I'm not sure how to approach this proof? any ideas
Let $A$ be a set of intervals of the real line any two of which are disjoint - in other words, if $(a,b)$ and $(x,y)$ are distinct elements of $A$ then $(a,b)\cap(x,y)=\emptyset$. Prove that A is countable. *(Use the fact that $\mathbb Q$ is countable)
Set of intervals on the real line
2
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elementary-set-theory
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2How many intervals in $A$ can contain the given rational number? – 2012-03-22
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5Hint: You can assign to each $(a,b) \in A$ an element $q_{(a,b)} \in (a,b) \cap \mathbb Q$. This gives a map $f:A \to \mathbb Q$. What can you say about $f$? – 2012-03-22
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0@martini the function f is surjective? i still don't understand what you mean by q(a,b)∈(a,b)∩Q. Thanks – 2012-03-22
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0No, $f$ is never surjective. By $q_{(a,b)} \in (a,b) \cap \mathbb Q$ I mean the following: $(a,b) \subseteq \mathbb R$ is a non-empty intervall, so there is a rational number $q_{(a,b)}$ with $a < q_{(a,b)} < b$, i. e. it is contained in the intersection of $(a,b)$ with $\mathbb Q$, so $q_{(a,b)} \in (a,b) \cap \mathbb Q$ (if the index to $q$ is confusing you, for different intervalls there will be different rational numbers [as your intervalls are disjoint]. $f$ is now defined by $f\bigl((a,b)\bigr) := q_{(a,b)}$. – 2012-03-22
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0List the rationals as $r_1,r_2,\dots$. For any of our intervals $(a,b)$, let $f((a,b))$ be the least integer $i$ such that $r_i$ is in $(a,b)$. – 2012-03-22
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0@martini Thanks for clearing that up. Therefore f is injective function? – 2012-03-22
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0@AndréNicolas thanks – 2012-03-22
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0Yes. And now, since there is an injective $f\colon A \to \mathbb Q$ and the fact you gave in your question ... – 2012-03-22
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0@martini A is countable? thanks allot. – 2012-03-22
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0@Jenn Yes, $A$ is countable. – 2012-03-22
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1A similar question: [Every collection of disjoint non-empty open subsets of $\mathbb R$ is countable?](http://math.stackexchange.com/questions/75781/every-collection-of-disjoint-non-empty-open-subsets-of-mathbbr-is-countable). – 2012-06-25