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The sin(1/x) curve is notoriously a subset of the plane which is connected but not path-connected, because of its "singularity" at the origin. I think we can make another curve which is like a fractal based on that curve, which roughly speaking has such singularities everywhere, and is still connected but has no nontrivial paths at all.

Definitions

We start with $w: \Bbb{R} \to \Bbb{R}, x \mapsto 0$ (for $x \le 0$), $\sin(1/x)$ (for $x > 0$). This function is left continuous, and right continuous except at $0$. There's a decreasing sequence $x_i \to 0$ with $f(x_i) = f(0) = 0$, and also a decreasing sequence $x'_i \to 0$ with $f(x'_i) = 1$. Clearly $w[\Bbb{R}] = [-1, 1]$.

Pick some countable dense set $D \subset \Bbb{R}$, $D = \{d_i\}_0^\infty$ with the $d_i$ distinct, and some $(\epsilon_i)_0^\infty$ with all $\epsilon_i > 0$ and $\Sigma_{j > i} \epsilon_j < (1/3) \epsilon_i$. For each $i$, define $f_i: \Bbb{R} \to \Bbb{R}, x \mapsto \epsilon_i w(x - d_i)$. For any $x$, $\Sigma^\infty|f_i(x)| \le \Sigma^\infty \epsilon_i$, which is finite, so we can define $f = \Sigma f_i$. Also define $f_{< i} = \Sigma_{j < i} f_j$ and note that $f_{< i} \to f$ uniformly. Define $G$ to be the graph of $f$. Also, define $\pi_0: G \to \Bbb{R}$ to be projection onto the $x$-axis.

Claim. $G$ is connected and totally path-disconnected.

Proof. See Lemma 1 and Lemma 5 below. QED

Question. Does the proof below work?

Bonus questions. $G$ is kind of a "fake line": $\pi_0$ gives a bijective continuous map $G \to \Bbb{R}$; from the proof below, this map's inverse (although not continuous) preserves connected subsets; removing any point of $G$ separates it into two components, and in particular $G$ has empty interior. This makes it somewhat like the original construction of the pseudo-arc, which is totally path-disconnected but has some of the properties of an arc. The pseudo-arc is compact and homogeneous, which $G$ is not: $G$ appears to be locally compact nowhere, and locally connected precisely off the singularities. Does the pseudo-arc have any other properties similar to $G$? For example, what happens if we remove a point from it? Are there other examples of nontrivial planar spaces which are connected and totally path-disconnected? (I see none in pi-base.)

Lemma 0. $f$ is not right continuous at any $d_i$.

Proof. (Near $d_i$, $f$ is a continuous function plus a singularity plus a small perturbation.) Take $f_{> i} = \Sigma_{j > i} f_j$ and note $f = f_{< i} + f_i + f_{> i}$. Each $f_j$ is continuous off $d_j$, so $f_{< i}$ is continuous at $d_i$: get $\delta > 0$ with $|f_{< i}(x) - f_{< i}(d_i)| < (1/6)\epsilon_i$ for $|x - d_i| < \delta$. Also, $|f_{> i}(x)| < (1/3)\epsilon_i$ for all $x$. $f_i$ is just a shifted and scaled copy of $w$, so get a decreasing sequence $x_s \to d_i$ with $f_i(x_s) = \epsilon_i$ for all $s$. Now for sufficiently large $s$,

$$\begin{align} & |f(x_s) - f(d_i)| = \\ & |f_i(x_s) - f_i(d_i) + f_{< i}(x_s) - f_{< i}(d_i) + f_{> i}(x_s) - f_{> i}(d_i)| \ge \\ & (\epsilon_i - 0) - |f_{< i}(x_s) - f_{< i}(d_i) + f_{> i}(x_s) - f_{> i}(d_i)| \ge \\ & \epsilon_i - |f_{< i}(x_s) - f_{< i}(d_i)| - |f_{> i}(x_s)| - |f_{> i}(d_i)| \ge \\ & \epsilon_i(1 - 1/6 - 1/3 - 1/3) = \\ & (1/6)\epsilon_i. \end{align}$$

QED

Lemma 1. $G$ is totally path-disconnected.

Proof. (A path in the graph means continuity.) Suppose $\alpha: I \to G$ is a path. $\alpha[I]$ is connected and compact, so $\pi_0 \alpha[I]$ is some interval $[a_0, a_1]$. Take the restriction $g: \alpha[I] \to [a_0, a_1]$: it's bijective and continuous, from a compact to a Hausdorff space, so its inverse $g$ is also continuous. But $g$ is just $f$ restricted to $[a_0, a_1]$, and by Lemma 0, $f$ is not continuous on any nontrivial interval. So $a_0 = a_1$ and $\alpha[I]$ is a point. QED

Lemma 2. Suppose $X \subseteq \Bbb{R}$, and $\phi_i: X \to M$ are maps to some metric space $(M, d)$, with $\phi_i \to \phi$ uniformly. If all $\phi_i$ are left (respectively, right) continuous at $x_0 \in X$, then $\phi$ is too.

Proof. Given $\epsilon > 0$, get $i_0$ with $d(\phi_i(x), \phi(x)) < (1/3)\epsilon$ for all $i \ge i_0$ and all $x$, and get $\delta > 0$ with $d(\phi_{i_0}(x), \phi_{i_0}(x_0)) < (1/3)\epsilon$ for all $x \in (x_0 - \delta, x_0] \cap X$ (respectively, $[x_0, x_0 + \delta) \cap X$). Then for all such $x$, $d(\phi(x), \phi(x_0)) \le d(\phi(x), \phi_{i_0}(x)) + d(\phi_{i_0}(x), \phi_{i_0}(x_0)) + d(\phi_{i_0}(x_0), \phi(x_0)) < \epsilon$. QED

Lemma 3. Suppose $U \subseteq G$ is open in $G$ and $(x, f(x)) \in U$. Then for some $\delta > 0$, $\pi_0^{-1}(x - \delta, x] \subseteq U$.

Proof. Get $\epsilon > 0$ with the open ball $B((x, f(x)), \epsilon) \cap G \subseteq U$. Get $\epsilon' > 0$ with $B(x, \epsilon') \times B(f(x), \epsilon') \subseteq B((x, f(x)), \epsilon)$. All $f_{< i}$ are left continuous, so from Lemma 2, $f$ is too. So get $\delta_0 > 0$ with $f(t) \in B(f(x), \epsilon')$ for all $t \in (x - \delta_0, x]$. Now take $\delta = \min\{\delta_0, \epsilon'\}$. QED

Lemma 4. For any $x$, there's a decreasing sequence $x_s \to x$ with $f(x_s) \to f(x)$.

Proof. This is clear for all $x \notin D$, since $f$ is continuous at such $x$ by Lemma 2. So suppose $x = d_i$. Get a suitable $x_s \to x$ with $f_i(x_s) \to f_i(x)$. By Lemma 2, $f - f_i$ is continuous at $x$, so the same $(x_s)$ works for $f_i + (f - f_i) = f$. QED

Lemma 5. For any closed interval $X \subseteq \Bbb{R}$, $\pi_0^{-1}(X)$ is connected.

Proof. $|X| \le 1$ is trivial, so suppose $|X| \ge 2$ and $\pi_0^{-1}(X)$ has an open partition $A \cup B$. Note that $\{\pi_0(A), \pi_0(B)\}$ is a partition of $X$. After swapping $A$ and $B$ as needed, we can get $a \in \pi_0(A)$ and $b \in \pi_0(B)$ with $a < b$. Take $p = \inf\{x: (x, b] \subseteq \pi_0(B)\}$. $p$ exists and $p < b$ by Lemma 3, $p$ is finite because $a < b$, and $p \in \pi_0(X)$ because $X$ is closed. In fact $p \in \pi_0(A)$ by Lemma 3. So for some $\epsilon > 0$, we have $(B(p, \epsilon) \times B(f(p), \epsilon)) \cap G \subseteq A$. But also, by Lemma 4, there's a decreasing sequence $(p_s)$ in $X$ with $\pi_0^{-1}(p_s) \to \pi_0^{-1}(p)$. So for some $s_0$, $p_{s_0} - p < \min\{\epsilon, b - p\}$, which means $p_{s_0} \in \pi_0(A)$ and $p_{s_0} \in (p, b]$. But this contradicts the definition of $p$. QED

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  • 1
    $\begingroup$ Haven't had a chance to look over your construction but there is Cantor's Leaky tent: en.wikipedia.org/wiki/Knaster%E2%80%93Kuratowski_fan [The fact that removing one point makes it totally disconnected immediately implies it is already totally path-disconnected.] $\endgroup$
    – M W
    Nov 14, 2023 at 21:53
  • $\begingroup$ topology.pi-base.org/spaces/S000125/properties is in the pi-Base but the totally path-disconnected property is missing. $\endgroup$ Nov 15, 2023 at 2:59
  • $\begingroup$ Likely because most of its properties came from Steen/Seebach's Counterexamples, which similarly leaves this property open. $\endgroup$ Nov 15, 2023 at 3:05
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    $\begingroup$ Clarification on previous comment- this implication also uses that $\mathbb R^2$ is $T_1$. math.stackexchange.com/q/4807248/1210477 $\endgroup$
    – M W
    Nov 15, 2023 at 8:26
  • 2
    $\begingroup$ Thanks to the discussion in mathse:4807248, S125 now appears in OP's search. $\endgroup$ Nov 17, 2023 at 3:25

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