If $U$ and $V$, having $U \cap V = \emptyset$, are open in $\mathbb R$ with the standard metric, then there exists $c \notin U \cup V$

Question

If $U$ and $V$, having $U \cap V = \emptyset$, are open in $\mathbb R$ with the standard metric, then there exists $c \notin U \cup V$

Proof:

Let $A = [a,b] \cap U$ where $a \in U$ and $b \in V$. Since $a \in A$ and $a \in U$, $A$ is non-empty and for all $x \in A$, $x \leq b$ so $A$ is bounded above. Thus $c = \text{sup} A$ exists.

If $c \in V$ we have $B(c,r) \subset V$ for some positive $r$. Now $t < c$ for $t \in A$, and if $t < x < c$, $x \in A$. Thus for every positive $r$, some there is some $t \in A$ for which $t \in B(c,r)$. This especially means $t \in U$ so $U \cap V$ is nonempty which is a contradiction. Thus $c \notin V$.

If $c \in U$ we have $B(c,r) \subset U$ for some positive $r$. Now $t \leq c$ for all $t \in A$. If $c \in [a,b]$ then for all $v \in (c,b]$ we have $v \in V$, which implies $v \in B(c,r)$ for any positive $r$. Thus $c \notin [a,b]$ and $b < c$. Thus for all positive $r$, $b \in B(c,r)$ which is a contradiction. Thus $ c \notin U$ and this implies $c \notin U \cup V$. $\blacksquare$

Are there any major errors or unnecessarily complicated arguments?

Answer 1

The proof is mostly correct, and in my opinion not unnecessarily complicated. There is just one small error:

If $c\in[a,b]$ then for all $v\in(c,b]$ we have $v\in V$, which implies $v\in B(c,r)$ for any positive $r$.

You cannot really conclude that $v\in V$, but if there would be a $v\in (c,b]$ such that $v\notin V$ then you would have $v\notin U\cup V$, giving you the result.

So you could argue it's not really an error but just a behind the scenes 'w.l.o.g.'.