# Supremum of a set

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#1
Say you have a set A. Its supremum is 1 and its infimum is -2.
You also have a set B, which is defined as B = {-a : a ∈ A}.

So for all values b in the set B, b = -a.
It's clear that 2 is an upper bound for the set B by a bit of rearranging and using inequalities.

How do I show that 2 is the least upper bound (i.e. supremum) of the set B?

Thanks!
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#2
Bumping this post
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2 years ago
#3
(Original post by NathCave)
Bumping this post
You posted this in the wrong section I believe hence why you're not getting any replies. Anyway, the proof is similar to what you have done for the first bit (sort of like going the other direction). If u is an upper bound for A then a ≤ u and - u ≤ -a for all a in A. Thus -u is a lower bound for B so -u ≤ -2 or 2 ≤ m. Hence sup(A) = 2.
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2 years ago
#4
I've moved this to the maths study help forum for you
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