Show a connected metric space (X,d) is $\epsilon$-chainable for $\epsilon >0$

$\epsilon$-chainable Definition

(X,d) is $\epsilon$-chainable if given any two points $a,b \in X$, $\exists$ a finite set $\{ x_0,x_1,...,x_n\}$ of points in X with $x_0 = a$, $x_n=b$ and $d(x_{n-1},x_n)<\epsilon$, $\forall$ n

Clopen Definition

A set is clopen $iff$ its boundary is empty

Connected Metric Definition

Let $(X,d)$ be a metric space. It is connected $iff$ it has no proper subsets which are clopen

The following are equivalent:

  • $X$ is connected

  • The only subsets of $X$ that are clopen are $\emptyset$ and $X$

  • $\nexists$ non-empty, disjoint, open sets $A$ and $B$ such that $X=A\bigcup B$

  • $\nexists$ non-empty, disjoint, closed sets $A$ and $B$ such that $X=A\bigcup B$


Let $(X,d)$ be a metric space, and let $S$ be any subset of $X$. Then $S$ is closed $iff$ $S^c$ is open.

So how would someone show this exactly?

Let the set $A$ be the set of all elements in $(X,d)$ that can be reached from $a\in X$ in a finite number of steps with length less than $\epsilon$.

Does this mean that there should be no isolated points? (an isolated point would be at distance infinity from any other point, or distance zero from itself, since $\epsilon >0$ we don't want this?)

Knowing that $X$ is connected $\implies$the only subsets of X that are clopen are $\emptyset$ and $X$

Also, by construction of A;

A is the set of all $\epsilon$-chains between any element and $a$

And, so $A^c = \emptyset$ if $A$ was $\epsilon$-chainable (because A has to have a path from all elements to a, for this to be true, the complement wouldn't have anything in it)

Leaving only $A$ as the remaining clopen set

Does this make any sense? Or should an alternative be used, and how to formalize it.


  • $\begingroup$ Your idea is good: first fix $\varepsilon$ and $a$, and define $A$ as you make. Show that $A$ is open, closed, and non-empty to conclude by connectedness. $\endgroup$ – Davide Giraudo Nov 6 '12 at 22:19
  • $\begingroup$ A set is clopen $iff$ its boundary is empty. Show that and use the following: A is non-empty because there needs to exist at least two points to have a path $\endgroup$ – user42538 Nov 6 '12 at 22:33
  • $\begingroup$ Edit: Theorem: Let $(X,d)$ be a metric space, and let $A$ be any subset of $X$. Then $A$ is closed $iff$ $A^c$ is open. However $A^c$ is closed and open. $\endgroup$ – user42538 Nov 6 '12 at 23:10
  • $\begingroup$ See also: math.stackexchange.com/q/49661 $\endgroup$ – Martin Sleziak Aug 31 '16 at 10:41
  • $\begingroup$ For reference, this is also an exercise in Joshi's Intro to General Topology p. 148. $\endgroup$ – LucasSilva Feb 10 '18 at 1:01

Fix $a\in X$ and $\varepsilon>0$. Define an $\varepsilon$-chain between $p$ and $p'\in X$ as a $n$-uple for some integer $n$ of the form $(x_1=p,x_2,\dots,x_{n-1},x_n=p)$ where $d(x_j,x_{j+1})<\varepsilon$ for all $j$. $$B:=\{b\in X,\mbox{we can find an }\varepsilon\mbox{-chain between }a\mbox{ and }b\}.$$ If we can show that $B=X$, we will be done.

  • $B$ is not empty, as $a\in B$ (take $n=1$, $a=x_2$).
  • $B$ is open. Let $b\in B$. We show that $B(b,\varepsilon)\subset B$. Let $n\in\Bbb Z_{\geqslant 1}$, $x_1,\dots,x_n$ as in the definition of $B$. Let $b'\in B(b,\varepsilon)$. Then consider $y_j=x_j$ if $1\leqslant j\leqslant n$ and $y_{n+1}=b'$ to see that $b'\in B$.
  • $B$ is closed. Let $b \in \overline B$. By definition, we can find an element $b'$ such that $d(b,b')<\varepsilon$ and $b'\in B$. If $(a,x_1,\dots,x_{n-1},b')$ is an $\varepsilon$-chain, so is $(a,x_1,\dots,x_{n-1},b',b)$, so $b\in B$.
  • $\begingroup$ Ahhh yes. Thank you. And every finite set is closed. $\endgroup$ – user42538 Nov 6 '12 at 23:59

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