How to show $-\sup(-A)=\inf(A)$?

Question

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.

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 $$ -\sup(-A) \leq a < -\sup(-A) + \epsilon $$ — Ben Grossmann, Sep 26 '14 at 14:44
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. — Martin Sleziak, Sep 26 '14 at 15:27
In fact both are equal, see http://math.stackexchange.com/q/392129/11994. — MarnixKlooster ReinstateMonica, Apr 28 '16 at 09:44

score 4 · Answer 1 · answered Sep 26 '14 at 14:45

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 $\inf(A)$.

score 2 · Answer 2 · answered Sep 26 '14 at 14:47

2

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.

answered Sep 26 '14 at 14:47

lhf

221,500

2

Or, you could call it an order-preserving map $(\mathbb{R},<)\rightarrow(\mathbb{R},>)$ - i.e. going from the order to its dual – Milo Brandt Dec 13 '14 at 01:59

Milo Brandt · Answer 3 · 2014-12-13T02:11:08.647

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 $-\inf(-B)=\sup(B)$.

score 0 · Answer 4 · answered Dec 13 '14 at 01:52

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, $\alpha = \inf(A) = -\sup(-A) = -(-\alpha) = \alpha$.

score 0 · Answer 5 · edited Oct 19 '20 at 18:33

0

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

edited Oct 19 '20 at 18:33

kimchi lover

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answered Oct 12 '17 at 17:29

Sharlotta Neimor

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How to show $-\sup(-A)=\inf(A)$?

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