Domain theory: Definition of isomorphisms

Question

Consider the following definitions of CPOs, monotonicity and continuity: The pair $(M, \leq)$ is called CPO (complete partial order), if

• $M$ is a set,
• $\leq$ is a partial order on $M$,
• there exists a least element according to $\leq$ in $M$ (called bottom, $\bot$),
• for every chain $m_0, m_1, m_2, \dots$ in $M$ its supremum $\bigsqcup \{m_0, m_1, m_2, \dots\}$ (least upper bound) exists in $M$.

(where a chain is a sequence $m_0, m_1, m_2, \dots$ of elements in $M$ such that for all $i, j \in \mathbb{N}$ $m_i \leq m_j \vee m_j \leq m_i$ holds)

A function $f \colon M \to N$ between two CPOs $(M, \leq)$ and $(N, \sqsubseteq)$ is called monotone if for all $a, b \in M$

$$a \leq b \implies f(a) \sqsubseteq f(b)$$

holds.

A function $f \colon M \to N$ between two CPOs $(M, \leq)$ and $(N, \sqsubseteq)$ is called continuous, if it is already monotone and for all chains $m_0, m_1, m_2, \dots$ we have

$$f\left(\bigsqcup \{m_0, m_1, m_2, \dots\}\right) = \bigsqcup \{f(m_0), f(m_1), f(m_2), \dots\}.$$

I read in a book using above definitions that two CPOs $(M, \leq)$ and $(N, \sqsubseteq)$ are isomorphic if there exists a function $f \colon M \to N$ such that

• $f$ is bijective
• $f$ and $f^{-1}$ are continuous (and therefore monotone).

I understand the demand of $f$ and $f^{-1}$ being monotone, so that they preserve the order.

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Question: Why is this not sufficient? Why do we also demand $f$ to be continuous? Why its inverse function too? What are examples where a violation of one property would result in a strange view of equivalence? Another question: Is it possible to drop one property, because the other would imply it? I'm thinking about to demand only $f$'s continuity and to show that $f^{-1}$'s continuity follows – however I was not able to do so, or to come up with a counter example.

Edit: I suggest a »domain theory« tag.