I am reading John K. Hunter's lecture notes and in chapter 5, there is a theorem about every set in $\mathbb{R}$ being connected if and only if it is an interval. The proof is somewhat intricate, but I have only one question about a possible typo at the end of the proof. To understand the proof, however, the definitions of disconnected (i.e. not connected) and interval need to be given.
Definition 5.58. A set of real numbers $A \subset \mathbb{R}$ is disconnected if there are disjoint open sets $U, V \subset \mathbb{R}$ such that $A \cap U$ and $A \cap V$ are nonempty and
$$
A=(A \cap U) \cup(A \cap V).
$$
…
The condition $A=(A \cap U) \cup(A \cap V)$ is equivalent to $U \cup V \supset A$. If $A$ is disconnected as in the definition, then we say that the open sets $U,V$ separate $A$.
Definition 5.62. A set of real numbers $I\subset \mathbb{R}$ is an interval if $x,y\in I$ and $x<y$ implies that $z\in I$ for every $x<z<y$.
I think this definition can be restated as follows $$(\forall x,y \in I)(\forall z \in \Bbb R)((x<z) \land (z<y) \implies z\in I).$$ The negation is $$(\exists x, y \in I)(\exists z \in \Bbb R)((x < z) \land (z < y) \land (z \not\in I)).$$
Now to the theorem,
Theorem 5.63. A set of real numbers is connected $\iff$ it is an interval.
The author proves both directions by contraposition, but for brevity, I have only included the $\Longleftarrow$ direction.
Proof. …
To prove the converse, suppose that $I \subset \mathbb{R}$ is not connected. We will show that $I$ is not an interval. Let $U, V$ be open sets that separate $I$. Choose $a \in I \cap U$ and $b \in I \cap V$, where we can assume without loss of generality that $a<b$. Let
$$
c=\sup (U \cap[a, b]).
$$
We will prove that $a<c<b$ and $c \notin I$, meaning that $I$ is not an interval. If $a \leq x<b$ and $x \in U$, then $U \supset[x, x+\delta)$ for some $\delta>0$, so $x \neq \sup (U \cap[a, b])$. Thus, $c \neq a$ and if $a<c<b$, then $c \notin U$.If $a<y \leq b$ and $y \in V$, then $V \supset(y-\delta, y]$ for some $\delta>0$, and therefore $(y-\delta, y]$ is disjoint from $U$, which implies that $y \neq \sup (U \cap[a, b])$. It follows that $c \neq b$ and $c \notin U \cap V$, so $a<c<b$ and $c \notin I$, which completes the proof.
My question concerns "$c \notin U \cap V$". It's trivially true that $c \notin U \cap V=\emptyset$ ($U$ and $V$ are disjoint, since $I$ is not connected). Maybe it should be $c \notin U \cup V$? Then certainly $c\notin I$, since by definition of disconnectedness, $I\subset U\cup V$.
Best Answer
This does seem to be a typo; it should be $c\notin U\cup V$ instead of $c\notin U\cap V$. The previous two arguments show that $c\notin U$ and $c\notin V$, and from this the author concludes that $c\notin U\cup V$.