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In a comment of Fractal of the topologist's sine curve is connected and totally path-disconnected? M W asserts that the existence of a dispersion point, a point for which the removal of results in a totally disconnected space, implies that the space is totally path disconnected (even before the removal). Why must this be the case?

Steven Clontz
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3 Answers3

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This is not true in general. One needs to assume $X$ is at least $T_1$ in addition.

Ulli has already given a $T_0$ counterexample and a proof for Hausdorff spaces.

Let me add a proof in the case $X$ is only $T_1$.

In this case, suppose $x_0$ is a dispersion point of $X$, so $X\backslash \{x_0\}$ is totally disconnected. Suppose by contradiction $\gamma\colon [0,1]\to X$ is a nonconstant path in $X$. Then since $\{x_0\}$ is closed, $\gamma^{-1}(\{x_0\})$ is closed.

Since $\gamma^{-1}(\{x_0\})$ is closed, its complement is a disjoint union of open intervals, and by total disconnectedness of $X\backslash \{x_0\}$, $\gamma$ is constant on each complementary interval. But then the preimage of any point $x\in \gamma([0,1])\backslash \{x_0\}$ is open, hence not closed, (since $\gamma$ is nonconstant and $[0,1]$ is connected), contradicting the $T_1$ property of $X$.

Remark

We cannot replace $T_1$ by any weaker condition here, since if $X$ is not $T_1$, then for some distinct $x,y\in X$ we have $x\in \overline{\{y\}}$, at which point $x$ and $y$ can be connected by the path in Ulli's counter-example: $$\gamma(t)= \begin{cases} x & t=0\\ y & t\neq 0. \end{cases} $$ On the other hand, the existence of a dispersion point $x_0$ already implies $X\backslash \{x_0\}$ is $T_1$, so that verifying the $T_1$ condition in specific cases reduces to establishing that $x_0\notin \overline{\{x\}}$ and $x\notin \overline{\{x_0\}}$ for arbitrary $x\neq x_0$.

M W
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    @M W: oh yes, about your remark, it didn't come to my mind: $X$ totally path disconnected $\Rightarrow$ T1 – Ulli Nov 15 '23 at 12:00
  • Thanks to both of you - this T1 result is a little tighter so I suppose it gets the green checkmark. And I should have thought to query https://topology.pi-base.org/spaces?q=t_0%2B~Totally+Path+Disconnected%2BHas+a+Dispersion+Point before asking the question; thanks to you both we'll soon have a theorem showing https://topology.pi-base.org/spaces?q=t_1%2B~Totally%20Path%20Disconnected%2BHas%20a%20Dispersion%20Point is impossible. – Steven Clontz Nov 15 '23 at 22:20
  • @MW In the last paragraph, without extra assumption it does seem to be true that the complement of the dispersion point singleton is Hausdorff. For example one can take a non-Hausdorff totally disconnected space $Y$ (https://topology.pi-base.org/spaces?q=Totally%20Disconnected%2B~T2) and construct a space $X$ by adding a point $p$ and taking the open sets of $X$ to be the unions of an open set in $Y$ with ${p}$, together with the empty set. $p$ is then a dispersion point but its complement is not Hausdorff. – PatrickR Nov 16 '23 at 20:20
  • @PatrickR yes, I changed the post earlier today for that reason [earlier I had falsely been thinking totally disconnected meant Hausdorff]- perhaps you are looking at a previous version? – M W Nov 16 '23 at 20:28
  • Ah, yes. I was indeed looking at an older version. – PatrickR Nov 16 '23 at 23:45
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    As far as I can tell, these results can be generalized as follows. Say that a space is path-explodable, if there exists a point $x_0 \in X$ such that $X \setminus {x_0}$ is totally pathdisconnected. Then the proof in this answer works similarly to show that a path-explodable T1 space is totally path-disconnected. However, a totally path disconnected space is also path-explodable and T1, so there is equivalence. It is also possible to generalize the concepts and equivalence from T1 to R0 once singletons are replaced with trivial subsets in all concepts. – kaba May 03 '24 at 03:07
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This is true, if $X$ is T2:

Let $x_0$ be the dispersion point, and asssume $A \subset X$ is path-connected, $|A| \ge 2$. Then $x_0 \in A$. Then, by T2, there is a $B \subset A$ such that $x_0 \in B$ and $B$ is homeomorphic to $[0, 1]$. Then $x_0$ is dispersion point of $B$ as well, which is a contradiction, since $[0, 1]$ has no dispersion point.

If $X$ is only T0, this is false:

Let $X \neq \emptyset$ be an arbitrary set, $x_0 \in X$. $\tau := \{U \subset X: x_0 \in U\} \cup \{\emptyset\}$ is a topology on $X$ with dispersion point $x_0$. Of course, $X$ is T0.
$X$ is path-connected: If $x \in X$, then $f: [0, 1] \rightarrow X, \space f(r) = \left\{\begin{array}{ll} x, \text{ if } r = 0 \\ x_0, \text{ if } r \neq 0 \\ \end{array} \right. \space $ is a path.

Remark

In case of T2, we also have the following elegant conclusions:

$X$ has a dispersion point $\Rightarrow$ $X$ is SG $\Rightarrow$ $X$ is punctiform (i.e., the only non-empty, connected, compact subsets are the singletons) $\Rightarrow$ $X$ is totally path disconnected

However, this does not hold in case of T1, as the one-point compactification of the rationals shows. Hence, for the T1 case, M W's proof is definitely relevant.

Ulli
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The following theorem generalizes those in previous answers:

Theorem: If a T1 space is covered by two totally pathdisconnected subsets, one of which is closed, then the space is totally pathdisconnected.

This theorem also generalizes the claim; note that totally disconnected implies totally pathdisconnected.

It can be proved by using similar ideas as in the previous answers. The most important property that is used in the proofs is that an open subset of a real interval is a disjoint union of (pathconnected) (subspace-)open intervals.

Aside: a similar result does not hold for totally disconnected spaces

kaba
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