# Fractal of the topologist's sine curve is connected and totally path-disconnected?

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

• 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.]
– M W
Nov 14, 2023 at 21:53
• topology.pi-base.org/spaces/S000125/properties is in the pi-Base but the totally path-disconnected property is missing. Nov 15, 2023 at 2:59
• Likely because most of its properties came from Steen/Seebach's Counterexamples, which similarly leaves this property open. Nov 15, 2023 at 3:05
• Clarification on previous comment- this implication also uses that $\mathbb R^2$ is $T_1$. math.stackexchange.com/q/4807248/1210477
– M W
Nov 15, 2023 at 8:26
• Thanks to the discussion in mathse:4807248, S125 now appears in OP's search. Nov 17, 2023 at 3:25