can a one-element set $L$ have a top and bottom element? A top element $u_2$ satisfies $a\leq u_2$ for all $a$ in $L$; similarly, a bottom element $u_1$ satisfies $u_1\leq a$ for all $a$ in $L$. If we have a one element subset, then $u_1=u_1$, so $u_1$ is both a top and bottom element. True?
 A: I've often seen these "top elements" referred to as suprema, and the "bottom elements" as infima, so I'll use that terminology going forward. We'll let $\le$ denote the partial ordering going forward.

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*An element $a \in S$ is a supremum if $s \le a$ for every $s \in S$.

*An element $b \in S$ is an infimum if $a \le s$ for every $s \in S$.

Note that we do not use "for every $s \in S \setminus \{a\}$" (or $S\setminus \{b\}$) here -- that is, it has to hold for every element of $S$, including the supremum/infimum itself! (We also note, somewhat more subtly, there is no tenet that says these two need be distinct elements.)
In that light, it of course makes sense that you could have a singleton set $L := \{a\}$ with both a supremum and infimum; you need only give it the proper partial ordering $\le$ (which would likely be more difficult to not do considering there's only one element in $L$).
For instance, let $\le$ be, quite literally, the "less than or equal to" relation, with $a$ some real number. Then $a \le \ell$ for all $\ell \in L$ (namely $a \le a$), so $a$ is an infimum of the set. Likewise, $\ell \le a$ for all $\ell \in L$ (namely $a \le a$), so $a$ is also a supremum of the set. So as a result, $\inf(L) = \sup(L) = a$. While somewhat silly to think about, this is totally fine and mostly just a consequence of working with a really simple example.
