Let $X$ be a connected compact metric space. Given a positive $\varepsilon$ and two points $x,y\in X$ we write $x\sim_\varepsilon y$ if there exists a sequence $C_1,\dots,C_n$ of connected subsets of diameter $<\varepsilon$ in $X$ such that $x\in C_1$, $y\in C_n$ and $C_i\cap C_{i+1}\ne\emptyset$ for all $i<n$. It is clear that $\sim_\varepsilon$ is an equivalence relation on $X$. The equivalence class $[x]_\varepsilon:=\{y\in X:y\sim_\varepsilon x\}$ will be called the $\varepsilon$-connected component of $x$.

Problem. What can be said about the $\varepsilon$-connected components of $X$? In particular, is each $\varepsilon$-connected component $[x]_\varepsilon$ dense in $X$? Is each $\varepsilon$-connected component $\sigma$-compact?

Added in Edit. It can be shown that each $\varepsilon$-connected component $[x]_\varepsilon$ is $\sigma$-compact. To prove this fact, fix a countable base $\mathcal B$ of the topology of $X$, which is closed under finite unions and let $\mathcal B_\varepsilon$ be the subfamily of $\mathcal B$ consisting of basic sets of diameger $<\varepsilon$. It can be shown that each connected subset of diameter $<\varepsilon$ is contained in the connected component of some set $\bar B$ with $B\in\mathcal B_\varepsilon$. Now fix an enumeration $\{B_n\}_{n\in\omega}$ if the family $\mathcal B_\varepsilon$. Given any point $x\in X$ let $C_0=\{x\}$ and for every $n\in\mathbb N$ let $C_n$ be the union of $C_{n-1}$ with all connected components of $\bar B_n$ that intersect $C_{n-1}$. It can be shown that the sets $C_n$ are compact and $[x]_\varepsilon=\bigcup_{n\in\omega}C_n$.

  • $\begingroup$ Isn't $[x]_\varepsilon$ always $\sigma$-compact since the closure of a connected set is connected and has the same diameter? $\endgroup$ Aug 14, 2018 at 15:20
  • $\begingroup$ Oh sorry that in no way implies it's a union of countably many compact sets, just that it's the union of some family of compact sets. $\endgroup$ Aug 14, 2018 at 15:25
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    $\begingroup$ Let $X$ be the union of the vertical interval $0\times[-2,2]$ together with a snake $\{\,(x,\sin\tfrac1x)\mid 0<x\le 1\}$. Note that $[x]_\varepsilon$ is not dense in $X$ for small enuf $\varepsilon$. $\endgroup$ Aug 14, 2018 at 18:12
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    $\begingroup$ The pseudo-arc is another example where $[x]_\varepsilon$ is not dense for small enough $\varepsilon$ (but the reasons are a little more complicated -- basically, you use the fact that the pseudo-arc is hereditarily indecomposable to show that $[x]_\varepsilon$ has diameter at most $2\varepsilon$.). $\endgroup$
    – Will Brian
    Aug 14, 2018 at 20:15
  • $\begingroup$ After all these counterexamples, I have modified the question in this MO-post: mathoverflow.net/questions/308309/…? $\endgroup$ Aug 14, 2018 at 20:55

1 Answer 1


I believe $[x]_\varepsilon$ is always $\sigma$-compact. First consider a slightly modified definition where we replace 'diameter $<\varepsilon$' with 'diameter $\leq \varepsilon$' and call the corresponding set $[x]_{\leq\varepsilon}$.

Note that $[x]_\varepsilon = \bigcup_{\delta < \varepsilon}[x]_{\leq\delta}$, since being in $[x]_\varepsilon$ is witnessed by a finite sequence of sets whose diameters must have a uniform bound below $\varepsilon$. So we also have $[x]_\varepsilon = \bigcup_{n}[x]_{\leq(\varepsilon -2^{-n})}$ so if we can show that $[x]_{\leq\varepsilon}$ is always $\sigma$-compact, then we'll have that $[x]_\varepsilon$ is always $\sigma$-compact as well.

So the second claim, as mentioned in the comments, is that we can restrict our attention to closed $C_i$ in the definition. This is because the closure of a connected set is connected and has the same diameter.

Fix a $\varepsilon > 0$. For any closed set $F$, let $F^\ast = \bigcup\{G:G\text{ closed, connected, }G\cap F\neq \varnothing\text{, diam}(G)\leq\varepsilon\}$. Now notice that $\{G:G\text{ closed, connected, }G\cap F\neq \varnothing\text{, diam}(G)\leq\varepsilon\}$ is a closed subset of hyperspace (connectedness, having non-empty intersection with $F$, and having diameter $\leq \varepsilon$ are all closed conditions in hyperspace). This implies that $F^\ast$ is also a closed set, since the union of a closed family of sets is closed (since $X$ is compact).

Then some definition chasing tells us that $[x]_{\leq\varepsilon}$ is $\{x\}\cup\{x\}^\ast\cup\{x\}^{\ast\ast}\cup\dots$, so that in particular $[x]_{\leq \varepsilon}$ is a $\sigma$-compact set.

Therefore $[x]_\varepsilon$ is $\sigma$-compact as well.

As for the density question, I suspect that you can use the boundary bumping theorem to show that $[x]_\varepsilon$ always contains the composant of $x$, which would imply that it's always dense since composants are always dense.

The topologist's sine curve gives an example of a non-dense $\varepsilon$-connected component. Specifically the $\varepsilon$-connected component of any point in the line segment is not dense for sufficiently small $\varepsilon$.

  • $\begingroup$ After your two comments I have realized how to prove the $\sigma$-compectness of $\varepsilon$-connected components and after writing this argument (in Edit) have seen your answer. Now I do not know what to do (of course I can delete my comment in Edit). $\endgroup$ Aug 14, 2018 at 15:49
  • $\begingroup$ Your argument for the $\sigma$-compactness is very nice and clear. Thank you. Concerning the density of $\varepsilon$-connected components I am still not convinced. Could you provide a more precise argument? $\endgroup$ Aug 14, 2018 at 15:53
  • $\begingroup$ Actually now I think there might be a counterexample to denseness. I'm still thinking about it. $\endgroup$ Aug 14, 2018 at 16:02
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    $\begingroup$ Oh of course, the topologist's sine curve is a counterexample for sufficiently small $\varepsilon$. $\endgroup$ Aug 14, 2018 at 16:06
  • $\begingroup$ Perfect! (But not what I expected). Then please add this counterexample to your answer and I shall accept it. $\endgroup$ Aug 14, 2018 at 16:19

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