# Compact and connected set in $\mathbb{R}^2$ which is locally path-connected at only one point?

Given a compact and connected set $$A$$ in $$\mathbb{R}^2$$. Can it be locally path-connected at only one point? And can it be locally path-connected at every point except one?

My thought

First I thought of the topological sine curve, which is compact and connected but not locally path-connected. But it doesn't solve the problems. Then I tried to construct a space that is locally path-connected at every point except one as $$A=\{(t\cos x,t\sin x):\cos x\in \mathbb{Q} , t\in[0,1]\}$$ But this is not compact.

Do such spaces exist? Any hints would be highly appreciated.

• Idea for loc. path connected at only one point: Hawaiian earring with rational radii – G. Chiusole Apr 6 '20 at 8:34
• Cone over Cantor set is compact, path connected and locally path connected only at a single point (the top vertex). But I'm not sure about locally path connected except a single point. – freakish Apr 6 '20 at 8:36
• @G.Chiusole Hawaiian earring is locally path connected at every point, regardless of radii. – freakish Apr 6 '20 at 8:37
• I meant a union over circles in $\mathbb{R}^2$ with radius $q$ and centered at $q/2$ for all $q \in \mathbb{Q}$, so in particular at $(1,0)$ (and all others except $(0,0)$) this should not be path connected. But I just realized that this space is far from compact. – G. Chiusole Apr 6 '20 at 8:43
• A trivial example for the first is the singleton: It is obviously compact, connected and path connected, and since it consists of only one point, it trivially is only locally path connected at one point. – celtschk Apr 6 '20 at 10:53

freakish's example checks out, so credit to him. I'll just elaborate on his comment.

Let $$C \hookrightarrow \{0\} \times \mathbb{R} \subseteq \mathbb{R}^2$$ the ternary cantor set. and let $$\mathcal{C}(C)$$ be the cone over it. That is

$$\mathcal{C}(C) = (C \times I)/C \times \{1\} \subseteq \mathbb{R}^2~~.$$

Then $$\mathcal{C}(C)$$ is the quotient of a compact Hausdorff space by collapsing a closed subspace and thus also compact (and Hausdorff). The cone over any space is contractible (to the tip) and thus connected, but $$\mathcal{C}(C)$$ is locally path connected only at the tip $$t := \pi( C \times \{1\})$$ with $$\pi$$ being the projection $$C \times I \rightarrow \mathcal{C}(C)$$.

Locally path connected at the tip: Let $$U$$ be a neighborhood of $$t \in \mathcal{C}(C)$$. Then since $$\mathcal{C}(C)$$ carries the subspace topology so there is some $$\varepsilon > 0$$ s.t. $$B_{\varepsilon}(t) \cap \mathcal{C}(C) \subseteq U$$. This neighborhood is contactible via the contraction on $$\mathcal{C}(C)$$ restricted to $$B_{\varepsilon}(t) \cap \mathcal{C}(C) \subseteq U$$.

Nowhere else locally path connected: Let $$x = (x_1, x_2) \in C \times I$$ with $$x_1 \neq 1$$ and let $$\varepsilon > 0$$ s.t. $$1 \not\in (x_1 - \varepsilon, x_1 + \varepsilon)$$. Then consider the open neighborhood $$(x_1 - \varepsilon, x_1 + \varepsilon) \times C$$ of $$x$$. Since $$\{ 1 \} \times C \cap (x_1 - \varepsilon, x_1 + \varepsilon) \times C = \emptyset$$, applying $$\pi$$ to this set yields a homeomorphism of an open subset of $$\mathcal{C}(C)$$ and $$(x_1 - \varepsilon, x_1 + \varepsilon) \times C$$. The latter is not locally path connected since $$C$$ is nowhere locally path connected.

At least subspaces reduced to a single point satisfy the requirements of the question.

• As far as I am concerned, if $A$ is locally path connected at $a$, it means that for any open subset containing $a$, there is a smaller open set containing $a$, which is path-connected in the subspace topology. I can't see how to show that for any open subset containing $b$, there is a smaller open set containing $b$ which is path-connected in the subspace topology. Do I miss some point? – Chiquita Apr 6 '20 at 7:34
• You're right and I'll modify my answer accordingly. However, a subspace reduced to a single point still answer your requirements. – mathcounterexamples.net Apr 6 '20 at 7:43