To continue, let me give a silly trivial example that illustrates the point I just made. I gave this example in [an answer to Daniel Fava in the Lecture 6 thread](https://forum.azimuthproject.org/discussion/comment/16703/#Comment_16703):

For any poset \$$A\$$ whatsoever, the identity function \$$1_A : A \to A\$$ has a left and right adjoint, namely itself. This is easy to check straight from the definition:

$$a \le a \textrm{ if and only if } a \le a .$$

If you compute these adjoints using the formulas above, you see that it only requires sets of the form

$$\\{ a \in A : \; a \le b \\}$$

to have greatest lower bounds - and such a set indeed does, namely the element \$$b\$$. Similarly, only sets of the form

$$\\{b \in A: \; a \le b \\}$$

need have least upper bounds - and such a set indeed does, namely the element \$$a\$$.

So, I'm claiming

$$\bigwedge \\{ a \in A : \; a \le b \\} = b$$

and

$$\bigvee \\{b \in A: \; a \ge b \\} = a$$

whenever \$$A\$$ is any poset.

That said, if someone gave me a puzzle whose answer was \$$a\$$, and I said the answer was \$$\bigvee \\{b \in A: \; a \ge b \\} \$$, we'd have to say my answer wasn't the best available, because I failed to simplify it as much as possible.