Comparing topologies in plane I am learning Munkres's Topology on my own. I have some confusions.
Let $I = [0,1]$. Consider the following topologies
I want to prove the followings:
Let, $\mathbb{R}_d$ is $\mathbb{R}$ equipped with discrete topology. Then,

*

*The product topology on $I \times I$  is strictly coarser than the topology inherits as a subspace of $\mathbb{R}_d \times \mathbb{R}$.


*The dictionary order topology on   $I \times I$ is strictly coarser than the topology inherits as a subspace of $\mathbb{R}_d \times \mathbb{R}$.
My attempt for 1. : A basis element of product topology on $I \times I$ is $$((a, b) \cap [0, 1]) \times ((c, d) \cap [0, 1]) $$ and a basis element of the topology inherits as a subspace of $\mathbb{R}_d \times \mathbb{R}$ is $$\{x\} \times ((c, d) \cap [0, 1])$$. We see that $$((a, b) \cap [0, 1]) \times ((c, d) \cap [0, 1]) \ = \bigcup_{x \in (a, b) \cap [0, 1]} \{x\} \times ((c, d) \cap [0, 1]) $$
Hence, open in $\mathbb{R}_d \times \mathbb{R}$. For strictness, consider $\{\frac{1}{2}\} \times (\frac{1}{3}, \frac{2}{3})$. Clearly, $\{\frac{1}{2}\} \times (\frac{1}{3}, \frac{2}{3})= \{\frac{1}{2}\} \times ((\frac{1}{3}, \frac{2}{3})\cap [0,1])$., so it is open in the subspace topology, but not open in the product topology.
Am I correct so far? Please let me know if you find any wrong.
But, I can not prove 2.  Till now I have attempted for 2. :
A basis element of the order topology is $(a \times b, c\times d)$. I have written that
$$(a \times b, c\times d) = \{a\} \times (b,1) \cup \bigcup_{a<x<c} (\{x\} \times (0,1)) \cup \{c\} \times (0,d)$$ Is the above correct?
Please correct me.
 A: For 1: $\Bbb R_d \times \Bbb R$ contains all sets that are product open in $\Bbb R \times \Bbb R$ and so on the subset $I \times I$ we have the same relation, and the product topology on $I \times I$ has the subspace topology wrt the latter. The inclusion is strict because $\{0\} \times I$ is not open in the product topology on $I \times I$ but is open in the subspace topology from $\Bbb R_d \times \Bbb R$, e.g.
For 2: a basic neighbourhood of $0 \times 0$ in $I \times I$ (order top.) is $[0 \times 0, 0 \times b) = (\{0\} \times (-1,b)) \cap (I \times I)$. We can write similar equations for other open intervals or sets of the form $(1\times a, 1 \times 1]$, showing that the order topology on $I \times I$ is coarser than the inherited topology from $\Bbb R_d \times \Bbb R$. Strictness can be witnessed as the open set $\{\frac12\} \times I$ (inherited from $\Bbb R_d \times \Bbb R$) is not open in the order topology as $\frac12 \times 0$ is not an interior point of it.
A: Expanding on the comment,  the first one looks fine.  The second only works as a basis for when $a<c$,  when $a=c$ then your set looks like $\{a\}\times (b,d)$ (along with the whole space as needed for a topology.
These are clearly open in the $\mathbb{R}_d \times \mathbb{R}$ topology,  as all singletons times open intervals are open, as those are the basis elements for each topology.
The converse is a bit trickier, MOST singletons times an interval are open in the dictionary order.  $\{0\}\times [0,1]$ and $\{1\}\times [0,1]$ are going to be your missing ones, as the only way to generate a singleton out of the dictionary order is if $a=c$,  and you can't get the whole interval $[0,1]$ with any choice of $b,d$, the best you can get is $b=0,d=1$ where you get $(0,1)$
