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Let $P,Q$ be posets and endow them with the interval topology $\tau_i(P)$ and $\tau_i(Q)$ respectively. Is it true that if $f: P\to Q$ is order-preserving, then it is continuous, and vice versa?

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    $\begingroup$ In view of the comments at the linked question, this interval topology agrees with the usual topology in the case $P=Q=\mathbb R$, and here there are trivial counterexamples for both of the proposed implications. $\endgroup$
    – Andreas Blass
    Commented Aug 27, 2015 at 7:25

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Both implications are incorrect in general.

Consider the real interval $I=[0,1]$. Then the interval topology on $I$ is just the topology coming from the Euclidean metric on $I$. Let $f: I\to I$ be defined as $f(1) = 1$ and $f(x) = 0$ for $x\in I\setminus \{0\}$. Then $f$ is clearly order-preserving, but not continuous.

For a continous function $f:I\to I$ that is not order-preserving, let $f(x) = x$ for $x\in [0,0.5]$ and $f(x) = 1 - x$ for $x \in I\setminus [0, 0.5]$.

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