Difficult analysis problem involve supremum and function concepts

[anelys]

Homework Statement

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

Homework 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 $$u \le v$$

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]

Homework Statement

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]

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]

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.