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For a metric space $X$, show that if $F \subset X$ is closed and connected then for every pair of points $a, b \in F$ and each $\varepsilon > 0$ there are points $z_0, z_1, ... , z_n \in F$ with $z_0 = a$, $z_n = b$ and $d(z_{k−1},z_k) < \varepsilon$ for
$1 \le k \le n$. Is the hypothesis that $F$ be closed needed?

Actually, I tried to prove it by contradiction, i.e. let there exists distinct $a, b \in F$ and there does not exist any finite set of points in F such that the above condition holds . Then, $\exists \varepsilon>0$ such that for any finite set of points $\{a, z_0, z_1, ..., z_n\} \in F$, there is a positive integer $k<n$ with $d(z_{k−1},z_k) > \varepsilon$ , then for any $z \in F$, $z \ne a,b$ ; either $d(a,z) < \varepsilon$ or $d(a,z) \ge \varepsilon$, now as $F$ is connected, there is a $z$ with $d(a,z) = d(b,z) = \varepsilon$, then I can't approach .

Alex Ravsky
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Rabi Kumar Chakraborty
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1 Answers1

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We don’t need the assumption that the set $F$ is closed, because we can restrict our consideration to the case when $X=F$ (and is non-empty). Fix any $\varepsilon>0$. Define an equivalence relation $\sim$ on the set $F$ as follows. For each $x,y\in F$ put $x\sim y$ iff there exist $z_0, z_1, ... , z_n \in F$ with $z_0 = x$, $z_n = y$ and $d(z_{k−1},z_k) < \varepsilon$ for $1 \le k \le n$. Let $K$ be any equivalence class of the relation $\sim$ and $x\in K$ be any point. If $y\in F$ and $d(x,y)<\varepsilon$ then $y\in K$, so the set $K$ is open in $F$. Since the partition of the set $F$ into the equivalence classes is a partition into open (in $F$) sets, each of these classes is also closed. But, since the set $F$ is connected, this is possible only when there is only one equivalence class. This means that any two points $a,b\in F$ are equivalent, that is the condition holds.

Alex Ravsky
  • 106,166
  • By saying that you can "restrict" to the case $X=F$, aren't you in fact using the fact that $F$ is closed? You need to say that open/closed balls are open/closed in $F$, and you therefore need that $F$ is open/closed. – Keen-ameteur Jul 22 '19 at 20:43
  • @Alex Ravsky, Sir, it is a question of J B Conway's book and there is a question against this problem that "Is the hypothesis F is closed is needed?" Actually, I think the hypothesis is not needed . – Rabi Kumar Chakraborty Jul 22 '19 at 21:26
  • @Alex Ravsky, Sir, can you provide me an example of a subset F which is not connected but closed and satisfies this property ??? – Rabi Kumar Chakraborty Jul 22 '19 at 21:29
  • @Keen-ameteur I don’t use closedness or opennes of $F$ in $X$ because both connectivity of $F$ and the property depend “only” on the metric space $F$. I added to the answer that I’m considering open sets in $F$. – Alex Ravsky Jul 22 '19 at 22:49
  • @RabiKumarChakraborty As I showed in my answer, the closedness of $F$ is not needed. An example of a subset $F$ of the plane $\Bbb R^2$ (endowed with the standard topology) which is not connected but closed and satisfies this property is a union $F={(x,y)\in\Bbb R^2: xy=\pm 1}$ of two hyperbolas. – Alex Ravsky Jul 22 '19 at 22:57