Set $A$ interval in $\mathbb{R}\implies$ connected

Question

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?

Answer 1

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 $w\notin I$.

Answer 2

The easiest case is when the interval $A$ is open. Consider the least upper bounds of elements in $U\cap A$ and $V\cap A$, one of which (say $U\cap A$) must be in $A$ since $A$ is an interval and since we have $U\cap A \neq \varnothing \neq V \cap A$ and $U\cap V \cap A = \varnothing$. Since $U\cap A$ is open, this least upper bound cannot be in $U\cap A$ (as no neighborhood of it is in $U\cap A$), so by $A = (U\cap A) \cup (V\cap A)$ it is in $V\cap A$. But then $V\cap A$ contains some neighborhood of the least upper bound for $U\cap A$, so intersects $U\cap A$, a contradiction. Thus $A$ is connected. Can you see how to modify this for other intervals?

Answer 3

Let's call the interval $I$. Let $a \in I$. Without loss of generality, we shall assume $a \in U$. Now, $U=\{x\in I: x \in U, x\geq a\}\cup\{x\in I: x \in U, x\leq a\}$ is not all of $I$, so (without loss of generality) there exists $b>a$ with $b \notin U$. Let $z$ be the infimum of all such $b$. You can show that $z$ cannot be contained in either $U$ or $V$.

Answer 4

Suppose that an open interval $(a,b)$ is covered by open sets $U , V$, and $U \cap (a,b) \neq \emptyset \neq V \cap (a,b)$. (Note that we may assume that $V , U \subseteq (a,b)$.) We will show that $U \cap V \neq \emptyset$.

If either $U \subseteq V$ or $V \subseteq U$ we are done. Therefore we may assume that $U \nsubseteq V$ and $V \nsubseteq U$. Take any $x \in U \setminus V$. Note that $U$ itself is a union of countably many pairwise disjoint open intervals, say $U = \bigcup_{i=1}^\infty (c_i,d_i)$.[*] Thus there is an $i$ such that $c_i < x < d_i$. It can easily be shown that neither $c_i$ nor $d_i$ belong to $U$.

Since $U \subsetneq (a,b)$ it follows that either $a < c_i$ or $d_i < b$. Without loss of generality assume that $a < c_i < x < b$. Then $c_i \in V$. But as $V$ is open there is an $\epsilon > 0$ such that $(c_i - \epsilon , c_i + \epsilon ) \subseteq V$. Since $x \notin V$ it follows that $\epsilon \leq x - c_i$. It then follows that $c_i + \frac{\epsilon}{2} \in V$ (by the choice of $\epsilon$) and $c_i + \frac{\epsilon}{2} \in U$ (by choice of $c_i$ and observation of $\epsilon$, above), and so $U \cap V \neq \emptyset$.

The case for closed and half-open intervals can be similarly made.

[*] If you do not want to appeal to this characterisation (which can be found in, e.g., Royden's text), we can get around it by defining $c = \inf \{ z \in \mathbb{R} : [z,x] \subseteq U \}$ and $d = \sup \{ z \in \mathbb{R} : [x,z] \subseteq U \}$. It then follows that $(c,d) \subseteq U$ but $c \notin U$ and $d \notin U$. Why? Consider $c$:

• Given $c < y < x$, by definition of $c$ there is a $c \leq z < y$ such that $[z,x] \subseteq U$, which means that $y \in U$. Therefore $(c,x) \subseteq U$.
• If $c \in U$, then since $U$ is open there is a $\epsilon > 0$ such that $( c-\epsilon , c + \epsilon ) \subseteq U$, but then it would follow that, e.g., $[ c - \frac{\epsilon}{2} , x ] \subseteq U$, contradicting our definition of $c$.

The case for $d$ is analogous, and by choice we have that $x \in U$.

Answer 5

Here's another way to look at it. WLOG, assume that $I = (a,b) = U \cup V$, where $U$ and $V$ are nonempty, open and disjoint. $U$ is open so can be written as a disjoint union of open intervals. Let $(c,d)$ be one such interval, and WLOG assume $d\not= b$. Therefore $d\in V$, but this implies that $U$ and $V$ aren't disjoint.

To finish it off notice that a separation of $[a,b]$, $[a,b)$ or $(a,b]$ clearly separates $(a,b)$ which we showed is impossible.