Why does the existence of a dispersion point imply total path disconnectedness?

Question

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?

Answer 1

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 $x\neq x_0$.

Answer 2

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.