# $X$ is bounded iff $\sup(X)$ and $\inf(X)$ are finite ($X\subseteq\mathbb{R}, X\neq\emptyset$)

I'm working on a proof of the following theorem:

"Let $X$ be a non-empty subset of $\mathbb{R}$; then $X$ is bounded iff $\sup(X)$ and $\inf(X)$ are finite".

Def.(Bounded sets) $X\subseteq\mathbb{R}$ is said to be bounded if $X\subset [-M,M]$ for some real number $M>0$.

Now, I've managed to prove the rightward implication, namely that if $X$ is bounded then $\sup(X)$ and $\inf(X)$ are finite but I'm stuck on the leftward one. For that part I thought about setting $G:=\max\{|\sup(X)|,|\inf(X)|\}$ and show that $X\subset[-G,G]$ but while it's easy to prove that $x\leq G\ \forall x\in X$ I haven't been able to show that $-G\leq x\ \forall x\in X$.

So, I would appreciate any hint/comment about how to finish this last part of the proof.

• You may wish to note that $\inf(X)\geq -G$, by your choice of $G$. – SamM Mar 8 '17 at 21:15
• @SamM that's exactly what I want to prove. – lorenzo Mar 8 '17 at 21:42

By definition, $G\geq|\inf(X)|$, and since $G\geq 0$, we must have $-G\leq \inf(X)$. Thus, since for all $x\in X$, we must have $\inf(X)\leq x$ (because that's how $\inf(X)$ is defined), we must have $-G\leq\inf(X)\leq x$ or $-G\leq x$.
$$G\geq|\inf(X)|\implies -G\leq \inf(X)$$ Firstly, if $\inf(X)$ is negative (that is, $|\inf(X)|=-\inf(X)$), then we have $$G\geq |\inf(X)|\implies G\geq -\inf(X)\implies -G\leq\inf(X)$$ if on the other hand $\inf(X)$ is positive or $0$ (that is, $|\inf(X)|=\inf(X)$), then $$G\geq |\inf(X)|\implies G\geq \inf(X)\implies -G\leq-\inf(X)\leq\inf(X)$$ so in both cases, $-G\leq \inf(X)$.
• You mean $G \ge |\inf X|$... – Reinstate Monica Mar 8 '17 at 21:24
• @vrugtehagel why is it $G\leq |\inf (X)|$? as the commenter above noted, shouldn't it be $G\geq |\inf(X)|$? – lorenzo Mar 8 '17 at 21:27
• @vrugtehagel could you also explain to me why $-G\leq \inf(X)$? I mean $G\geq |\inf(X)|\Rightarrow -G\leq -|\inf(X)|$; from here, by definition of absolute value, to get rid of the $-$ sign in front of $|\inf (X)|$ shouldn't one assume that $\inf(X)$ is negative..? Why should one assume such a thing? – lorenzo Mar 8 '17 at 21:37