Proof using definitions of limit

[roam]

Hi,

Here's my question;

(a). Let l ? IR be the least upper bound of a nonempty set Sof real numbers. Show that for each ?>0 there is an x?S such that x>l-?

I start off by deriving a contradiction;
Statement: "for each ?>0 there is an x?S such that x>l-?"
Suppose it's NOT true. Then there is an epsilon such that x ? l-? for all x in S.

Can l really be lub(S)? I don't understand how to prove these...


(b). If f is continious at a and f(a)>0 show that there exists a ?>0 such that f(x)>0 for all x?(a-?,a+?).

I have no idea on how to show this...

I'd appreciate your guidance.

Regards.

 

[stapel]

(a). Let l ? IR be the least upper bound of a nonempty set Sof real numbers. Show that for each ?>0 there is an x?S such that x>l-?

What is the character after the lower-case ell? What is the definition of "IR"? ("R" is "the reals", and "I" is "the integers", but I've never seen "IR" before...?) What is your book's definition of "least upper bound"?

(b). If f is continious at a and f(a)>0 show that there exists a ?>0 such that f(x)>0 for all x?(a-?,a+?).

(My first instinct Suppose not. Then, for all positive delta (for all d > 0), there is some x in (a - d, a + d) with f(x) < 0. Pick any delta d > 0, and let x[sub:1yxmata2]d[/sub:1yxmata2] be an x in (a - d, a + d) such that f(x[sub:1yxmata2]d[/sub:1yxmata2]) < 0.

See if applying the definition of continuity at "a" to the above supposition is useful.

Eliz.

 

[pka]

(a). Let l ? IR be the least upper bound of a nonempty set Sof real numbers. Show that for each ?>0 there is an x?S such that x>l-?

(b). If f is continious at a and f(a)>0 show that there exists a ?>0 such that f(x)>0 for all x?(a-?,a+?).

First a language lesson.
The statement that x is an upper bound for S means: \(\displaystyle \left( {\forall y \in S} \right)\left[ {y \leqslant x} \right]\)
The statement that z is not an upper bound for S means: \(\displaystyle \left( {\exists t \in S} \right)\left[ {z < t} \right]\)
The statement that x is the least upper bound for S means, that x is an upper bound and no number less than x is an upper bound.

Suppose that \(\displaystyle x = \mbos{lub}(S)\) then \(\displaystyle \varepsilon > 0 \Rightarrow \quad - \varepsilon < 0 \Rightarrow \quad x - \varepsilon < x\) .

So x is the least upper bound and \(\displaystyle x - \varepsilon < x\) apply the above.

Because \(\displaystyle f(a) > 0\) use \(\displaystyle \varepsilon > \frac{{f(a)}}{2}\) in the definition of continuity at a.

\(\displaystyle \left({\exists\delta>0}\right)\left[{\left|{x-a}\right|<\delta}\right]\Rightarrow\quad\left|{f(a)-f(x)}\right| < \frac{{f(a)}}{2}\)