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Does there exist a locally connected space which can be covered by two non-locally-connected closed subsets?

Notes

kaba
  • 2,881
  • Take a singleton space. Your two proposed non-locally connected sets don't exist. – Brevan Ellefsen Jun 26 '24 at 14:24
  • My problem statement was ambiguous. The question is whether there exists a space like that, not whether every locally connected space has that property. Fixed the question. – kaba Jun 26 '24 at 14:28

1 Answers1

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Yes. Let

$$ \begin{aligned} X & = \{(x, x \sin(1/x)) : x \in (0, 1]\} \cup \{(0, 0)\}, \\ \mathcal{T} & = \mathcal{T}_{\mathbb{R}}^2|X, \\ V_0 & = \{(x, y) \in X : y \geq 0\}, \\ V_1 & = \{(x, y) \in X : y \leq 0\}. \end{aligned} $$

Then $V_0, V_1$ are closed, cover $X$, and fail to be locally connected at $(0, 0)$, but still $\mathcal{T}$ is locally connected. The same example works for locally path-connected too.

kaba
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  • What does $\mathcal{T}_{\mathbb{R}}^2|X$ mean? Do you just mean that $X$ should have the subspace topology from $\mathbb{R}^2$? – diracdeltafunk Jun 26 '24 at 20:37
  • Yes, I mean the usual subspace topology of the Euclidean topology on $\mathbb{R}^2$. – kaba Jun 26 '24 at 20:38