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

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?

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I am not sure why do $(1)$ and $(2)$ come about? But, a set is not connected if it can be written as union of two disjoint non-empty open sets. –  user21436 Feb 24 '12 at 20:49
How do you know there exists such a $z$? –  Nate Eldredge Feb 24 '12 at 20:51
@KannappanSampath (1) and (2) are part of my definition of not connected. –  Emir Feb 24 '12 at 20:52
@NateEldredge Is it because $U$ and $V$ are open? –  Emir Feb 24 '12 at 20:53
That might be related, but you would have to prove it. In general I think it may be false. ($U$ and $V$ are open sets, but they don't have to be open intervals!) –  Nate Eldredge Feb 24 '12 at 20:58

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 $w\notin I$.

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Btw, I'm using the definition that a space is connected if is not the union of two nonempty separated sets, which is equivalent to saying the space is the not the union of two disjoint nonempty open sets. If you don't already have this equivalence, I have some explanation of this in Exercise 2.20 of this on page 18. I think it's right. –  yunone Feb 28 '12 at 6:16
It is a good explaination but only proved that open interval instead of interval. Are you intended to adopt the same method to other cases? BTW, you make the solution yourself??? that's pretty cool. –  Mathematics Nov 17 '12 at 6:11
@Mathematics Notice that open, closed, and half-open intervals all satisfy that property, so it should work for all cases. And thanks, I did write those solutions up, with some help of course. –  yunone Nov 17 '12 at 6:15

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 (1) and (2). 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 (3) 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?

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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 $V$.

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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 $x \in U$.

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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.

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