Real Analysis - Infimum and Supremum Proof

In summary, the conversation discusses the problem of proving that if a set D has two functions f and g that satisfy the condition f(x) ≤ g(y) for all x,y ∈ D, then f(D) is bounded above and g(D) is bounded below, and that sup f(D) ≤ inf g(D). The conversation also includes a helpful Lemma and its proof that simplifies the argument for proving sup f(D) ≤ inf g(D).
  • #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.

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!
 
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  • #2
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
 
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  • #3
Maybe just a nitpick (as what you mean is obvious), but I would rephrase the following:
zigzagdoom said:
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)
 
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  • #4
lordianed said:
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
Samy_A said:
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
ll
zigzagdoom said:
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##.
 
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  • #7
Ray Vickson said:
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
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
 

1. What is the definition of infimum and supremum in real analysis?

The infimum of a set of real numbers is the greatest lower bound of the set, meaning it is the largest number that is less than or equal to all the numbers in the set. The supremum of a set of real numbers is the least upper bound of the set, meaning it is the smallest number that is greater than or equal to all the numbers in the set.

2. How do you prove the existence of infimum and supremum for a set of real numbers?

In order to prove the existence of infimum and supremum for a set of real numbers, one must use the completeness axiom of the real numbers. This axiom states that every non-empty set of real numbers that is bounded below has an infimum, and every non-empty set of real numbers that is bounded above has a supremum.

3. Can a set have more than one infimum or supremum?

No, a set can have at most one infimum and at most one supremum. If a set has more than one infimum, then all the numbers in the set would have to be infima, which is not possible. Similarly, if a set has more than one supremum, then all the numbers in the set would have to be suprema, which is also not possible.

4. How do you prove that a number is the infimum or supremum of a set?

To prove that a number is the infimum or supremum of a set, one must show that it satisfies the definition of infimum or supremum. For infimum, this means proving that the number is less than or equal to all the numbers in the set and that there is no other number that is smaller. For supremum, this means proving that the number is greater than or equal to all the numbers in the set and that there is no other number that is larger.

5. Can the infimum or supremum of a set be included in the set?

Yes, the infimum or supremum of a set can be included in the set. This is possible if the number is the minimum or maximum of the set, as it is both the smallest and largest number in the set and therefore satisfies the definition of infimum and supremum.

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