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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

**and**

*f*:*D*→ ℝ**. If for**

*g*:*D*→ℝ**∀**

*then*

**x,y ∈ D**,**f(x) ≤ g(y),***is bounded above and*

**f(D)****is bounded below.**

*g(D)*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***is bounded above and**

*f(x)***is bounded below.**

*g(y)*Now, let

**and let**

*supf(D)=A***. Then**

*infg(D)=B**and*

**∀ x ∈ D, f(x) ≤ A****.**

*∀ y ∈ D, g(y) ≥ B*Assume: ¬

*and*

**( A ≤ B ) ↔ A>B***. But if*

**A ≠ B***, then*

**A>B****.**

*supf(D)>infg(D)*This means

**which in turn means**

*∃ g(D) ≤ supf(D)***.**

*∃ f(D)>g(D)*But this is a contradiction as

*. So it follows*

**∀ x,y ∈ D, f(D) ≤ g(D)***That is*

**A ≤ B.**

*sup f(D) ≤ inf g(D)*.Thanks!