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Let $A$ and $B$ two subsets of $\mathbb{R}$ such that : $A \cup B = \mathbb{R}$ and $A \cap B = \emptyset$.

We suppose that $A$ has a supremum that we denote $\alpha$ , and $B$ has an infimum that we denote $\beta$.

Show that $\alpha = \beta$

My attempt :

By definition of the sup :

$\forall \epsilon > 0 : \alpha + \epsilon \notin A$

So :

$\forall \epsilon > 0 : \alpha + \epsilon \in B \ \Longrightarrow \forall \epsilon > 0 : \alpha + \epsilon \geq b$

Hence $ \alpha \geq \beta $

I've been stuck for the past hours on how to achieve the second part of the inequality (i.e $\alpha \leq \beta$ )

Any help is appreciated.

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    $\begingroup$ This is not true without further hypotheses: consider $A = (-\infty, 0) \cup \{1\}$ (and $B = \mathbb{R} \setminus A$). $\endgroup$
    – Naïm Favier
    Commented Jul 7 at 19:16

1 Answer 1

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You're having trouble with this because it's false. Let $B=\{0 \} \cup (1, \infty).$ Let $A= \Bbb R \setminus B$. Then $\sup A=1$ but $\inf B = 0$. For this to be true, you need the additional condition that $A$ is closed downward and $B$ is closed upward.

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  • $\begingroup$ What does "closed downward" mean? $\endgroup$
    – Adam Rubinson
    Commented Jul 7 at 19:48
  • $\begingroup$ does that mean B should be an interval of the form $[\gamma , +\infty)$ $\endgroup$
    – Adam Boussif
    Commented Jul 7 at 19:53
  • $\begingroup$ @AdamRubinson $A$ is closed downward if $(x \in A \land y \lt x) \Rightarrow y \in A$. $\endgroup$
    – Robert Shore
    Commented Jul 7 at 20:26

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