# A review of supremum infimum.

1. Oct 30, 2006

### MathematicalPhysicist

i need to show that when A is a subset of B and B is a subset of R (A B are non empty sets) then: infB<=infA<=supA<=SupB

here's what i did:
if infA is in A then infA is in B, and by defintion of inf, infB<=infA.
if infA isnt in A then for every e>0 we choose, infA+e is in A and so infA is in B, so infA+e>=infB, we can find e'>0 such that infA+e>=infB+e'>infB
so we have: infA-infB>=e'-e, let e=e'/2 then we have infA-infB>e'/2>0 so we have infA>infB. (is this method correct?).
obviously supA>=infA by defintion.
i think that the same goes for supA and supB, with suitable changes.

Last edited: Oct 30, 2006
2. Oct 30, 2006

### StatusX

You don't need to use an e>0. Go back to the original definition of inf in terms of just $\leq$.

3. Oct 30, 2006

### MathematicalPhysicist

how exactly to use it?
i mean if a is in A then a is in B, infA<=a so for every c<=a infA>=c
because a is in B also then a>=infB, but because infA is the greatest lower bound then we have infB<=infA, correct?

4. Oct 30, 2006

### StatusX

Everything there sounds right except for that thing with c which I didn't follow.