- #1
zigzagdoom
- 27
- 0
Hi Guys,
I am self teaching myself analysis after a long period off. I have done the following proof but was hoping more experienced / adept mathematicians could look over it and see if it made sense.
Question:
Suppose D is a non empty set and that f: D → ℝ and g: D →ℝ. If for ∀x,y ∈ D, f(x) ≤ g(y), then f(D) is bounded above and g(D) is bounded below.
Furthermore, sup f(D) ≤ inf g(D).
N/A
Assume that there exists a set such as in the question.
Since f(x) ≤ g(y) ∀ x,y ∈ D , f(x) is bounded above and g(y) is bounded below.
Now, let supf(D)=A and let infg(D)=B . Then ∀ x ∈ D, f(x) ≤ A and ∀ y ∈ D, g(y) ≥ B .
Assume: ¬ ( A ≤ B ) ↔ A>B and A ≠ B. But if A>B , then supf(D)>infg(D) .
This means ∃ g(D) ≤ supf(D) which in turn means ∃ f(D)>g(D) .
But this is a contradiction as ∀ x,y ∈ D, f(D) ≤ g(D) . So it follows A ≤ B. That is sup f(D) ≤ inf g(D).
Thanks!
I am self teaching myself analysis after a long period off. I have done the following proof but was hoping more experienced / adept mathematicians could look over it and see if it made sense.
Homework Statement
Question:
Suppose D is a non empty set and that f: D → ℝ and g: D →ℝ. If for ∀x,y ∈ D, f(x) ≤ g(y), then f(D) is bounded above and g(D) is bounded below.
Furthermore, sup f(D) ≤ inf g(D).
Homework Equations
N/A
The Attempt at a Solution
Assume that there exists a set such as in the question.
Since f(x) ≤ g(y) ∀ x,y ∈ D , f(x) is bounded above and g(y) is bounded below.
Now, let supf(D)=A and let infg(D)=B . Then ∀ x ∈ D, f(x) ≤ A and ∀ y ∈ D, g(y) ≥ B .
Assume: ¬ ( A ≤ B ) ↔ A>B and A ≠ B. But if A>B , then supf(D)>infg(D) .
This means ∃ g(D) ≤ supf(D) which in turn means ∃ f(D)>g(D) .
But this is a contradiction as ∀ x,y ∈ D, f(D) ≤ g(D) . So it follows A ≤ B. That is sup f(D) ≤ inf g(D).
Thanks!