I am looking at the Penguin Dictionary of Mathematics $4$th edition edited by David Nelson, studying it as a course of self-study.
Let $L = (S, \vee, \wedge, \preccurlyeq)$ be a lattice.
The definition of the bottom of $L$ is defined as either:
- the smallest element of $S$ w.r.t. $\preccurlyeq$
or:
- the identity element of $S$ w.r.t. $\vee$, where $\vee$ is the join operation.
Let us denote $\bot$ as the bottom of $L$.
Now, looking at the abovementioned Penguin Dictionary of Mathematics, the definition of an atom of $S$ is given as an element $A$ of $S$ such that:
- $\forall B \in S: B \prec A \implies B = \bot$
where $B \prec A$ denotes that $B \preccurlyeq A$ such that $B \ne A$.
But where does that leave $\bot$ itself?
By definition, there are no elements $B$ of $S$ such that $B \prec \bot$.
Hence vacuously:
- $B \prec \bot \implies B = \bot$
and so, vacuously, $\bot$ is indeed an atom of $L$.
Is it a commonplace to need to specify that an atom also needs to be specified as to be an element of $S$ which is not the bottom?
That is, is this an omission (that is, a mistake) in Penguin Dictionary of Mathematics that the reader needs to be aware of?
Or is the bottom of a lattice also classed as an atom?