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I should prove or give a counterexample for:

$\partial (\bigcup_{i \in \mathbb{N}} A_i) \subset \bigcup_{i \in \mathbb{N}} (\partial A_i)$

Where $\partial$ is the boundary and (X,d) is a metric space with $A_{i} \subset X $ for each $i \in \mathbb{N}$

I cannot make up a counterexample for finite sets $A_{i}$ but I think there might be one for infinite ones, or am I wrong?

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@User3123: Your $U_{i}$ should be $A_{i}$ i think –  anonymous Nov 21 '10 at 12:18
    
Yep, I changed it. –  Listing Nov 21 '10 at 12:26

2 Answers 2

up vote 1 down vote accepted

you can take the metric space $\mathbb{R}$ and the sets $A_i = \{\frac{1}{i}\}$. the boundary of $\bigcup A_i = \{\frac{1}{i} \mid i\in \mathbb{N}\}=A$ is $A\cup \{0\}$ while $\partial A_i = A_i$ so $\bigcup \partial A_i = A $

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Let $X = \{ 1/i , i \in \mathbb N \}$ considered as a metric sub-space of $\mathbb R$. Let $A_i = \{ 1/i\}$. The RHS is empty whereas LHS is the set $\{ 0\}$.

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