In order topology, are "continuous bijection is increasing or decreasing" and "space is connected" equivalent? Let $\langle X, \prec \rangle$ be a linear order, and $\langle X, \tau \rangle$ be the induced order topology.
Are the following two statements equivalent?


*

*Every continuous (in the sense of the topology) bijection is either increasing or decreasing (in the sense of the ordering).

*The space $X$ is connected (in the sense of the topology).

 A: They are not equivalent, even if you require $X$ to have more than two points.  For instance, suppose $X$ is a linear order such that every point of $X$ has a distinct uncountable cofinality from below and cofinality $\omega$ from above.  (Such a linear order can be constructed by inductively adding new points to give the requisite cofinalities from below in an induction of length $\omega$; see my first example in this answer.)
Such an $X$ cannot be Dedekind-complete (otherwise it would have points with cofinality $\omega$ from below), so it is disconnected in the order topology.  However, I claim the only continuous injection $X\to X$ is the identity map.  Indeed, suppose $f:X\to X$ is a continuous injection and $f(x)=y\neq x$ for some $x\in X$.  Let $\kappa$ be the cofinality of $x$ from below and let $\lambda$ be the cofinality of $y$ from below.  Let $(x_\alpha)_{\alpha<\kappa}$ be a strictly increasing sequence approaching $x$ from below and let $(y_\beta)_{\beta<\lambda}$ be a strictly increasing sequence approaching $y$ from below.  Also let $(y^n)_{n<\omega}$ be a strictly decreasing sequence approaching $y$ from above.
By our choice of $X$, $\kappa\neq \lambda$.  First, suppose $\kappa>\lambda$.  By continuity of $f$, for each $\beta<\lambda$ there exists $\gamma_\beta<\kappa$ such that $x_\alpha>y_\beta$ for all $\alpha>\gamma_\beta$.  Also, for each $n<\omega$, there exists $\gamma^n<\kappa$ such that $x_\alpha<y^n$ for all $\alpha>\gamma^n$.  Let $\gamma$ be the supremum of all the $\gamma_\beta$s and $\gamma^n$s.  Since $\kappa$ is regular and $\kappa>\lambda$, $\gamma<\kappa$.  But then for $\alpha>\gamma$, we have $f(x_\alpha)=y$, since $y$ is the only element of $X$ which is greater than every $y_\beta$ and less than every $y^n$.  This contradicts injectivity of $f$.
Now suppose $\kappa<\lambda$.  Replacing $(x_\alpha)$ with a cofinal subsequence, we may assume either $f(x_\alpha)<y$ for all $\alpha$ or $f(x_\alpha)>y$ for all $\alpha$.  In the case $f(x_\alpha)>y$, we can use the same argument as the previous paragraph (using just the $\gamma^n$s) to reach a contradiction.  In the case $f(x_\alpha)<y$, note that for every $\alpha$ there exists $\beta_\alpha<\lambda$ such that $f(x_\alpha)<y_{\beta_\alpha}$.  Since $\kappa<\lambda$, the supremum $\beta$ of all these $\beta_\alpha$s is less than $\lambda$.  But then $f(x_\alpha)<y_\beta$ for all $\alpha$.  This means $(f(x_\alpha))$ does not converge to $y$, contradicting continuity of $f$.

If you have difficulty understanding the example above, here's the gist of it.  You can have a linear order $X$ which is disconnected but highly asymmetrical, so that every point of $X$ "looks different" topologically.  This asymmetry makes $X$ very "rigid", so there are no nontrivial continuous bijections $X\to X$ at all, so rather vacuously they are all increasing or decreasing.
A: I don't think $(1)\implies (2)$ in general. The following is a counterexample.
Let $X=\{0,1\}$ and $\tau:=\{\emptyset,\{0\},\{1\}, X\}$ (i.e., $\tau$ is the subspace topology that $X$ inherits as a subspace of $\mathbb{R}$ equipped with usual (order) topology). Then observe that every continuous bijection from $\langle X,\tau\rangle$ to $\langle X,\tau\rangle$ is order preserving (or order reversing). However the space $\langle X,\tau\rangle$ is not connected because $\{1\}$ is a non-trivial clopen set in the topological space.
