Given basis of an ordered square seems to be only open intervals

Question

The basis given here What is the Basis of an Ordered Square? seems to be different from the basis given in Topology by James Munkres, if any:

From Chapter 14

From Chapter 16

It seems like the topology that defines the ordered square will be generated by a basis of open intervals. I believe I will not get the answers to the following exercise (also in this question) if my basis consists of only open intervals because of the issues with $0 \times 0$ and $1 \times 1$.

Which of the following explains what's going on?

1. Munkres implies dictionary order topology that forms the ordered square is generated by a basis of open intervals, and Munkes is right.

2. Munkres implies dictionary order topology that forms the ordered square is generated by a basis of open intervals, and Munkes is wrong.

3. Munkres does not imply that the dictionary order topology that forms the ordered square is generated by a basis of open intervals. Instead (insert explanation here).

4. Munkres actually does not well-define the dictionary order topology that forms the ordered square.

5. Other

Answer 1

The ordered square is the set $[0,1] \times [0,1]$ in the order topology from the order that the square inherits from $\mathbb{R} \times \mathbb{R}$ with the lexicographic order (the lexicographically ordered plane).

Munkres already spent some time warning the reader that this is not the same topology that $[0,1] \times [0,1]$ gets as a subspace of the ordered plane.

In general, if $(X,<)$ is an ordered set with order topology $\mathcal{T}_<$ and $A \subseteq X$ we can consider two topologies on $A$: the subspace topology $\mathcal{T}_<|_A = \{O \cap A: O \in \mathcal{T}_<\}$ that $A$ inherits from $(X,\mathcal{T}_X)$ and the order topology that $A$ gets when we restrict the order first to $A$ and then generate the topology from that restricted order.

And these topologies concide when $A$ is order-convex: $\forall a_1 \in A, a_2 \in A: \forall b \in X: (a_1 < b < a_2 ) \to b \in A$.

Even in the reals you can see this phenomenon: Take $A = [0,1) \cup \{2\}$. As a subspace of the reals (ints order topology), $2$ is an isolated point, i.e. $\{2\}$ is open, while in the restricted order topology $2$ lies "just above" $[0,1)$ and in the restricted order topology the set is homeomorphic to $[0,1]$. Note that $A$ is not order convex.

Also $A = [0,1]\times [0,1]$ is not order convex in the ordered plane: e.g. $(0,1), (1, 0)$ are both in $A$, while $(0,2)$ which lies between them lexicographically is not.

Munkres does take the square in its (restricted) order topology, not the subspace topology, so the subbase for its topology (as in any order topology) is the set of all open segments both up and down, and the derived base has the two special cases involving $(0,0) = \min(A)$ and $(1,1) = \max(A)$. So not only open intervals, also sets like $[(0,0), (0,t))$ and $((1,t), (1,1)]$ are basic open.

Answer 2

Number Three (3).

We know

• that standard topology on $\mathbb R$ is the order topology on $\mathbb R$ and

• that the standard topology on $\mathbb R^2$ is not the order topology on $\mathbb R^2$.

But actually the dictionary topology on $\mathbb R^2$ is the order topology on $\mathbb R^2$:

Observe that Munkres says the dictionary topology on $\mathbb R^2$ aka order topology on $\mathbb R^2$ can be generated by a basis of open intervals because $\mathbb R^2$ has no $\max$ or $\min$. Thus, there is no reason to conclude that a topology that is on a subset of $\mathbb R^2$ with a $\max$ or $\min$ and that is inherited from the dictionary topology of $\mathbb R^2$, again, aka order topology on $\mathbb R^2$ (and that is not inherited as a subspace) can be generated by a basis of open intervals. Instead, the subset's dictionary topology's basis goes back to the definition of order topology:

The basis does not consist of only open intervals.