# Connectedness problem: sequences of points with distances at most $\varepsilon$

Here's problem 6.1.D (a), page $359$ from Engelking's book, stuck with it for a while.

Verify that if a space $X$ with the topology induced by a metric $p$ is connected, then for every pair $x,y$ of points of $X$ and any $\varepsilon >0$ there exists a finite sequence $x_{1},x_{2},..,x_{k}$ of points of $X$ such that $x_{1}=x$, $x_{k}=y$ and $p(x_{i},x_{i+1})<\varepsilon$ for $i=1,2,..,k-1$.

• Needless to say, the converse is false. – PseudoNeo Jul 5 '11 at 18:52
• @PseudoNeo: what is a counterexample? – user10 Jul 5 '11 at 19:14
• Try the rationals in [0,1]. This is extremally disconnected but it enjoys this "lilypad property". Could adding completeness to the hypothesis give a converse? – ncmathsadist Jul 5 '11 at 19:21
• @ncmathsadist: compactness is enough (if disconnected, the two pieces must be a positive distance apart), but completeness is not (for example, the union of the x axis and the curve $xy=1$ in the plane). – Chris Eagle Jul 5 '11 at 19:49
• – Martin Sleziak Aug 31 '16 at 10:40

Hint. Fix a point $x\in X$. Now put $$Q = \{y\in X| \exists x_1, x_2, \cdots ,x_n\in X \hbox{ with } p(x_k,x_{k+1}) < \epsilon\}.$$ Show that $Q$ is both open and closed in $X$. Your result will follow right away.
This is a typical application of the chain lemma from my answer here. Use the open cover by balls of radius $\varepsilon$, and use the centres plus intersection points.