questions about subspace topology, order topology and convex subset

Question

I am learning Munkres' s topology, and have some questions about subspace topology, order topology and convex subset.

I have read 3 similar questions here [ref]:One question on subspace topology and order topology, Subspace topology and order topology, and Subspace topology and order topology (2).

Example 3 in section 16, let $I=[0,1]$, the set $\{1/2\}\times (1/2,1]$ is open in $I\times I$ in the subspace topology, but not in the order topology.
Proof: $\{1/2\}\times (1/2,1]= \{1/2\} \times(1/2,2) \cap[0,1] \times [0,1]$, and $\{1/2\} \times(1/2,2)$ is open in $R\times R$, so it is open in subspace topology.

The definition. Let X be a set with a simple order relation; assume X has more than one element. Let $\mathcal{B}$ be the collection of all sets of the following types:
(1) All open intervals (a,b) in X.
(2) All intervals of the form $[a_{0},b)$, where $a_{0}$ is the smallest element (if any) of X.
(3) All intervals of the form $(a,b_{0}]$, where $b_{0}$ is the largest element (if any) of X.
The collection $\mathcal{B}$ is a basis for a topology on X, which is called order topology.

the point $1/2 \times 1$ is not the largest element of $I\times I$, so $\{1/2\}\times (1/2,1]$ is not open in order topology in $I\times I$.
Q.E.D

Question:
(1). $\{1\}\times (1/2,1]$ is open in subspace topology and order topology in $I\times I$, correct? $(1\times 1/2, 1\times 1]$ is the interval of the form $(a,b_{0}]$, correct?

The definition. Given an ordered set X, let us say that a subset Y of X is convex in X if for each pair of points $a<b$ of Y, the entire interval $(a,b)$ of points of X lies in Y. Note that intervals and rays in X are convex in X.

(2). $[0,1]$ is a convex subset in R. But why $I\times I$ is not convex in $R\times R$ ? How to prove this from the definition?
(3). The convex subset in $I\times I$ ?
(4). The intervals and rays in $R\times R$ ?

Answer 1

You only demonstrated that $\{\frac{1}{2}\} \times (\frac{1}{2},1]$ is not a basic element (you still need to show it's not an open interval though). But open sets are not the only open subsets, all unions of them are, so to show it is indeed not open in the order topology (as $(\frac{1}{2},1)$ is not the maximal element as you say): assume that there is some open interval (these form a local base for that point, after all), say $((a,b)(c,d))$ such that $$(\frac{1}{2}, 1) \in ((a,b), (c,d)) \subseteq \{\frac{1}{2}\} \times (\frac{1}{2}, 1]$$ an deduce a contradiction. (hint: we must have $c > \frac{1}{2}$, and find some point between $(\frac{1}{2},1)$ and $(c,d)$ that is not in the right hand set). So the set $\frac{1}{2} \times (\frac{1}{2}, 1]$ shows that the subspace topolgoy with respect to ordered $\mathbb{R}^2$ is not the same as the order topology from the restricted order, which is the whole point fo the example. If $(X,<)$ is an ordered space and $C \subseteq X$ is (order-)convex then these two topologies do coincide on $C$, hence the question on convexity of $I \times I$.

So as to 1.: The set $\{1\} \times (\frac{1}{2},1]$ is an open-closed interval with the maximum as upper limit, so indeed open in the order topology. It is also open as a subspace from $\mathbb{R} \times \mathbb{R}$ in that same order. As to 2.: $(0,1) < (0,2) < (1,0)$ and $(0,2) \notin I \times I$,s o $I \times I$ is not convex. I'll leave you to think about 3. and 4. Draw some pictures.