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Difficult analysis problem involve supremum and function concepts

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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:
[tex]\forall x \in A x \le u[/tex]
if [tex]v[/tex] is an upper bound of A, then [tex]u \le v[/tex]

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.
 

Answers and Replies

  • #2
jbunniii
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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 [itex]c[/itex], just some random point between [itex]f(b)[/itex] and [itex]f(a)[/itex]? What is known about [itex]f[/itex]? Is it monotonically decreasing? Is it continuous?
 
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There must be some information missing. What is [itex]c[/itex], just some random point between [itex]f(b)[/itex] and [itex]f(a)[/itex]? What is known about [itex]f[/itex]? 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).
 
  • #4
jbunniii
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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.
 

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