Supremum and Infimum absolute values

Question

Let $X \subset \mathbb{R}$. Assume $X$ contains positive and negative numbers and assume $X$ is nonempty and bounded. Show $\sup(|X|) = -\inf(X)$ if $\sup(X) < \sup(|X|)$ where $|X| = \{|x| : x \in X\}$.

Any hints on where to even begin? We know $\inf(X) < 0 $ so -$\inf(X) > 0$.

Answer 1

After your observation we have that if $\sup(X)<\sup(|X|)$ than $\sup(|X|)$ must be different from $\sup(X)$, so it must taken by a negative $x\in X$, in particular the "biggest" in modulus, that is $x=\inf(X)$.

Answer 2

An alternative statement of this theorem is that $\sup|X|=\max(-\inf X,\sup X)$. Then it must equal one of the two values, so if $\sup|X|\ne\sup X$ then $\sup|X|=-\inf X$.

We can substitute $-\inf X=\sup(-X)$, because $x\mapsto-x$ is an order isomorphism from $(\Bbb R,\le)$ to $(\Bbb R,\ge)$.

To show $\max(\sup(-X),\sup X)\le\sup|X|$, we need $\sup(-X)\le\sup|X|$ and $\sup X\le\sup|X|$, which is true because the sets are "pointwise" less, that is $x\le|x|$ and $-x\le|x|$ for each $x\in X$, which implies that each upper bound of $|X|$ is an upper bound of $X$ and $-X$.

Conversely, to show $\sup|X|\le\max(\sup(-X),\sup X)$ we need $|x|\le\max(\sup(-X),\sup X)$ for all $x\in X$. If $x\ge 0$, then $|x|=x\le\sup X\le\max(\sup(-X),\sup X)$; and if $x\le 0$ then $|x|=-x\le\sup(-X)\le\max(\sup(-X),\sup X)$.

Answer 3

Let's try to treat this problem by focusing on the most complex part, expanding the various definitions, and then simplifying, and let's see where that leads us.$% \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\Ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow} \newcommand{\abs}[1]{\left| #1 \right|} %$

Working in $\;\mathbb R\;$, our task is to prove for non-empty bounded $\;X\;$ that $$ \tag{0} \sup(X) \lt \sup(\abs{X}) \;\then\; \sup(\abs{X}) = -\inf(X) $$ with the additional assumptions that $\;\sup(X) > 0\;$ and $\;\inf(X) < 0\;$.

Now, the simplest definitions of $\;\sup\;$ and $\;\inf\;$ I know are that, for all $\;z\;$, \begin{align} \tag{1} \sup(S) \le z &\;\equiv\; \langle \forall s : s \in S : s \le z \rangle \\ \tag{2} z \le \inf(T) &\;\equiv\; \langle \forall t : t \in T : z \le t \rangle \\ \end{align} for any non-empty upper-bounded $\;S\;$ and non-empty lower-bounded $\;T\;$. And to prove equalities through this definitions, properties like the following are useful: $$ \tag{3} x = y \;\equiv\; \langle \forall z :: x \le z \;\equiv\; y \le z \rangle $$ which says that two numbers are the same if they have the same upper bounds.

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Looking at $\Ref{0}$, we see that the most complex expression there is $\;\sup(\abs{X})\;$. So let's investigate its upper bounds: for any $\;z\;$, we calculate $$\calc \sup(\abs{X}) \le z \op\equiv\hint{expand definition $\Ref{1}$ of $\;\sup\;$} \langle \forall s : s \in \abs{X} : s \le z \rangle \op\equiv\hint{expand definition of $\;\abs{X}\;$} \langle \forall s : \langle \exists x : x \in X : s = \abs{x} \rangle : s \le z \rangle \op\equiv\hint{logic: merge quantifiers} \langle \forall s,x : x \in X \land s = \abs{x} : s \le z \rangle \op\equiv\hint{logic: one-point rule} \langle \forall x : x \in X : \abs{x} \le z \rangle \op\equiv\hint{expand definition of $\;\abs{x}\;$} \langle \forall x : x \in X : -z \le x \land x \le z \rangle \op\equiv\hint{logic: distribute $\;\land\;$ over $\;\forall\;$} \langle \forall x : x \in X : -z \le x \rangle \;\land\; \langle \forall x : x \in X : x \le z \rangle \tag{*} \op\equiv\hint{definition $\Ref{2}$ of $\;\inf\;$; definition $\Ref{1}$ of $\;\sup\;$} -z \le \inf(X) \;\land\; \sup(X) \le z \op\equiv\hint{arithmetic} -\inf(X) \le z \;\land\; \sup(X) \le z \op\equiv\hint{property of $\;\max\;$} \max \left\{-\inf(X), \sup(X) \right\} \le z \endcalc$$ Therefore, by $\Ref{3}$, this longish but straightforward calculation shows that $$ \tag{5} \sup(\abs{X}) \;=\; \max \left\{-\inf(X), \sup(X) \right\} $$ And from this, with the definition of $\;\max\;$, $\Ref{0}$ immediately follows.

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Note how until $\Ref{*}$, the above calculation is only about expanding definitions and simplifying. The rest is directly inspired by $\Ref{0}$.

Finally, note that we need did not use the fact that "$\;X\;$ contains positive and negative numbers", so that $\;\sup(X) > 0\;$ and $\;\inf(X) < 0\;$.