[Math] Totally disconnected metric space

general-topologymetric-spaces

I want to prove that a certain metric space is totally disconnected.

In a metric space context this is the same as saying that every connected component is a singleton.

I think another way of proving that a space is TD is proving that there is a proper, nonempty open and closed set. Is that right?

Please let me know any alternative equivalent definitions you might now.

Cheers!

Best Answer

The following two facts concerning totally disconnected spaces should (separately) help you demonstrate that your space is totally disconnected.

Fact 1: Every T$_1$-space with a basis consisting of clopen sets (i.e., every zero-dimensional space) is totally disconnected.

proof. Suppose that $A \subseteq X$ contains at least two points, and let $x , y \in A$ be distinct. By assumption there is a clopen $U \subseteq X$ such that $x \in U$ and $y \notin U$. But then $U$ and $X \setminus U$ witness that $A$ is not a connected subset of $X$. $\quad\Box$

Fact 2: Every product of (nonempty) totally disconnected spaces is totally disconnected.

proof. Suppose that $X_i$ is totally disconnected for all $i \in I$, and let $A \subseteq \prod_{i \in I} X_i$ contain at least two points. Then there must be a $j \in I$ such that $A_j = \{ x_j : x = ( x_i )_{i \in I} \in A \}$ contains at least two points. As $X_j$ is totally disconnected, there are open $U_j , V_j \subseteq X_j$ such that $U_j \cap A_j \neq \emptyset \neq V_j \cap A_j$ and $U_j \cap V_j \cap A_j = \emptyset$ and $A_j \subseteq U_j \cup V_j$. Let $$U = {\textstyle \prod_{i \neq j}} X_i \times U_j; \quad V = {\textstyle \prod_{i \neq j}} X_i \times V_j.$$ Then $U , V$ are open subsets of $\prod_{i \in I} X_i$, $U \cap A \neq \emptyset \neq V \cap A$, $U \cap V \cap A = \emptyset$ and $A \subseteq U \cup V$. Thus $A$ is not a connected subset of $\prod_{i \in I} X_i$. $\quad\Box$

Either of these should be useful (but especially the second) because your space appears to be a subspace of $\{ 1 , \ldots , k \}^{\mathbb{N}}$ taking $\{ 1 , \ldots , k \}$ to be discrete, and then taking the product topology. (Also, total disconnectedness is a hereditary property of topological spaces.)

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