# 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 $A_{\varepsilon}$ the set of all points $z\in X$ which can be joined to $x$ with a sequence of points $p_{1},\ldots,p_{n}$ with $p_{1}=x$, $p_{n}=z$ and $d(p_{i},p_{i+1})<\varepsilon$. If $z\in A_{\varepsilon}$ then also $B_{\varepsilon}(z)\subset A_{\varepsilon}$ since given $w\in B_{\varepsilon}(z)$ we can simply add the point $p_{n+1}=w$ to the sequence $p_{1},\ldots,p_{n}$. This immediately shows that $A_{\varepsilon}$ is an open subset of $X$. On the other hand we can show that $A_{\varepsilon}$ is also closed. In fact suppose that $x_{n}\in A_{\varepsilon}$ and $x_{n}\to\bar{x}\in X$. Then there exists $k$ such that $\bar{x}\in B_{\varepsilon}(x_{k})$ and hence $\bar{x}\in A_{\varepsilon}$ by the property stated above. Since both $A_{\varepsilon}$ and its complementary set are open then, being $X$ connected, we conclude that $A_{\varepsilon}$ is either empty or its complementary set is empty. Clearly $x\in A_{\varepsilon}$ so we conclude that $A_{\varepsilon}=X$. Since this is true for all $\varepsilon>0$ 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 $A\cap B=\emptyset$. 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

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

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 $A\cap B=\emptyset$. So the claim is proven.

As a consequence, given $\varepsilon<\delta$ it is not possible to join a point of $A$ with a point of $B$. In fact in the sequence $p_{1},\ldots,p_{n}$ 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 $d(p_{i},p_{i+1})\geq\delta>\varepsilon$.

Title proof of characterization of connected compact metric spaces. ProofOfCharacterizationOfConnectedCompactMetricSpaces 2013-03-22 14:17:06 2013-03-22 14:17:06 paolini (1187) paolini (1187) 4 paolini (1187) Proof msc 54A05