# "In a bounded lattice the join is the least element and meet is the greatest element"... yet a join is the least upper bound I don't understand.

This page Wikipedia: lattice says it is vacuously true that for every $$a \in \{ \emptyset \}: x \leq a$$ and $$a \leq x$$ so then every element of the lattice is an upper and lower bound of the empty set; and that implies the join is the least element of the null set and the meet is the greatest element of the null set .

I don't know how to continue - if I look at the greatest and least element page there is a subset of a partial order with $$1$$, the meet, as the least element and $$3,4$$ hypothetically as greatest elements which is opposite of whatever is true for the null set there in the first page describing the lattice. anyway the join is supposed to be a greatest element as I always have read so I guess another interpretation for my question is why if the join is a least upper bound, in the null set the greatest element which every element is comparable to is the meet which compares to all the other elements?

• Note: $\{\varnothing\}$ is not the same thing as $\varnothing$. The former is a set whose only element is the empty set; the latter is the empty set. (An empty bag is not the same thing as a bag that has an empty bag inside) Commented May 28, 2021 at 1:42
• “This page”. What page? The government doesn’t let me read minds without a warrant, so you need to say where you are reading these things, so people can see what you refer to. Commented May 28, 2021 at 1:48
• That Wikipedia page does not say what you accuse it of saying. It says that the join of $\varnothing$ (not the join of $\{\varnothing\}$) is the least element of the whole lattice, not the least element of $\varnothing$.
– bof
Commented May 28, 2021 at 3:34

By definition, given a subset $$S$$ of the lattice $$L$$, the element $$s_0\in L$$ is “the join of $$S$$”, $$\bigvee S = s_0$$, if and only if two things happen:

1. For every $$x\in S$$, $$x\leq s_0$$; and
2. For every $$y\in L$$, if for every $$x\in S$$ we have $$x\leq y$$, then $$s_0\leq y$$.

Item 1 tells you “$$s_0$$ is greater than or equal to everything in $$S$$”. Item 2 tells you “$$s_0$$ is the smallest among all elements that are greater than or equal to everything in $$S$$.”

What happens when $$S$$ is the empty set? Item 1 is satisfied by every element $$z$$ in $$L$$, because the proposition “for every $$x\in \varnothing$$, $$x\leq z$$“ is true “by vacuity”: there is no $$x\in\varnothing$$ to falsify the statement, so the statement is true.

That means that the collection of all upper bounds of the empty set is the whole lattice. So... what does that mean for item 2? It means that for an element $$s_0\in L$$ to satisfy both 1 and 2 it must:

1. Well, 1 is automatically satisfied, so no problem there; and
2. For every $$z\in L$$, $$s_0\leq z$$.

Why? Because every $$z\in L$$ satisfies “for every $$s\in \varnothing$$, $$s\leq z$$”. So we must have $$s_0\leq z$$ for every $$z\in L$$.

So what elements of $$L$$ can satisfy that they are the join of the empty set? Only the least element of $$L$$. So $$\bigvee\varnothing = \mathbf{0}$$, where $$\mathbf{0}$$ is the least element of the lattice.

It may seem counterintuitive that the “supremum of $$S$$” is the smallest element of $$L$$; but remember that the supremum of $$S$$ is also the least upper bound. Since everything is an upper bound, the least upper bound is the smallest possible element.