Real Analysis - Infimum and Supremum Proof

  • #1
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.

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!
 

Answers and Replies

  • #2
23
7
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
 
  • #3
Samy_A
Science Advisor
Homework Helper
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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)
 
  • #4
27
0
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.
 
  • #5
27
0
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!
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
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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##.
 
  • #7
27
0
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
 
Last edited:
  • #8
27
0
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
 

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