What is the Relationship Between inf(S) and -sup(-S)?

  • Thread starter Thread starter SMA_01
  • Start date Start date
SMA_01
Messages
215
Reaction score
0
Prove inf(S)=-Sup(-S)??

Homework Statement




Let S,T be subsets of ℝ, where neither T nor S are empty and both Sup(S) and Sup(T) exist.

Prove inf(S)=-sup(-S).

Starting with =>

I let x=inf(S). Then by definition, for all other lower bounds y of S, x≥y.

I'm stuck at this point...

Any help please?

Thanks
 
Physics news on Phys.org


You have to show two things:

  • -x is an upper bound of -S
  • If y is another upper bound of -S, then -x\leq y

So, in order to prove that -x is an upper bound of -S. Take an element -s from -S and prove -s\leq -x. Why is that true?
 


I'm not following along, why would -s≤-x? Unless you mean if I multiply both side by (-1), then it would be x≤s? Or am I totally off on a tangent?
 


SMA_01 said:
I'm not following along, why would -s≤-x? Unless you mean if I multiply both side by (-1), then it would be x≤s? Or am I totally off on a tangent?

If -s \in -S, then s \in S. And by definition, x is a lower bound of S...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top