2)\cap I\times I$ which is in $\T_{1}$ but not in $\T_{2}.$ Indeed, suppose $I\times(0,1]\in\T_{2},$ then $(0,1)\in\mathcal{B}_{i}$ for some $B_{i}=\left\langle (a_{i},b_{i}),(c_{i},d_{i})\right\rangle \in\mathcal{B}_{2},$ which implies $\left(0,1\right)\prec\left(c_{i},d_{i}\right)$, which indicates $c_{i}>0$. Then we have $\left\langle \left(0,1\right),\left(c_{i},d_{i}\right)\right\rangle =\left[\left(0,c_{i}\right)\times\left[0,1\right]\right]\cup\left[\left\{ c_{i}\right\} \times[0,d_{i})\right]$ containing $\left(c_{i},0\right)$ which is not in $I\times(0,1].$ Hence, $\T_{2}$ and $\T_{1}$ are incomparable.

  • $\begingroup$ I would suggest that you add figures to help to visualize the relationship among these squares and their subsets. For example $I\times I$ is a square. $I\times I$ with point $P(t)=(0,t)$ glue to point $Q(t)=(1,1-t)$ $(0\leq t\leq 1)$ is a Mobius strip, etc. Thus people can understand your problem and solution faster. $\endgroup$ – mike Sep 11 '16 at 3:31
  • $\begingroup$ Another way to show that $\tau_2 \not \subset \tau_1$ is that $\tau_1$ has a countable base, which it inherits from a countable base for the usual topology on $\mathbb R^2.$ So if $F$ is a family of pairwise disjoint $\tau_1$-open sets then $F$ is countable. But $ \{ \; \{x\} \times (1/3,2/3) :x\in [0,1]\}$ is an uncountable pairwise disjoint $\tau_2$-open family, $\endgroup$ – DanielWainfleet Jan 13 '17 at 3:31

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