# simply connectedness [closed]

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 $$i$$?

## 1 Answer

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):

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.