Michael wrote:

> I can see how the repaired \$$F(x)\$$ doesn't preserve arbitrary joins since \$$4 \leq x \leq 6\$$ and \$$6 \leq x\$$ both map to 2.

Right. We can use the adjoint functor theorem to check an a candidate to see if it is really an adjoint.

> But for \$$G\$$ I am not sure but maybe it's not preserving the top element which is \$$\infty\$$ and instead setting 6 as the top element?

This is also right. Here's the meet \$$G\$$ is not preserving: \$$\bigwedge \varnothing = \top\$$. The idea is that arbitrary joins follow the rule \$$\left(\bigwedge A\right) \wedge \left(\bigwedge B\right) = \bigwedge (A \cup B)\$$. Using that rule it's pretty easy to prove that \$$\bigwedge \varnothing = \top\$$.

By the adjoint functor theorem if \$$G\$$ has a left adjoint then \$$G \left( \bigwedge \varnothing\right) = \bigwedge \varnothing\$$. In the case of \$$\lbrace 0,1,2\rbrace\$$, \$$\bigwedge \varnothing = 2\$$. So we want \$$G ( 2) = \bigwedge \varnothing\$$. But \$$\bigwedge \varnothing = \top\$$, and we know there's no top element of \$$[0,\infty)\$$. So there we see that's the meet \$$G\$$ is not preserving.