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.

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.