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Can you verify my proof to the statement "If $A$ is bounded in $\mathbb{R}$, then $\inf(A) \leq \sup(A)$"?

Solution: Since $A$ is bounded in $\mathbb{R}$, there is $0 < B \in \mathbb{R}$ such that $\left| A \right| \leq B$ or $-B \leq A \leq B$. Hence, $B$ and $-B$ are upper bound and lower bound of $A$, respectively. By the Completeness Axiom, $b=\sup(A)$ and $-b=\inf(A)$ both exist. Thus, the following also holds $$\inf(A) =-b \leq A \leq b = \sup(A).$$ Therefore, $\inf(A) \leq \sup(A)$

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    $\begingroup$ What is the meaning of $$\inf(A) =-b \leq A \leq b = \sup(A)$$? $\endgroup$
    – Kavi Rama Murthy
    Apr 27 at 12:26
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    $\begingroup$ By definition, for all $x\in A$, $\inf(A)\leq x\leq \sup(A)$... $\endgroup$
    – Surb
    Apr 27 at 12:26
  • $\begingroup$ Doesn't this hold for any $A\neq\emptyset$ ? Observe that $\sup(A) < \inf(A)$ means that all upper bound on $A$ are strictly less than all lower bounds on $A$, if $a$ is a lower bound and $b$ a upper bound then for all $x\in A$, $a\leq x\leq b < a$ which is false for any $x\in A$, therefore $A=\emptyset$. $\endgroup$
    – P. Quinton
    Apr 27 at 12:27
  • $\begingroup$ It depens which sense you give to the comparison of $-\infty$ and $\infty$. @P.Quinton $\endgroup$
    – nicomezi
    Apr 27 at 12:35
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    $\begingroup$ No, this is not a good proof. If $b=\sup A$, it is not guaranteed that $-b=\inf A$. Just take a set that is not symmetrical to the origin, say: $A=(-1, 2), \sup A=2, \inf A=-1$. Also, I think the problem should specify that $A$ is nonempty, because an empty set is bounded but has no infimum (or alternatively has infimum $+\infty$) and has no supremum (or alternatively has supremum $-\infty$). $\endgroup$
    – Stinking Bishop
    Apr 27 at 12:58

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