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.


  • $\begingroup$ thx, corrected already $\endgroup$ – chango Feb 7 '13 at 12:53
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    $\begingroup$ The characterisation you provide in the third paragraph is again just for disconnectedness. $\endgroup$ – user642796 Feb 7 '13 at 12:56
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    $\begingroup$ Can you sketch here that certain metric space? $\endgroup$ – Berci Feb 7 '13 at 13:03
  • $\begingroup$ $\Sigma^+_A$ that is the space of all sequences taking values from 1 to k and such that $A_{x_k x_k+1}=1$ for all $k$, where $A$ is a $k \times k$ matrix of zeroes and ones. The metric is $d_\theta(x,y)= \theta^{n(x,y)}$ where $n(x,y)$ is the highest number such that the sequences coincide in all previous positions and $0<\theta<1$. $\endgroup$ – chango Feb 7 '13 at 13:23
  • $\begingroup$ That’s a subspace of a Cantor space, so it’s zero-dimensional and therefore totally disconnected. $\endgroup$ – Brian M. Scott Feb 7 '13 at 22:07

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.)


Instead of totally disconnected, your definition in fact is that of disconnected, and your equivalent definition is correct. A space is connected if and only if the only closed and open sets are the empty set and the whole space.


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