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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.

Chiquita
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  • Idea for loc. path connected at only one point: Hawaiian earring with rational radii – G. Chiusole Apr 06 '20 at 08:34
  • 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 06 '20 at 08:36
  • @G.Chiusole Hawaiian earring is locally path connected at every point, regardless of radii. – freakish Apr 06 '20 at 08:37
  • 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 06 '20 at 08:43
  • 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 06 '20 at 10:53

2 Answers2

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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.

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

mathcounterexamples.net
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  • 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 06 '20 at 07:34
  • 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 06 '20 at 07:43