Exercise on joins on order (Introduction to Lattices and Order)

Question

I'm reading a book "Introduction to Lattices and Order" by Davey and Priestley and I'm stuck on exercise 3 of p57. You should be able to find the book here, but I'll describe it anyway.

Then the questions ask to find the join ∨ and meet ∧ of various subsets. These are the ones motivating this post:

4. $d \vee e$

5. $c \vee ( d \vee e)$

6. $\bigvee\{c, d, e\}$

for 4. it's clear that the join does not exist as $j$ and $l$ seems both valid candidates.

for 5. I was first tempted to say that as the previous question has no meet, this one as no meet as well. However, observe that $c \vee (d \vee e = (c \vee d ) \vee e$. Note that $c \vee d$ does not exist as well because it might be $f$ or $k$, so my answer is that this meet does not exist as well. But, if we consider the same order where we remove the element k, what happens? It seems that we might be able to say that $c \vee ( d \vee e) = f \vee e = j$, even if $d \vee e$ does not exist?

for 6. I am not sure how to interpret the question. I am tempted to say that it is equal to $(c \vee e) \vee d = j \vee d = top$, and that does not feels right. From the associativity of meet I don't understand how 6. is different from 5.

Answer 1

You are correct in saying that $d \vee e$ doesn't exist, and the reason is that the set of their common upper bounds, $\{d,e\}^u$ is $\{j,\ell,m,\top\}$, and $j$ and $\ell$ are minimal among these, so there is no least upper bound.
(This was more or less your argument.)

As for $c\vee (d \vee e)$, I would say it doesn't exist because, as we saw above, $d\vee e$ doesn't exist.

But $\bigvee\{c,d,e\}$ exists and it's $j$. Indeed, $\{c, d, e\}^u=\{j,m,\top\}$, and $j$ is the least of them.
(Notice that, contrary to what you wrote, $j \vee d = j$ because $d \leq j$.)