Let B={-x : x ϵ A}, A a non-empty subset of R. Show B is bounded below

**juanma101285** · November 28th 2011, 03:08 PM

Hi, I have the following problem... I know the second bit of it, but I have no clue how to prove the first bit. I know B is bounded below (it is basically set A but with opposite signs, so the supremum of A will be the lowest number of set B, its infimum), but I don't know how to express it mathematically. Thanks for any help you can give me!

"Let A be a non-empty subset of R that is bounded above. Let

B = {-x : x ϵ A}.

Show that B is bounded below. How are sup A (supremum of A) and inf B (infimum) related?"

I don't know how to prove that B is bounded below, but for the question "How are sup A (supremum of A) and inf B (infimum) related?", I have the following:

sup A = -inf B.

**Plato** · November 28th 2011, 03:53 PM

Originally Posted by juanma101285

"Let A be a non-empty subset of R that is bounded above. Let
B = {-x : x ϵ A}.
Show that B is bounded below. How are sup A (supremum of A) and inf B (infimum) related?" sup A = -inf B.

If $\alpha=\sup(A)$ then if $t\in B$ then $-t\in A$ .
So $-t\le\alpha$ or $-\alpha\le t$ .
Thus $-\alpha\le\beta=\inf(B)$

Suppose that $-\alpha<\beta$ then $\alpha>-\beta$ so $\left( {\exists x \in A} \right)\left[ { - \beta < x \leqslant \alpha } \right]$
What is wrong with that?

**juanma101285** · November 29th 2011, 06:41 AM

Thanks for the message!

Ermm... so that would be a contradiction because that would mean x is sup(A) and -x is inf(B)??

**Plato** · November 29th 2011, 06:50 AM

Originally Posted by juanma101285

Thanks for the message!

Ermm... so that would be a contradiction because that would mean x is sup(A) and -x is inf(B)??

Well it means $-x\in B$ and $-x<\beta~.$
Can that happen?

**HallsofIvy** · November 29th 2011, 07:05 AM

Let a be an upper bound of A. Let x be any member of B. Then x= -y for some y in A. Then y< a. If follows that -y= x> -a. Thus, -a is a lower bound for B.

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