Real Analysis - Infimum and Supremum Proof

[zigzagdoom]

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.

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!

 

[lordianed]

Hello zigzagdoom, sorry that I'm not the adept Mathematician you are looking for, but I hope my points make sense. I would perhaps phrase the second sentence of your argument this way:
Since $D$ is non-empty, then the images of $f$ and $g$ are non-empty. Furthermore, as $f(x) \leq g(y)$ for all $x,y \in D$, then any element in the image of $f$ is a lower bound of $g(D)$, and any element in the image of $g$ is an upper bound of $f(D)$; thus, $g(D)$ is bounded below, and $f(D)$ is bounded above.

edit: codomain vs image

 

[Samy_A]

Maybe just a nitpick (as what you mean is obvious), but I would rephrase the following:

This means ∃ g(D) ≤ supf(D) which in turn means ∃ f(D)>g(D)

$f(D)$ and $g(D)$ are not numbers, but sets, so I would write something like:

∃ y ∈ D: g(y) < supf(D) which in turn means ∃ x ∈ D: f(x)>g(y)

 

[zigzagdoom]

Hello zigzagdoom, sorry that I'm not the adept Mathematician you are looking for, but I hope my points make sense. I would perhaps phrase the second sentence of your argument this way:
Since $D$ is non-empty, then the images of $f$ and $g$ are non-empty. Furthermore, as $f(x) \leq g(y)$ for all $x,y \in D$, then any element in the image of $f$ is a lower bound of $g(D)$, and any element in the image of $g$ is an upper bound of $f(D)$; thus, $g(D)$ is bounded below, and $f(D)$ is bounded above.

edit: codomain vs image

Thank you, very helpful indeed.

 

[zigzagdoom]

Maybe just a nitpick (as what you mean is obvious), but I would rephrase the following:
$f(D)$ and $g(D)$ are not numbers, but sets, so I would write something like:

∃ y ∈ D: g(y) < supf(D) which in turn means ∃ x ∈ D: f(x)>g(y)

Thanks for the clarification!

 

[Ray Vickson]

ll

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.

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!

I think you can simplify and clean up your sup f <= inf g argument a bit, if you first prove a helpful little Lemma: if $f(x) \leq A$ for all $x \in D$ then $\sup_{x \in D} f(x) \leq A$ (and a similar result applies to the inf when $g(y) \geq B$ for all $y \in D\:$).

Using the Lemma we have that $f(x) \leq g(y)$ for all $x,y \in D$ implies $\sup_D f \leq g(y)$ for any $y \in D$; then the Lemma again gives $\sup_D f \leq \inf_D g$.

 

[zigzagdoom]

ll

I think you can simplify and clean up your sup f <= inf g argument a bit, if you first prove a helpful little Lemma: if $f(x) \leq A$ for all $x \in D$ then $\sup_{x \in D} f(x) \leq A$ (and a similar result applies to the inf when $g(y) \geq B$ for all $y \in D\:$).

Using the Lemma we have that $f(x) \leq g(y)$ for all $x,y \in D$ implies $\sup_D f \leq g(y)$ for any $y \in D$; then the Lemma again gives $\sup_D f \leq \inf_D g$.

Thanks a lot.

For proof of the Lemma here is my attempt (by symmtery similar for B):

Lemma:
If f(x) ≤ A ∀ x ∈ D then supf(x) ≤ A.

Proof of Lemma (by contradiction):
Assume that f(x) ≤ A ∀ x ∈ D and supf(x) > A. Then we have that f(x) ≤ supf(D) and supf(x) > supf(D)

Let K be the set consisting of all f(x) , then supf(D) is an upper bound of K.

Since supf(x) is the least upper bound of K, and supf(x) > supf(D) , ∃ f(x) ∈ K such that f(x) ≥ supf(D) .

But this is a contradiction as supf(D) is an upper bound of A.Therefore ¬supf(x) > A ↔ supf(x) ≤ A

 

[zigzagdoom]

For completeness, updated proof attempt, building on the interpreted guidance of those on PF:

Proof Strategy 1
Assume that there exists a set such as D.

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 ∃ x ∈ D such that g(D) ≤ supf(D) which in turn implies ∃ y ∈ D such that f(D) > g(D)

But this is a contradiction as ∀ x,y ∈ D, f(D) ≤ g(D) . So it follows A ≤ B.
Proof Strategy 1
Assume that there exists a set such as D.

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 .

Lemma 1:
If f(x) ≤ A, ∀ x ∈ D then supf(x) ≤ A.

Using Lemma 1, since f(x) ≤ g(y) ∀ x,y ∈ D → supf(x) ≤ g(y), ∀y∈D

Applying Lemma 1 again, since supf(x) ≤ g(y), ∀y∈D → supf(x) ≤ inf g(y), since infg(y) ≤ g(y), ∀y∈D