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)<r\}.

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 $d(x,y)<r_{0}$ 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 $d(x,y)<r_{0}$ and therefore $y\in B(x,r_{0})\subset\bigcup_{r\in R}B(x,r)$ , which completes the proof. $\square$

Proposition. 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}|<r_{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 $a<c<b$ 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 $a<b$ 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
Canonical name	IntervalsAreConnected
Date of creation	2013-03-22 18:32:49
Last modified on	2013-03-22 18:32:49
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	5
Author	joking (16130)
Entry type	Example
Classification	msc 54D05