# 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

$$ |

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 $$ and thus $$, so $y\in B(x,\mathrm{sup}(R))$.

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

Proposition^{}. The space of real numbers is connected.

Proof. Assume that $U,V\subseteq \mathbb{R}$ are open subsets of $\mathbb{R}$ such that $U\cap V=\mathrm{\varnothing}$ and $U\cup V=\mathbb{R}$. Furthermore assume that $U\ne \mathrm{\varnothing}$ and take any ${x}_{0}\in U$. Then (since $U$ is open) there is ${r}_{0}\in \mathbb{R}$ such that the open ball

$$ |

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 $$. 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 $a\notin U$ or $b\notin U$. 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 $(a-\u03f5,b+\u03f5)$ is contained in $U$ (for some $\u03f5>0$), but $(a,b)$ was maximal. Contradiction^{}.

Without loss of generality we can assume that $b\notin U$. Then $b\in V$, because $U\cup V=\mathbb{R}$. But then (since $V$ is open) there is $c\in \mathbb{R}$ such that $$ and $c\in V$. Thus $U\cap V\ne \mathrm{\varnothing}$. Contradiction. Therefore $R$ is unbounded^{}.

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

$$\mathbb{R}=\bigcup _{n=1}^{\mathrm{\infty}}B({x}_{0},{a}_{n})\subseteq U$$ |

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

Corollary. For any $a,b\in \mathbb{R}$ such that $$ 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 |