Prob. 7 (a), Sec. 24, in Munkres' TOPOLOGY, 2nd ed: Any order preserving surjective map between ordered sets is a homeomorphism

Question

Here is Prob. 7 (a), Sec. 24, in the book Topology by James R. Munkres, 2nd edition:

Let $X$ and $Y$ be ordered sets in the order topology. Show that if $f \colon X \to Y$ is order preserving and surjective, then $f$ is a homeomorphism.

My Attempt:

For any two elements $a$ and $b$ of $X$ such that $a <_X b$, we define the sets $(a, b)_X$, $[a, b)_X$, $(a, b]_X$, and $[a, b]_X$ as follows: $$ (a, b)_X \colon= \left\{ \ x \in X \ \colon \ a <_X x <_X b \ \right\}, \tag{1} $$ $$ [a, b)_X \colon= \left\{ \ x \in X \ \colon \ a \leqq_X x <_X b \ \right\} = (a, b)_X \cup \{ a \}, \tag{2} $$ $$ (a, b]_X \colon= \left\{ \ x \in X \ \colon \ a <_X x \leqq_X b \ \right\} = (a, b)_X \cup \{ b \}, \tag{3} $$ and $$ [a, b]_X \colon= \left\{ \ x \in X \ \colon \ a \leqq_X x \leqq_X b \ \right\} = (a, b)_X \cup \{ a, b \}. $$ All possible sets of types (1), (2), and (3) constitute a basis for the order topology on $X$, with the proviso that for sets of type (2) the element $a$ must only be the smallest element, if any, of $X$, whereas for sets of type (3) the element $b$ must only be the largest element, if any, of $X$.

Am I right?

Similarly, for any two elements $c$ and $d$ of $Y$ such that $c <_Y d$, we define $$ (c, d)_Y \colon= \{ \ y \in Y \ \colon \ c <_Y y <_Y d \ \}, \tag{4} $$ $$ [c, d)_Y \colon= \{ \ y \in Y \ \colon \ c \leqq_Y y <_Y d \ \} = (c, d)_Y \cup \{ c \}, \tag{5} $$ $$ (c, d ]_Y \colon= \{ \ y \in Y \ \colon \ c <_Y y \leq_Y d \ \} = (c, d)_Y \cup \{ d \}, \tag{6} $$ and $$ [c, d ]_Y \colon= \{ \ y \in Y \ \colon \ c \leqq_Y y \leqq_Y d \ \} = (c, d)_Y \cup \{ c, d \},$$ All possible sets of types (4), (5), and (6) constitute a basis for the order topology on $Y$, provided that for sets of type (5) the element $c$ must only be the smallest element, if any, of $Y$, whereas for sets of type (6) the element $d$ must only be the largest element, if any, of $Y$.

Let $x_1$ and $x_2$ be any two distinct elements of $X$. Then either $x_1 <_X x_2$ or $x_2 <_X x_1$; let's assume that $x_1 <_X x_2$. Then as $f$ is order preserving, so we can conclude that $f \left( x_1 \right) <_Y f \left( x_2 \right)$, which of course implies that $f \left( x_1 \right) \neq f \left( x_2 \right)$. Therefore $f$ is injective, and since $f$ is also given to be surjective, we can conclude that $f$ is bijective.

Am I right?

Now if $u$ and $v$ are any two elements of $Y$ such that $u <_Y v$, then there exist unique elements $a$ and $b$ of $X$ such that $u = f(a)$ and $v = f(b)$. Now if it were the case that $b \leqq_X a$, then the following would also hold. $$ v = f(b) \leqq_Y f(a) = u,$$ which is contrary to $u <_Y v$. Hence $a <_X b$.

Thus the inverse function $f^{-1} \colon Y \to X$ is also bijective and order preserving.

Am I right? Is this logic correct?

If $a$ and $b$ are any two elements of $X$ such that $a <_X b$, then $f(a) <_Y f(b)$. Moreover, if $y \in ( f(a), f(b) )_Y$ [Refer to (4) above], then by definition $y \in Y$ and $$ f (a) <_Y y <_Y f(b). $$ Moreover, the bijectivity of $f$ implies the existence of a unique element $x$ of $X$ such that $y = f(x)$, and, as $f$ is order-preserving, we can also conclude that $$ a <_X x <_X b,$$ which implies that $$ y \in f \left( (a, b)_X \right). $$ Therefore [Refer to (1) and (4) above.] we can conclude that $$ \left( \ f(a), f(b) \ \right)_Y \subset f \left( \ (a, b)_X \ \right). \tag{7} $$ Using the same logic [Refer to (2) and (5) above.] we can also conclude that $$ \left[ \ f(a), f(b) \ \right)_Y \subset f \left(\ [a, b)_X \ \right), \tag{8} $$ and [Refer to (3) and (6) above.] also that $$ \left( \ f(a), f(b) \ \right]_Y \subset f \left( \ (a, b ]_X \ \right). \tag{9} $$

Am I right?

Similarly, using (7), (8), and (9) together with the fact that $f^{-1}$ is also bijective and order preserving, we can show that if $c$ and $d$ are any two elements of $Y$ such that $c <_Y d$, then $$ \left( \ f^{-1}(c), f^{-1} (d) \ \right)_X \subset f^{-1} \left(\ (c, d)_Y \ \right), \tag{10} $$ $$ \left[\ f^{-1}(c), f^{-1} (d) \ \right)_X \subset f^{-1} \left( \ [ c, d)_Y \ \right), \tag{11} $$ and $$ \left( \ f^{-1}(c), f^{-1} (d) \ \right]_X \subset f^{-1} \left( \ (c, d ]_Y \ \right). \tag{12} $$

Am I right?

Let $U$ be an open set in $X$. We need to show that $f(U)$ is open in $Y$. Let $y \in f(U)$. Then there exists a unique element $x \in U$ such that $y = f(x)$. Now as $x \in U$ and $U$ is open in $X$, so there exists a set $B_x$ of one of the types (1), (2), or (3) above such that $$ x \in B_x \subset U. $$ Then $$ y = f(x) \in f \left( B_x \right) \subset f(U),$$

On the other hand, since $f$ is order preserving and since $B_x$ is one of the sets of the types (1) (2), or (3) above, we can conclude that $y = f(x)$ is in one of the sets on the left-hand-sides of (7), (8), or (9) and that particular set (again by virtue of (7), (8), and (9) above) in turn is contained in $f \left( B_x \right)$ and thence in $f(U)$, which shows that $f(U)$ is open in $Y$. Therefore $f$ is an open map. Hence $f^{-1}$ is continuous.

Is this argument correct?

Analogously, we can show that the bijective order-preserving map $f^{-1} \colon Y \to X$ is an open map; that is, for every open set $V$ in $Y$, the set $f^{-1}(V)$ is open in $X$. Therefore $f$ is continuous.

Is this proof correct? If so, then is my presentation clear enough too? If not, then where have I messed up?

Answer 1

You can forget about the basic elements, for continuity it suffices to consider the subbasic elements $L_a = \{x \in X: x <_X a\}$, the left set of $a \in X$, and $R_a = \{x \in X: a <_X x\}$, the right set of $a \in X$. The order topology is defined as the smallest topology that contains the subbase $\mathcal{S} = \{L_a,R_b : a,b \in X\}$.

$f: (X,<_X) \to (Y,<_Y)$ is continuous iff $f^{-1}[S]$ is open for all $S \in \mathcal{S}$.

Now to the proposition: The strict order preservingness of $f$ indeed gives injectivity for free and surjectiveness is given. So $f$ is a bijection, say with inverse $g: Y \to X$. $f$ strictly order preserving and a bijection implies that $g$ is also stricly order preserving. I agree with those arguments.

Now $f$ order preserving gives that $f^{-1}[L_a] = L_{g(a)}$: $x \in f^{-1}[L_a]$ iff $f(x) <_Y a$ iff $x= g(f(x)) <_X g(a)$ iff $x \in L_{g(a)}$.

Likewise $f^{-1}[R_a] = R_{g(a)}$. So $f$ order preserving bijection means $f$ continuous. No need for open intervals or special cases for maxima and minima.

As the same holds for $g$, both $f$ and its inverse are continuous so $f$ is a homeomorphism.

Answer 2

This is more or less obvious. If $f:\>X\to Y$ is order preserving then $x\ne x'$, hence WLOG $x<x'$ implies $f(x)<f(x')$, in particular $f(x)\ne f(x')$. Since $f$ is surjective by assumption it therefore is bijective. It follows that $X$ and $Y$ are isomorphic as ordered sets. Since the order topologies on $X$, resp., $Y$ are defined solely in terms of the order on these sets we may conclude that $X$ and $Y$ are also homeomorphic with respect to these topologies.