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Be $\{A_n\}_{n \in \mathbb{N}}$ a sequence of connecteds ones such that $A_n \cap A_{n+1} \neq \emptyset \ \forall n \in {N}.$ Prove that $\cup_{n\in {N}}A_n $ is connected.

I tried to define $B_n$ as being the union of $A_1$ to $A_n$. By induction, these $B_n$ are related (as we know?). We can assume that $B_1$ is not empty and herefore contains a point $"a"$. So, all $B_n$ are related and have a point in common, and therefore the union is related. But I cannot mathematically express this demonstration.

Andrew Fernand
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1 Answers1

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Take $X \subset \bigcup_{n \in \mathbb{N}}A_n$ open and closed. Then, for $n \in \mathbb{N}$, $X \cap A_n$ and $X^\complement \cap A_n$ are open subset of $A_n$ so $X \cap A_n$ is equal to $A_n$ or is empty.

Assume that $X \cap A_n = A_n$ for one $n \in \mathbb{N}$ : then $X \cap A_{n-1}$ and $X \cap A_{n+1}$ are nonempty so $X \cap A_{n-1} = A_{n-1}$ and $X \cap A_{n+1} = A_{n+1}$. By induction, $X = \bigcup_{n \in \mathbb{N}}A_n$.

If otherwise $X \cap A_n$ is empty for all $n$, then $X$ is empty.

Thomas Guegamian
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