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In $R^2$. I was thinking of two disjoint sets, $A = \{(x,y)| y \ge 1, x > 0\}$ and $B = \{(x,0); x \ge 0 \}$. So I'm not really sure how to construct an uncountable collection of sets such as these.

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  • $\begingroup$ Your $A$ isn't closed . . . $\endgroup$ Feb 25, 2017 at 0:57

4 Answers 4

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In a T1 space (of which a metric space like $\mathbb{R}^2$ is an example) every point set is closed, so it's enough to consider $\mathbb{R}^2$ as the disjoint union of its uncountably many points

$$ \mathbb{R}^2 = \bigcup_{x \in \mathbb{R}^2}\, \{x\}. $$

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    $\begingroup$ This is clearly the correct answer. $\endgroup$ Feb 25, 2017 at 1:08
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For example, in $\mathbb R$, let $\mathbb I$ be the irrationals. Each singleton $\{i\}\in \mathbb I$ is, of course, closed, and disjoint with every other irrational. So $\{\{i\}, i\in\mathbb I\}$ is an uncountable collection of disjoint closed sets.

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    $\begingroup$ . . . or just look at the reals. :P $\endgroup$ Feb 25, 2017 at 0:57
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How about lines?

Each line "$x=c$" (for $c$ some constant) is closed, and if $c\not=d$ then the lines $x=c$ and $x=d$ don't intersect.

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Here's one $$A:= \left\{(x,f(x)),f: \mathbb R \to \mathbb R,\:\: \: f:= \left\{ \begin {array}{ccc} =1 &if& x\in \mathbb Q\\ =0 & & o.w. \end{array} \right.\right\}$$

Note that each element is a singleton in $\mathbb R^2$. And are obviously disjoint. As the set of singltons in $\mathbb R$ is uncountable this implies that $A$ as a collection of uncountable disjoint sets.

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