# intervals are connected

We wish to show that intervals (with standard topology) are connected. In order to this, we will prove that the space of real numbers $\mathbb{R}$ is connected. First we need a lemma.

Let $(X,d)$ be a metric space. Recall that for $x\in X$ and $r\in\mathbb{R}^{+}$ we have

 $B(x,r)=\{y\in X\ |\ d(x,y)

Lemma. Let $(X,d)$ be a metric space and $R\subset\mathbb{R}^{+}$ such that $R$ is nonempty and bounded. Then for any $x\in X$ we have

 $\bigcup_{r\in R}B(x,r)=B(x,\mathrm{sup}(R)).$

Proof. Assume that $y\in\bigcup_{r\in R}B(x,r)$. Then there is $r_{0}\in R$ such that $d(x,y) and thus $d(x,y)<\mathrm{sup}(R)$, so $y\in B(x,\mathrm{sup}(R))$.

Now assume that $y\in B(x,\mathrm{sup}(R))$. Then $d(x,y)<\mathrm{sup}(R)$ and it follows (from the definition of supremum) that there is $r_{0}\in R$ such that $d(x,y) and therefore $y\in B(x,r_{0})\subset\bigcup_{r\in R}B(x,r)$, which completes the proof. $\square$

. The space of real numbers is connected.

Proof. Assume that $U,V\subseteq\mathbb{R}$ are open subsets of $\mathbb{R}$ such that $U\cap V=\emptyset$ and $U\cup V=\mathbb{R}$. Furthermore assume that $U\neq\emptyset$ and take any $x_{0}\in U$. Then (since $U$ is open) there is $r_{0}\in\mathbb{R}$ such that the open ball

 $B(x_{0},r_{0})=\{x\in\mathbb{R}\ |\ |x-x_{0}|

is contained in $U$. Consider the set

 $R=\{r\in\mathbb{R}^{+}\ |\ B(x_{0},r)\subseteq U\}.$

Thus $R$ is nonempty.

Assume that $R$ is bounded. Denote by $s=\mathrm{sup}(R)<\infty$. We can apply the lemma:

 $\bigcup_{r\in R}B(x_{0},r)=B(x_{0},s).$

Thus (due to the definition of $R$) $B(x_{0},s)$ is a maximal open ball (with the center in $x_{0}$) which is contained in $U$. Now

 $B(x_{0},s)=(a,b)$

for some $a,b\in\mathbb{R}$. Since $(a,b)$ is maximal then $a\not\in U$ or $b\not\in U$. Indeed, if both $a\in U$ and $b\in U$, then (since $U$ is open) small neighbourhoods of $a$ and $b$ are also contained in $U$, so $(a-\epsilon,b+\epsilon)$ is contained in $U$ (for some $\epsilon>0$), but $(a,b)$ was maximal. Contradiction.

Without loss of generality we can assume that $b\not\in U$. Then $b\in V$, because $U\cup V=\mathbb{R}$. But then (since $V$ is open) there is $c\in\mathbb{R}$ such that $a and $c\in V$. Thus $U\cap V\neq\emptyset$. Contradiction. Therefore $R$ is unbounded.

Take any unbounded sequence $(a_{n})_{n=1}^{\infty}$ from $R$. Then we have

 $\mathbb{R}=\bigcup_{n=1}^{\infty}B(x_{0},a_{n})\subseteq U$

and thus $U=\mathbb{R}$, so $V=\emptyset$. This completes the proof. $\square$

Corollary. For any $a,b\in\mathbb{R}$ such that $a intervals $(a,b)$, $[a,b)$, $(a,b]$ and $[a,b]$ are connected.

Proof. One can easily show that intervals are continous image of $\mathbb{R}$ and therefore intervals are connected.

Title intervals are connected IntervalsAreConnected 2013-03-22 18:32:49 2013-03-22 18:32:49 joking (16130) joking (16130) 5 joking (16130) Example msc 54D05