Infimum

Aug 2009
639
2
How do I prove that the inf(S) = -sup(-S) where -S={-s l s in S}?

I thought the inf is the greatest lower bound and hence it should be sup (-S) ?
 

HallsofIvy

MHF Helper
Apr 2005
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7,909
How do I prove that the inf(S) = -sup(-S) where -S={-s l s in S}?

I thought the inf is the greatest lower bound and hence it should be sup (-S) ?
First, I note you have posted several questions on the "Number Theory" forum which have nothing to do with "Number Theory". Questions involving limits, inf, sup, etc. should be under "Calculus" or "Analysis".

Second, suppose S= {x| x> 1}. Obviously, its inf is 1. Now, -S= {-x| x> 1}= {y| y< -1} (I multiplied both sides of x> -1 by -1 and then let y= -x). It should be obvious that sup -S= -1. That is, inf S= 1= - sup -S, not "sup -S".

Now, as to the proof. Let a= inf(S). Then a is the "greatest lower bound" on S. Since a is a lower bound, if x is in S then \(\displaystyle x\ge a\) and, since a is the greatest lower bound, if b is any lower bound on S, \(\displaystyle a\ge b\).

Let y be any member of -S. Then x= -y is in S and so \(\displaystyle x= -y\ge a\). Multiplying both sides of that by -1, \(\displaystyle -x= y\le -a\) so -a is an upper bound on S.

To show it is the least upper bound, suppose b is any upper bound on -S. That is, if y is in -S, \(\displaystyle b\ge y\). By definition of -S, For any x in S, x= -y for some y in -S. Since \(\displaystyle b\ge y\), \(\displaystyle -b\le -y= x\). Since x could be any member of S, -b is a lower bound on S. Since a is the greatest lower bound on S, \(\displaystyle a\ge -b\). Now, multiply both sides of that by -1 to get \(\displaystyle -a\le b\). That shows that -a is the least upper bound on -S, so inf(S)= -sup(-S).
 
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Aug 2009
639
2
ok i will take note that these type of questions should not be in the number theory forum. i thought that they are classified under this section.

sorry about that.