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Let $F\subset X$ be a closed and connected subset of the metric space $(X,d)$ then for every pair of points $a, b\in F$ and each $\epsilon>0$ there are points $a=z_0,z_1,\ldots z_n=b$ in $F$ such that $d(z_{k-1}-z_k)<\epsilon$ for $1\leq k\leq n$.

Is the hypothesis that $F$ is closed needed?

Gaston Burrull
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  • This seems not to be true. Take $F={x}=X$. – user29999 Aug 14 '12 at 21:43
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    This seems similar to the "finite chain condition" as a characterization of connected sets (closed or not). See the second answer to this question. – David Mitra Aug 14 '12 at 21:45
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    @user29999: The statement is true in your example: $n=0$, $z_0=x$, and there is no $k$ such that $1\le k\le 0$, so the final condition is vacuously satisfied. – Brian M. Scott Aug 14 '12 at 21:56
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    Am I missing someting, but doesn't this work: Fix a point a. Consider the set of points b for which such a chain exists for a fixed $\epsilon$; this set is open (everything in a $\epsilon$ ball around a point in the set is in the set) and closed (everything in a $\epsilon$ ball around a point not in the set must also not be in the set, or the original point would be in the set.)

    I'm scared to post this as an answer, because it seems wrong to me...

     – only Aug 14 '12 at 22:09
  • @only I think that you are right. – Gaston Burrull Aug 14 '12 at 22:14

2 Answers2

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$F$ can be any connected subset of $X$. Given $\epsilon>0$, let $\mathscr{U}=\{F\cap B(x,\epsilon/2):x\in F\}$, where $B(x,r)$ is the open $d$-ball of radius $r$ centred at $x$. Now apply the chain characterization of connectedness proved in this answer (mentioned by David Mitra in the comments) to conclude that for any $a,b\in F$ there is a finite sequence $x_0=a,x_1,\dots,x_n=b\in F$ such that $B(x_k,\epsilon/2)\cap B(x_{k+1},\epsilon/2)\ne\varnothing$ for $k=0,\dots,n-1$. Clearly $d(x_k,x_{k+1})<\epsilon/2+\epsilon/2=\epsilon$ for $k=0,\dots,n-1$.

Community
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Brian M. Scott
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You're right, this hypothesis is useless.

Let $F \subset (X,d)$ a connected subset. Let $\epsilon > 0$ and $a, b \in X$. I'm going to prove that an $\epsilon$-chain exists between $a$ and $b$. Let $\Omega \subset F$ the subset of all $\omega \in X$ such that an $\epsilon$-chain exists between $a$ and $\omega$.

$\Omega$ is clearly open: if $\omega \in \Omega$, then $B(x,\epsilon/2) \subset \Omega$. It is also closed (in $F$): if $\omega_n$ is a sequence of elements of $\Omega$ converging to $x \in F$, one of the $\omega_n$ is at distance $< \epsilon/2$ of $x$, and that implies $x \in \Omega$. The connectedness of $F$ then implies that $\Omega$ (which is nonempty: $a \in \Omega$) is all of $F$ : $F$ is $\epsilon$-connected for all $\epsilon >0$.

PseudoNeo
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