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
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).