How to Prove that l is the Least Upper Bound of a Set S of Real Numbers?

[roam]

Let $$l \in R$$ be the least upper bound of a nonempty set S of real numbers.
Show that for every $$\epsilon < 0$$ there is an $$x \in S$$ such that
$$x > l - \epsilon$$



I don't understand this question very well, I appreciate it if you could give me some hints.



l is the l.u.b on S, therefore it is greater than or equal to any $$s \in S$$

By the definition of the limit; |f(x)-l| < ε if 0 < |x-a| < δ

l-ε < x < l+ε

|f(x)-l| < ε

"?"

 

[Dick]

Suppose it's NOT true. Then there is an epsilon such that x<=l-epsilon for all x in S. Can l really be lub(S)?

 

[roam]

Thanks, so that's how you derive the contradiction.

But could you explain it in more detail please(especially the last bit)?

I appreciate that.

 

[Dick]

l is supposed to be the LEAST upper bound. If x<=l-epsilon for all x in S, then l-epsilon is ALSO an upper bound. What the order relation between l and l-epsilon?

 

[HallsofIvy]

Let $$l \in R$$ be the least upper bound of a nonempty set S of real numbers.
Show that for every $$\epsilon < 0$$ there is an $$x \in S$$ such that
$$x > l - \epsilon$$



I don't understand this question very well, I appreciate it if you could give me some hints.



l is the l.u.b on S, therefore it is greater than or equal to any $$s \in S$$

By the definition of the limit; |f(x)-l| < ε if 0 < |x-a| < δ

What limit and what f(x) are you talking about? There is no function nor limit of a function mentioned in the problem

l-ε < x < l+ε

|f(x)-l| < ε

"?"

Dick has given the hint you need: use proof by contradiction.