How a "lacuna" can be a complement of S and have a upper and a lower bound inside of S? The textbook says: "a lacuna of $S$ is a nondegenerate interval of $R$ (extended real line) without points of $S$ but having a lower bound and an upper bound in $S$" and "a gap of $S$ is a maximal lacunda of $S$."
But I don't get it.
How can an interval be, at same time, a complement of $S$ (in the sentence "without points of $S$) and have a lower and a upper bound in $S$?
In my mind, if a set "$A$" has a upper and a lower bound in a set "$B$", it means that $A \subset B$. (Am I wrong?)
 A: The interval $[10,11]$ on the real number line has many upper bounds.
The number $12,$ $13,$ $22,$ $1000003,$ and $15.394$ are all upper bounds of
$[10,11]$.
The number $11$ is also an upper bound, but has a special property:
it is the least upper bound.
So the first thing to keep in mind is that you must not think "least upper bound" every time you see the words "upper bound", and you must not think "greatest lower bound" every time you see the words "lower bound".
Now consider the set $S = [0,1] \cup [19,20].$
And consider the set $A = [10,11].$
We can see that no point belonging to $S$ is a member of  $A$.
That is what is meant when we say that "$A$ is a set without points of $S$".
Note that $1$ is a lower bound of $A$ and is in $S,$ while $19$ is an upper bound of $A$ and is in $S.$ Therefore $A$ is a lacuna of $S.$
Now observe that the complement of $S$ on the extended real line is
$\bar S = [-\infty,0) \cup (1,19) \cup (20,\infty].$
Obviously $\bar S \neq A$; but $A \subseteq \bar S.$
In fact, any lacuna of $S$ will necessarily be a subset of $\bar S,$
but the reverse is not true in general.
In particular, in this case $\bar S$ is not itself a lacuna of $S$ because the only lower bound of $\bar S$ is $-\infty,$ which is not in $S.$
In some special cases (such as $S = [-\infty, -1] \cup [1, \infty]$),
the complement of $S$ will be a lacuna of $S,$
but in the more general case it is not.
A: Simple example: consider $S = [-\infty, -1] \cup [1, \infty]$. Its complement is $(-1, 1)$, a nondegenerate interval without points of $S$, but it has an upper bound and a lower bound, $1$ and $-1$ contained in $S$. Moreover, $(-1, 1)$ is a gap of $S$.
