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.

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

Edit: let's look a little closer at the following:

$$G\geq|\inf(X)|\implies -G\leq \inf(X)$$ 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, $-G\leq \inf(X)$.

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    $\begingroup$ You mean $G \ge |\inf X|$... $\endgroup$ – Reinstate Monica Mar 8 '17 at 21:24
  • $\begingroup$ @vrugtehagel why is it $G\leq |\inf (X)|$? as the commenter above noted, shouldn't it be $G\geq |\inf(X)|$? $\endgroup$ – lorenzo Mar 8 '17 at 21:27
  • $\begingroup$ Of course. Thanks for correcting me. $\endgroup$ – vrugtehagel Mar 8 '17 at 21:29
  • $\begingroup$ @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? $\endgroup$ – lorenzo Mar 8 '17 at 21:37
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    $\begingroup$ @lorenzo I edited the post to explain that in a little more detail. I hope this clears things up! $\endgroup$ – vrugtehagel Mar 8 '17 at 22:09

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