# proof of characterization of connected compact metric spaces.

First we prove the right-hand arrow: if $X$ is connected then the property stated in the Theorem^{} holds. This implication^{} is true in every metric space $X$, without additional conditions.

Let us denote with ${A}_{\epsilon}$ the set of all points $z\in X$ which can be joined to $x$ with a sequence of points ${p}_{1},\mathrm{\dots},{p}_{n}$ with ${p}_{1}=x$, ${p}_{n}=z$ and $$. If $z\in {A}_{\epsilon}$ then also ${B}_{\epsilon}(z)\subset {A}_{\epsilon}$ since given $w\in {B}_{\epsilon}(z)$ we can simply add the point ${p}_{n+1}=w$ to the sequence ${p}_{1},\mathrm{\dots},{p}_{n}$. This immediately shows that ${A}_{\epsilon}$ is an open subset of $X$. On the other hand we can show that ${A}_{\epsilon}$ is also closed. In fact suppose that ${x}_{n}\in {A}_{\epsilon}$ and ${x}_{n}\to \overline{x}\in X$. Then there exists $k$ such that $\overline{x}\in {B}_{\epsilon}({x}_{k})$ and hence $\overline{x}\in {A}_{\epsilon}$ by the property stated above. Since both ${A}_{\epsilon}$ and its complementary set are open then, being $X$ connected, we conclude that ${A}_{\epsilon}$ is either empty or its complementary set is empty. Clearly $x\in {A}_{\epsilon}$ so we conclude that ${A}_{\epsilon}=X$. Since this is true for all $\epsilon >0$ the first implication is proven.

Let us prove the reverse implication. Suppose by contradiction^{} that $X$ is not connected. This means that two non-empty open sets $A,B$ exist such that
$A\cup B=X$ and $A\cap B=\mathrm{\varnothing}$. Since $A$ is the complementary set of $B$ and vice-versa, we know that $A$ and $B$ are closed too. Being $X$ compact we conclude that both $A$ and $B$ are compact sets.
We now claim that

$$\delta :=\underset{a\in A,b\in B}{inf}d(a,b)>0.$$ |

Suppose by contradiction that $\delta =0$.
In this case by definition of infimum^{}, there exist two sequences
${a}_{k}\in A$ and ${b}_{k}\in B$ such that $d({a}_{k},{b}_{k})\to 0$. Since $A$ and $B$ are compact, up to a subsequence we may and shall suppose that ${a}_{k}\to a\in A$ and ${b}_{k}\to b\in B$. By the continuity of the distance function we conclude that $d(a,b)=0$ i.e. $a=b$ which is in contradiction with the condition $A\cap B=\mathrm{\varnothing}$. So the claim is proven.

As a consequence, given $$ it is not possible to join a point of $A$ with a point of $B$. In fact in the sequence ${p}_{1},\mathrm{\dots},{p}_{n}$ there should exists two consecutive points ${p}_{i}$ and ${p}_{i+1}$ with ${p}_{i}\in A$ and ${p}_{i+1}\in B$. By the previous observation we would conclude that $d({p}_{i},{p}_{i+1})\ge \delta >\epsilon $.

Title | proof of characterization^{} of connected compact metric spaces. |
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Canonical name | ProofOfCharacterizationOfConnectedCompactMetricSpaces |

Date of creation | 2013-03-22 14:17:06 |

Last modified on | 2013-03-22 14:17:06 |

Owner | paolini (1187) |

Last modified by | paolini (1187) |

Numerical id | 4 |

Author | paolini (1187) |

Entry type | Proof |

Classification | msc 54A05 |