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 $U\eq\\emptyset$ 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 $a\ot\\in U$ or $b\ot\\in U$ . 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 $b\ot\\in U$ . 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 $U\\cap V\eq\\emptyset$ . 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.