Show that an interval in $\mathbb{R}$ is a connected set in $\mathbb{R}$.
Edit: A different proof attempt.
Let $A$ be an interval in $\mathbb{R}$ and let $a=\sup(A)$ and $b=\inf(A)$. Suppose that $A$ is not connected $\implies\exists U,V$ open in $\mathbb{R}$ s.t. $U$ and $V$ disconnect $A$. $U\cap A\neq\emptyset$ and $V\cap A\neq\emptyset$, so without loss of generality assume $a\in U$ and $b\in V$. I want to show that there exists a $z\in\partial U\cap\partial V$: $a<z<b$ and $z\not\in U,V$. Since $U,V$ are open, $U\cap\partial U=\emptyset$ and $V\cap\partial V=\emptyset$. $z\in\partial U\implies$ $B_r(z)\cap U^c\neq\emptyset$ for $r>0$ and since $U$ is open, $z\not\in U\implies z\in U^c$. By the same logic, $z\not\in V$ but since $U,V$ cover $A$, $z\not\in A$. But then $A$ is not an interval.
Am I off base here?
Best Answer
A subset $I\subset\mathbb{R}$ is connected iff it has the following property, if $x\in I$ and $y\in I$, and $x<z<y$, then $z\in I$. (I first learned this property from Rudin's Real Analysis.)
If there exist $x,y\in I$ and some $z\in(x,y)$ such that $z\notin I$, then $I=A_z\cup B_z$, where $A_z=I\cap(-\infty,z)$ and $B_z=I\cap (z,\infty)$. Since $x\in A_z$ and $y\in B_z$, then $A\neq\emptyset$ and $B\neq\emptyset$. Since $A_z\subset(-\infty,z)$ and $B_z\subset(z,\infty)$, they are separated, and so $I$ is not connected.
Conversely, suppose $I$ is connected, so there exist nonempty separated sets $A$ and $B$ such that $A\cup B=I$. Let $x\in A$ and $y\in B$, and say $x<y$. Set $z=\sup(A\cap[x,y])$. Now for a nonempty set of reals which is bounded above, the supremum is an element of the closure. So $z\in\bar{A}$ and $z\notin B$, since they are separated. Thus $x\leq z<y$. If $z\notin A$, then $x<z<y$ and $z\notin I$. If $z\in A$, then $z\notin\bar{B}$, so there is some $w$ such that $z<w<y$ and $w\notin B$. Then $x<w<y$ and $w\notin I$.