We know that components of locally connected sets are open. Now consider the space $$X=\bigcup\limits_{q \in \mathbb{Q}}^{} \{(x,y) \in \mathbb{R}^2; x^2+y^2=q^2 \}-\{(0,0) \}$$ Now this seems locally connected, since every point belongs to a circle and you can just take smaller and smaller arcs around that point. However the components are the circles $x^2+y^2=q^2$ which aren't open. Edit: $q \in \mathbb{Q}^{+}$ not $q \in \mathbb{Q}$
2026-03-29 14:55:54.1774796154
Where is my reasoning wrong?
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It’s not locally connected at all: any point on the circle has basic neighbourhoods that are intersection of open balls in the plane with $X$, and these will contain many points from other rational-radius circles. So your arcs aren’t neighbourhoods at all.
You treat the circles like they’re a topological sum, not as a subspace of the plane.