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This proof has surely been addressed before on here, but I'm writing the proof from scratch, without looking at other variants of it, and hoping that someone could critique it, as I'm still somewhat new to proof-writing and very new to analysis. All comments on content, style, elegance (especially lack thereof, in my case) and so forth are very much welcome and appreciated.

Theorem. Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all real numbers $-x$, where $x \in A$. Prove that $\inf A = - \sup \left(-A\right)$.

Proof. Let $A$ be a nonempty subset of $\mathbb{R}$, and $-A$ to consist of the negations, $-x$, of all $x \in A$. The infinium of $A$ has the properties that \begin{align*} \forall x \in A, \inf A \leq x \ \wedge \ \forall y\in A, \left(y \leq x \implies y \leq \inf A \right), \end{align*} i.e., $\inf$ is both a lower bound of $x$ and the greatest of such lower bounds. Multiplying these expressions through by $-1$, which yields elements in $-A$, gives \begin{align*} \forall -x \in -A, -\inf A \geq -x \ \wedge \ \forall -y \in -A, \left(-y \geq -x \implies -y \geq - \inf A \right), \end{align*} so $-\inf A$ is an upper bound of $x$ and the smallest of such upper bounds, in which case we have $-\inf A = \sup \left(-A\right)$. Multiplying this equation through by $-1$ yields our result, $\inf A = - \sup \left(-A\right)$.

How does this look?

Gregory J. Puleo
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2 Answers2

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Your mathematical definition of infimum is not quite right, although you describe it correctly in English.

"$\inf A$ is a lower bound of $A$" can be written as $(\inf A) \le x$, $\forall x \in A$, as you have written.

However, "$\inf A$ is the greatest of all lower bounds" should be written as something like $\forall y \in \mathbb{R}$, $(y \le x, \forall x \in A \implies y \le \inf A)$.

Otherwise, your general approach (negating terms and reversing inequalities) will work.

angryavian
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  • Thank you for this: this is very helpful. One quick follow-up question: would it be possible to write this as $\forall x \in A, (\inf A \leq x \wedge \forall y \in \mathbb{R}, y \leq x \implies y \leq \inf A)$? –  Aug 21 '18 at 17:05
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    @Matt.P I don't think you can "pull out" the "$\forall x \in A$" like that. For example if $A = (0, \infty)$ and $x = 4$, then for $y = 3$ you are saying $3 \le 4$ implies $3 \le \inf A$ which is not true. – angryavian Aug 21 '18 at 18:02
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Characteristic for $\inf A$ are:

  • $\forall a\in A\;[\inf A\leq a]$
  • $\forall b\in\mathbb R\;[\forall a\in A\;[b\leq a]\implies b\leq\inf A]$

Under the first bullet it is stated that $\inf A$ is a lower bound of $A$, and under the second bullet it is stated that for every lower bound $b$ of $A$ satisfies $b\leq\inf A$.

drhab
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