How to prove ordered square is compact Let $I = [0,1]$ and let $I^2_{0}$ be the ordered square obtain by topologizing $I^2$ using the dictionary ordering on $I^2$
Munkre in his book "Topology" writes in Example 5 on pg 193 that $I^2_0$ is compact, but how do I prove it?
 A: Theorem: A linearly ordered space $(X,<)$ is compact (in the order topology) iff every $A \subseteq X$ has a supremum (least upper bound). 
(Note that $\sup(\emptyset)$ (if it exists) must be a minumum for $X$, so $\min(X)$ exists in particular, as does $\max(X) = \sup(X)$ as well).
Some notation: the standard subbase for the order topology is given by all sets $L_x, x \in X$, where $L_p = \{x \in X: x < p \}$ together with all subsets $U_x, x \in X$, where $U_p = \{x \in X: x > p \}$, the upper and lower sets of $X$. 
Note that no finite family of sets of the form $L_x$ can cover $X$ (consider the maximum of the finitely many $x$ that are used, which is then never covered) and similarly no finite family of sets of the form $U_x$ can cover $X$, by considering the minimum of the used $x$ instead.
Suppose first that $X$ is compact. Let $A \subseteq X$ be any subset, and define $u(A)$, the set of upperbounds of $A$, as $u(A) = \{x: \forall a \in A: a \le x \}$. Suppose that $A$ has no supremum in $X$. By definition $\sup(A) = \min(u(A))$.
Now define $\mathcal{U} = \{L_x: x \in A\} \cup \{U_u: u \in u(A) \}$. 
Suppose $x \in X$ is in none of the $L_x$, $x \in A$. This means $x$ is not below any element of $A$, so $x$ is an upperbound of $A$, hence $x \in u(A)$, which means there is some $u \in u(A)$ such that $u < x$ (as $x$ cannot be $\min(u(A)$) , so $x \in U_u$ for that $u$. So either $x$ is in some $L_a$ or it must be in some $U_u$, and this shows that $\mathcal{U}$ is an open cover of $X$. So finitely many of them also cover $X$, say $\mathcal{U'} = \{L_a: a \in A'\} \cup \{U_{u'}: u' \in U'\}$, where $U',A'$ are finite. Neither can be empty, by the remark above, so let $a_0 = \max(A'), u_0 = \min(U')$. Then $X = L_{a_0} \cup U_{u_0}$. This means that $a_0 \in U_{u_0}$ (as $a_0$ is not covered by $L_{a_0}$) which means that $a_0 > u_0$, contradicting that $u_0 \in u(A)$. 
This means that $\sup(A)$ must indeed exist.
On the other hand, if $X$ has the sup-property then $X$ is compact:
It suffices by Alexander's subbase lemma to prove that a cover by subbasic elements has a finite subcover. So let $\{L_{x_i}:  i \in I\} \cup \{U_{y_j}: j \in J \}$ be a cover of $X$ by subbasic open sets. 
Let $p = \sup\{x_i: i \in I \}$. This $p$ cannot be in any $L_{x_i}, i \in I$, so it in some $U_{x_j}$. So $x_j < p$. This means that $p$ is not an upperbound for $\{x_i: i \in I \}$ (or it would be smaller than the smallest such upperbounds $p$), so for some $i \in I$, $x_j < x_i$. But then $L_{x_i}$ and $U_{x_j}$ together cover $X$, giving us the required finite subcover.
This completes the proof of the characterisation of compact ordered spaces.
Theorem: If $(X,<_X)$ and $(Y,<_Y)$ both have the sup-property, then so has $X \times Y$ in the lexicographic ordering $(x,y) <_l (x',y')$ iff $x < x'$ or ($x = x'$ and $y < y'$).
Let $A \subseteq X \times Y$. Define $A_0 = \{x: \exists y \in Y: (x,y) \in A\}$, the projection onto $X$. Then $p := \sup(A_0)$ exists. Then define $B_0 = \{ y \in Y: (p,y) \in A \}$, then $q = \sup(B_0)$ exists as well.
Then $(p,q) = \sup(A)$. Suppose it's not an upperbound for $A$, then we have
some $(a,b) \in A$ with $(p,q) <_l (a,b)$. This means that either $p < a$, but then $a \in A_0$ and $p$ wouldn't be an upper bound for $A_0$, so this cannot be the case. So $p = a$ and $q < b$. But then $b \in B_0$ and $q$ is not an upperbound for $B_0$, so this also cannot happen.
Suppose $(p',q')$ is an upperbound for $A$ (we want to see that $(p,q)$ is minimal among those), then suppose $a \in A_0$. Then $(a,b) \in A$ for some $b \in Y$, so $(a,b) \le (p',q')$. In particular $a \le p'$, so $p'$ is an upperbound for $A_0$ and so $p \le p'$ by minimality of $p$. If $p < p'$ would hold, we'd know that $(p,q) < (p',q')$ and we'd be done. So suppose $p = p'$. Now for every $b \in B_0$ we know $(p,b) \in A$, so $(p,b) \le (p',q')$. Together with $p = p'$ this implies $b \le q'$, so $q'$ is an upperbound for $B_0$, and so again $q \le q'$. So in all we know that $(p,q) \le (p',q')$, as required.
Corollary: $I^2_0$ is compact, from the fact that $[0,1]$ is ordered and compact (you already know it, or see it from completeness of the reals).
