Proving Limsup h.w. Proofs: Monotonic Increasing Function

[aschwartz]

Homework Statement

1. A function f(x) is said to be monotonic increasing in A if for all x1, x2 ∈ A, x1≤x2 implies f(x1)≤f(x2).

Prove that if f(x) is monotonic increasing in R [f: R→R] and c is a cluster point of R then the limit of f(x) as x→c^{-} exists (might be +∞).

2. s(δ) = sup{f(x) :0<|x-c|<δ}

s(δ) is a monotonic decreasing function, hence based on previous result 1, lims(δ) as δ→0^{+} = L^{+}, which is defined to be the limsupf(x) as x→c.

Prove:

A. If L^{+} = limsupf(x) as x→c, then \exists a sequence x_{n}, such that as x_{n}→c, f(x_{n})→L^{+}.

B. If x_{n}→c and x_{n}≠c and f(x_{n})→L then L≤L^{+}.

C. Similarly define L^{-} = liminff(x) as x→c. [This is a monotonic increasing function.]

Prove that limf(x) as x→c = L if and only if L^{+} = L^{-} = L.


2. The attempt at a solution

Ok, so this is what I have so far - I was able to get #1, but got stuck with the proofs for #2.

1. If f(x) is monotonic increasing (decreasing) then limf(x) as x→c^{-} exists.
f: R→R
c ∈R

Case 1: (Proving a lefthand limit exists for a monotonic increasing function)

Let L = sup{f(x) : x<c}. We want to show that (\forall\epsilon>0) (\exists\delta) (\forallx ∈ R) (0<c-x<δ ⇒ |f(x) - L|<\epsilon.

Consider the interval (L - <\epsilon, L). \existsx_{1}, such that x_{1}<c and L - \epsilon<f(x_{1})≤L [because otherwise L is not the sup, but L - \epsilon would be the sup! So therefore, f(x_{1}) must exist in between those two numbers.]

Let \delta = c - x_{1}.

\forallx ∈ R if 0<c-x<δ → 0<c-x<c - x_{1} ⇒ x>x_{1}, so since f is monotonic increasing f(x)>f(x_{1}).

Then, because L - \epsilon<f(x_{1})<f(x)≤L,
L - \epsilon<f(x)≤L and |f(x) - L|<\epsilon.


This is all I have so far! Any help or advice in how to solve part 2 of this problem would be greatly appreciated - thanks so much!

 

[aschwartz]

I would just like to note - I think the latex imaging might have messed this up, but all the epsilons in the problem should not be listed as superscripts, but should be aligned normally and the 1 by the x, should be a subscript. Also, I left out the other 3 cases for fully proving part 1, but the proof is the same - it pretty much follows...Thanks!