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anelys
Nov3-09, 11:42 AM
1. The problem statement, all variables and given/known data

f(a) > c > f(b)

A = { x : b > x > y > a implies f(a) > f(y) }

let u = sup(A)

show that f(u) = c

2. Relevant equations

I have no idea in particular, save for the definition of the supremum:
\forall x \in A x \le u
if v is an upper bound of A, then u \le v

3. The attempt at a solution

My intuition led me to attempt a proof by contradiction. If you let f(x*) = c, assume that x* < u to arrive at a contradiction. Then assume that x* > u to arrive at a contradiction. Then to conclude that x* must be u. I don't know how to do this, or even if I can/should be done.

jbunniii
Nov3-09, 12:04 PM
1. The problem statement, all variables and given/known data

f(a) > c > f(b)

A = { x : b > x > y > a implies f(a) > f(y) }

let u = sup(A)

show that f(u) = c


There must be some information missing. What is c, just some random point between f(b) and f(a)? What is known about f? Is it monotonically decreasing? Is it continuous?

anelys
Nov3-09, 12:30 PM
There must be some information missing. What is c, just some random point between f(b) and f(a)? What is known about f? Is it monotonically decreasing? Is it continuous?
There is nothing missing from the problem. I guess we can assume that it's continuous. It isn't necessarily a decreasing monotone function. Also, yes, c is any point between f(b) and f(a).

jbunniii
Nov3-09, 02:12 PM
There is nothing missing from the problem. I guess we can assume that it's continuous. It isn't necessarily a decreasing monotone function. Also, yes, c is any point between f(b) and f(a).

Well, if that's the case then it's not true.

Let f(x) = -x for all x.

Let a = 0, b = 1, c = -0.5.

Then A = (0, 1), u = sup(A) = 1, but f(u) does not equal c.