This question was inspired to me by Lukas Geyer’s recent question.

A positive answer to this question would easily entail a positive answer to Lukas’ question also, and a negative answer would probably be informative as well.

Let $T=]0,1[^2$ be the open unit square. Let $(S_k)_{k\geq 1}$ be a countable family of open segments in $T$. What we call a cycle is a finite sequence $M_1,M_2, \ldots ,M_r$ of distinct points in $T$, not all on a line (thus $r>2$) such that all the segments $]M_1M_2[,]M_2M_3[, \ldots ,]M_{r-1}M_r[$, and also $]M_rM_1[$, are each included in one of the segments $S_k$, and likewise all the endpoints $M_1,M_2, \ldots ,M_r$ are each in one of the $S_k$. Formally, for each $i\in{\mathbb Z}/r{\mathbb Z}$ there is an index $k(i)$ with $]M_iM_{i+1}[ \subseteq S_{k(i)}$ and there is an index $l(i)$ such that $M_i \in l(i)$.

What we call a cut is a finite sequence $M_1,M_2, \ldots ,M_r$ of distinct points such that the first and the last, $M_1$ and $M_r$, are on the boundary of $T$, and the intermediate points $M_2,M_3, \ldots ,M_{r-1}$ are in $T$, and all the segments $]M_1M_2[,]M_2M_3[, \ldots ,]M_{r-2}M_{r-1}[, ]M_{r-1}M_r[$, are each included in one of the segments $S_k$, and likewise all the intermediate points $M_2,M_3, \ldots ,M_{r-1}$ are each in one of the $S_k$.

So the question is as follows : assume that there are no cycles or cuts. Does it follow that

$$ Z=T \setminus \bigg(\bigcup_{k\geq 1} S_k \bigg) $$

is necessarily path-connected ?

Update 16:15 : The definitions have been corrected so as to avoid the uninteresting counterexample explained in Dan Shved’s answer.

Update 19:10 : Dan Shved’s second counterexample strikes a much stronger blow to my hopes, because it shows a family $(S_k)$ such that $T \setminus (\cup_{k \geq 1} S_k)$ is not path-connected, and yet $T\setminus (\cup_{1 \leq k \leq n} S_k)$ is for every finite $n$. So there is no “finite” sufficient condition.


First of all, with the current definition of a cycle the whole situation is impossible, because there is always at least one cycle. Just take two points $M_1$ and $M_2$ on one of the segments, and sequence $M_1,\,M_2$ will be a cycle.

However, if we use the intuitive notion of a cycle, then the answer to your question will be negative. In fact, there is even a finite family $(S_k)$ that has no cuts nor cycles (in the intuitive sense) and such that $Z$ isn't connected.

Take $$ S_1 = \left]\left(0,\frac{1}{2}\right), \left(\frac{1}{2}, \frac{1}{2}\right)\right[,\\ S_2 = \left]\left(\frac{1}{2},0\right), \left(\frac{1}{2}, \frac{1}{2}\right)\right[,\\ S_3 = \left]\left(\frac{1}{4},\frac{1}{4}\right), \left(\frac{3}{4}, \frac{3}{4}\right)\right[. $$

They cut the square into two pieces.

Update: Here is a less trivial example that really has no cuts, even with the improved definition.

Take $S_1$ and $S_2$ from the example above. Instead of $S_3$, we will add a countable family of segments $(S_k)_{k \ge 3}$, specified by these formulas for each $n \ge 1$: $$ S_{2n+1}=\left]\left( \frac{1}{4} , \frac{1}{2}+\frac{1}{2n+2} \right), \left( \frac{1}{2} + \frac{1}{2n+1}, \frac{1}{2}+\frac{1}{2n+2}\right)\right[,\\ S_{2n+2}=\left]\left( \frac{1}{2}+\frac{1}{2n+2} ,\, \frac{1}{4} \right), \left( \frac{1}{2}+\frac{1}{2n+2} ,\, \frac{1}{2} + \frac{1}{2n+1}\right)\right[. $$ Let us show that with this family $(S_k)_{k \ge 1}$ excluded from $T=]0,1[^2$ there is no way to escape from $U=\left]0,\frac{1}{2}\right[^2 \cup \left\{\left(\frac{1}{2},\frac{1}{2}\right)\right\}$.

It will be convenient to use an additional mapping $g:T \to \mathbb{R}$, $(a,b) \to \max \{a,b\}$. It is easy to check that $g$ is continuous. Also, $U$ is exactly the set of those $M \in Z$ that satisfy the inequality $g(M)\le \frac{1}{2}$.

Suppose there is a path, i.e. a continuous map, $f\colon [0,1] \to Z$, and for some $t_0 \in [0,1]$ we have $f(t_0) \in U$. Let us show that there exists an $\varepsilon >0$ such that for every $t \in B_\varepsilon(t_0)$: $f(t) \in U$. (By $B_\varepsilon(t_0)$ I mean an open ball in $[0,1]$, i.e. the set $\{ t |\,t \in [0,1]\, \& \, |t - t_0| < \varepsilon \}$).

OK, now there are two possibilities: either $g(f(t_0)) < \frac{1}{2}$ or $g(f(t_0)) = \frac{1}{2}$. In the former case our statement follows directly from the continuity of $g \circ f$. Therefore, we may assume that $g(f(t_0)) = \frac{1}{2}$. Obviously, this means that $f(t_0)=\left(\frac{1}{2},\frac{1}{2}\right)$. Since $f$ is continuous, there exists an $\varepsilon > 0$ such that for every $t \in B_\varepsilon(t_0)$ we have $f(t) \in \left]\frac{1}{4},\frac{3}{4}\right[^2$.

Suppose that there exists a $t_1 \in B_\varepsilon(t_0)$ such that $f(t_1) \not \in U$. This means that $g(f(t_1)) > \frac{1}{2}$. Then there is a natural $n$ such that $g(f(t_1)) > \frac{1}{2} + \frac{1}{2n+2}$. And so we have $g(f(t_0)) < \frac{1}{2} + \frac{1}{2n+2}$, $g(f(t_1)) > \frac{1}{2} + \frac{1}{2n+2}$. By the intermediate value theorem there exists a $t_2 \in [t_0,t_1]$ such that $g(f(t_2)) = \frac{1}{2} + \frac{1}{2n+2}$. Also remember that $f(t_2) \in \left]\frac{1}{4},\frac{3}{4}\right[^2$. It follows that $f(t_2)$ lies on one of the segments $S_{2n+1}$ or $S_{2n+2}$. This is a contradiction.

So, we have proved that for every $t_0 \in [0,1]$ if $f(t_0) \in U$, then there exists a $\varepsilon > 0$ such that $f(t) \in U$ for every $t \in B_\varepsilon(t_0)$. This means that the set $X = \{t \in [0,1] |\, f(t) \in U\}$ is open in $[0,1]$. It is also quite easy to see from the continuity of $f$ that $X$ is closed in $[0,1]$. Since $[0,1]$ is connected, we have that $X$ is either $\varnothing$ or $[0,1]$. So no path can connect a point from $U$ to a point from $Z\setminus U$.

  • $\begingroup$ Thanks for the feedback. I don’t believe your example destroys my ideas completely though, all they need is a little bit of tidying. $\endgroup$ – Ewan Delanoy Dec 3 '12 at 15:13
  • $\begingroup$ @Yes, this answer is purely technical. BTW, with the updated definition of cycles (with $r>2$) there is still always at least one cycle. Instead of $2$ points on one of the segments we can just take $3$ points on one segment, and this will be a cycle :( $\endgroup$ – Dan Shved Dec 3 '12 at 16:25
  • $\begingroup$ Aargh! I must correct this also. $\endgroup$ – Ewan Delanoy Dec 3 '12 at 16:41
  • $\begingroup$ About your last counterexample : I guess that any path passing through $(1/2,1/2)$ must meet one of the $S_{2k}$ or $S_{2k+1}$, but I’m not completely convinced. Can you elaborate a little please ? $\endgroup$ – Ewan Delanoy Dec 3 '12 at 16:49
  • $\begingroup$ @EwanDelanoy I've added a proof of the "no escape" statement to my answer. $\endgroup$ – Dan Shved Dec 3 '12 at 17:45

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