# Prove that if A is bounded in $\mathbb{R}$, then $\inf(A) \leq \sup(A)$

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, $$\inf(A) \leq \sup(A)$$

• What is the meaning of $$\inf(A) =-b \leq A \leq b = \sup(A)$$? Apr 27 at 12:26
• By definition, for all $x\in A$, $\inf(A)\leq x\leq \sup(A)$...
– Surb
Apr 27 at 12:26
• 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 $A=\emptyset$. Apr 27 at 12:27
• It depens which sense you give to the comparison of $-\infty$ and $\infty$. @P.Quinton Apr 27 at 12:35
• 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 $-\infty$). Apr 27 at 12:58