Are an element and its image under a Galois connection linearly ordered?

Question

Let $(L,\leq)$ be a lattice, and $G$ a self-adjoint antitone Galois connection on it: this means it is a map from $X$ to itself such that $x\leq G(y)$ if and only if $y\leq G(x)$.

An element $x\in X$ is closed if $G(G(x))=x$. We can then easily equip the subset of closed elements of $L$ with a lattice structure as well; denote the domain of this lattice by $G(L)$.

Given $x\in G(L)$, it is true that either $x\leq G(x)$ or $G(x)\leq x$? In other words, is $\{x,G(x)\}$ linearly ordered?

I feel this is an easy result, to which the answer is yes, but I tried to recall a proof and could not find one. This leads me to believe I could be misremembering something and perhaps I'm wrong, and in that case I would appreciate a counterexample, if one is readily available.

---

Update: as shown below, I was quite naïve to believe this property would hold in general, the obvious Galois connection on a powerset being a counterexample.

For completeness' sake, let me present one example of the Galois connections I was thinking of when I thought of this question: let $\mathbb{Z}$ have its usual order, and define $G(n)=-n$; of course, $n\leq -m$ iff $m\leq -n$. Every element of $\mathbb{Z}$ is seen to be closed, and either $n\leq G(n)$ or $G(n)\leq n$. Notice that in this case obviously the property would hold, as the initial lattice is linearly ordered.

Answer 1

I initially missed that set in question is a lattice. Here's the updated answer; below is the previous (simpler) answer for general partial orders.

The answer is no. Take $X=\{\bot,a,b,\top\}$, where $\bot$ and $\top$ are the smallest and largest elements, respectively, while $a$ and $b$ are unordered (that is, $X$ is isomorphic to the power set lattice of $\{a,b\}$). Take $G$ to be the bijection that swaps $a$ and $b$, and swaps $\bot$ and $\top$. You can check on a case-by-case basis that this is an antitone Galois connection.

Now, $\{a,G(a)\}$, that is, $\{a,b\}$ is unordered.

---

What follows assumes $X$ is a general partial order and not a lattice.

The answer is no. Take $X=\{a,b\}$ where $a$ and $b$ are unordered. Take $G(a)=b$ and $G(b)=a$. It is a Galois connection, but $\{x,G(x)\}$ is unordered for any $x$.

---

More generally, take any (antitone, in your case) Galois connection between partial orders $X$ and $Y$ with $F:X\rightarrow Y$ and $G:Y\rightarrow X$. Form the disjoint union $Z=X\sqcup Y$ with the copies of $X$ and $Y$ unordered inside $Z$. Now define $H:Z\rightarrow Z$ by $H(x)=F(x)$ for $x\in X$ and $H(y)=G(y)$ for $y\in Y$. Effectively we've taken a Galois connection between two sets and just merged the sets to make it a connection on one set.

Now, $\{z,H(z)\}$ is always unordered.

Answer 2

You already have a good answer, by the user lisyarus, but here's another one, very classical indeed.
Moreover, that example is a particular case of this one (the smallest possible).

Suppose $L$ is a Boolean algebra, and $G(x)=x'$.
This gives an antitone Galois connection, as it is easy to check.
Now, the only examples of Boolean algebras in which $\{x,x'\}$ is an ordered set are when they have at most two elements.
The example lisyarus gave, is a Boolean algebra with four elements, where the map $G$ thus defined is the complement.

So you have infinitely many counterexamples to your conjecture.