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.