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Let $\emptyset\neq A\subseteq\mathbb{R}$ a bounded set. Consider $-A=\{-a:a\in A\}$. I want to prove that $-\sup(-A)=\inf(A)$.

It is easy to see that $-\inf(A)$ is an upper bound of $-A$, so $\sup(-A)\le -\inf(A)$, then $-\sup(-A)\ge \inf(A)$.

How can we prove that $-\sup(-A)\le \inf(A)$?

Thanks.

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    $\begingroup$ Don't you want to say, "Consider $-A = \{-a: a \in A \}$? $\endgroup$ – graydad Sep 26 '14 at 14:42
  • $\begingroup$ So, you've shown that $-\sup(-A)$ is a lower bound for $A$. Can you show that this lower bound is "tight"? That is, try to show that for any $\epsilon$, there is an $a$ such that $$ -\sup(-A) \leq a < -\sup(-A) + \epsilon $$ $\endgroup$ – Omnomnomnom Sep 26 '14 at 14:44
  • $\begingroup$ Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. You can find more tips for choosing a good title here. $\endgroup$ – Martin Sleziak Sep 26 '14 at 15:27
  • $\begingroup$ In fact both are equal, see math.stackexchange.com/q/392129/11994. $\endgroup$ – Marnix Klooster Apr 28 '16 at 9:44
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You should work directly from the definitions of $\sup$ and $\inf$. That is, prove that if $x = \inf(A)$ then $-x \ge b$ for all $b \in {-A}$ (i.e. $-x$ is an upper bound for $-A$) and that if $y \ge b$ for all $b \in {-A}$ then $y \ge {-x}$ (that is, $-x$ is a least upper bound). This verifies that $-x$ is the sup of $-A$, and its proof uses the (similar) definition of $\inf(A)$.

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The map $x \mapsto -x$ is an order-inverting bijection $\mathbb R \to \mathbb R$.

It sends lower bounds for $A$ to upper bounds for $-A$, and vice-versa, hence the result.

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    $\begingroup$ Or, you could call it an order-preserving map $(\mathbb{R},<)\rightarrow(\mathbb{R},>)$ - i.e. going from the order to its dual $\endgroup$ – Milo Brandt Dec 13 '14 at 1:59
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You can round out the argument by using a nice symmetry. In particular, you already have that: $$-\sup(-A)\geq \inf(A)$$ and easily, by the same reasoning, that $$-\inf(-B)\leq \sup(B)$$ Now, if we set $A=-B$, then we get, from the first inequality, that: $$-\sup(B)\geq \inf(-B)$$ then negating both sides: $$\sup(B)\leq -\inf(-B)$$ but we can simply tack on the second inequality $$\sup(B)\leq -\inf(-B)\leq \sup(B)$$ which implies $-\inf(-B)=\sup(B)$.

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My approach starts out similar to @CliveNewstead's but then I let $-\gamma = \sup(-A)$ and used the fact that the infimum is unique.

Since $A$ is nonempty and bounded below, $A = \{a:a\in A\}$ and $\inf(A) = \alpha$. Now, $-A = \{-a:a\in A\}$ is also nonempty. Since $\alpha$ is the infimum of $A$, $\alpha\leq a$ for all $a\in A$. By multiplying by $-1$, we get the following inequality $$ -\alpha\geq -a. $$ That is, $-\alpha$ is an upper bound of $-A$. Suppose $-\gamma = \sup(-A)$ and $\varepsilon > 0$. Then $-\gamma + \varepsilon\not\in -A$ $$ -\alpha\geq -\gamma + \varepsilon\geq -\gamma\geq -a $$ Again, by multiplying by negative one, we have $$ \alpha\leq \gamma - \varepsilon\leq\gamma\leq a $$ but $\gamma - \varepsilon\notin A$ so $\gamma$ is a lower bound of $A$ which would contradict the fact that $\alpha$ is the greatest lower bound of $A$. In order for $\gamma$ to be the lower bound, $\gamma = \alpha$ since the infimum is unique. So $-\alpha$ is the supremum of $-A$. Therefore, $\alpha = \inf(A) = -\sup(-A) = -(-\alpha) = \alpha$.

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Basiclly by using the definition of supremum and infinum we can eseally see that: $$ sup(A) \Leftarrow\Rightarrow \exists c_0 = min\{c|\forall a \in A: a \le c\} $$ $$ sup(-A) \Leftarrow\Rightarrow \exists c_1 = min\{c|\forall \bar a \in -A: \bar a \le c\} = min\{c|\forall a \in A: -a \le c\} $$ $$ -sup(-A) \Leftarrow\Rightarrow \exists c_2 = max\{c|\forall a \in A: a \ge c\} = inf(A) $$

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