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$A$ is a finite or countable union of pairwise disjoint intervals.

Discussion in 'Mathematics' started by user539807, Oct 8, 2018.

  1. user539807

    user539807 Guest

    Let $A\subseteq\mathbb{R}$ be non-empty and bounded. Also, $\forall x\in A$, $\exists\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subseteq A$. Prove that $A$ is a finite or countable union of pairwise disjoint open intervals.

    I'm aware that this question has many different proofs, but I'm interested in knowing how one would go about proving it in this particular way.

    For $q\in\mathbb{Q}\cap A$, define

    $\alpha(q):= \text{inf}\{x\in\mathbb{R}:(x,q]\subseteq A\}$; $\beta(q):= \text{sup}\{x\in\mathbb{R}:[q,x)\subseteq A\}$

    First, why do $\alpha$ and $\beta$ exist and $\alpha<\beta$? Then, how can one show that $I_q:=(\alpha(q),\beta(q))\subseteq A$? Lastly, how can one prove that

    • $\bigcup_{q\in\mathbb{Q}\cap A}I_q=A$
    • $\forall q, s\in\mathbb{Q}\cap A$, either $I_q=I_s$ or $I_q\cap I_s=\emptyset$

    Thanks in advance.

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