# Thread: Infimum theorems

1. ## Infimum theorems

good day everyone,.,i need ur help once again,.,.can u help me prove this??

infimum = alpha iff (-infinity, alpha) intersection S = null and

[alpha, alpha + epsilon) intersection S not equal to Null for all epsilon > 0.

sorry for the luck of symbols used,.,.thnx

2. ## Re: Infimum theorems

Originally Posted by aldrincabrera
good day everyone,.,i need ur help once again,.,.can u help me prove this??

infimum = alpha iff (-infinity, alpha) intersection S = null and

[alpha, alpha + epsilon) intersection S not equal to Null for all epsilon > 0.

sorry for the luck of symbols used,.,.thnx

The above seems to be pretty straighforward. What definition of "minimum of non-empty real set S" do you have?

Tonio

3. ## Re: Infimum theorems

clearly S is non-empty (let ε = 1, then [α,α+1) ∩ S is non-empty, so S is non-empty).

clearly S is also bounded below: if β < α, then for any s in S, β < s.

(if not, then since β is in (-∞,α), s would also be in (-∞,α), contradicting the fact that (-∞,α)∩ S = Ø).

so inf(S) exists. now show that both inf(S) < α and inf(S) > α violate one of your two conditions.

(hint: if inf(S) > α, use ε < inf(S) - α).

4. ## Re: Infimum theorems

Originally Posted by Deveno
clearly S is non-empty (let ε = 1, then [α,α+1) ∩ S is non-empty, so S is non-empty).

Uuh?? How can you know if S isn't given by the OP??

Tonio

clearly S is also bounded below: if β < α, then for any s in S, β < s.

(if not, then since β is in (-∞,α), s would also be in (-∞,α), contradicting the fact that (-∞,α)∩ S = Ø).

so inf(S) exists. now show that both inf(S) < α and inf(S) > α violate one of your two conditions.

(hint: if inf(S) > α, use ε < inf(S) - α).
.

5. ## Re: Infimum theorems

because....(from the original post):

"[alpha, alpha + epsilon) intersection S not equal to Null for all epsilon > 0."

i picked an epsilon, 1. 1 > 0...what's the problem?