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If $A_1,\ldots,A_n$ are simply connected open subsets of $X$ and $A_i\cap (\bigcup_{j\neq i} A_j)$ is simply connected for each $i$, then can we show $A_i \cap (\bigcup_{j>i} A_j)$ is still simply connected for each $i$?

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No, here is a counterexample.

Let $X = \mathbb{R}^2$, and consider

  • $A_1 = \{(x,y) : -1 < x < 1, -0.5 < y < 1\}$,
  • $A_2 = \{(x,y) : -1 < x < 1, -1 < y < 0.5\}$,
  • $A_3 = \{(x,y) : -1.5 < x < -0.5, -1 < y < 1\}$,
  • $A_4 = \{(x,y) : 0.5 < x < 1.5, -1 < y < 1\}$.

Or in a picture ($A_1$ is red, $A_2$ is green, $A_3$ is blue and $A_4$ is purple):

enter image description here

It is easy to check that this satisfies $A_i \cap \bigcup_{j \neq i} A_j$ for all $i$, but we have that $A_2 \cap \bigcup_{j > 2} A_j$ is disconnected.

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