Have scratchwork. Need to finish proof of sup E = inf U

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This completes the proof. In summary, we have shown that for a subset E of R that is not empty and bounded above, sup E = inf U, where U is the set of all upper bounds of E.
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pzzldstudent
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This problem was an in-class example that became assigned as additional homework. The scratchwork was done in class, but we ran out of time to write out the formal proof.

Claim: Let E be a subset of R, E is not empty and E is bounded above. Set U = {b in R: b is an upper bound of E}. Prove that sup E = inf U.

Scratchwork:
• U is not empty because E is bounded above
• If b is an upper bound of E (i.e. b is in U), then sup E ≤ b, by the definiton of supremum. Therefore sup E is a lower bound of U.
• To show it is an infimum: if ß > (what we think is the inf) and show there is an x in U such that ß > x ≥ (what we think is the infimum)
• Let ß > sup E. Need an upper bound of E (say b*) where ß > b* ≥ sup E
• Take b* = sup E

My attempt of a writeup of a proof: Let E be a subset of R, E is not empty, and E is bounded above. U is not empty because E is bounded above. If U is the set of all upper bounds, there is an element b that is an upper bound of E (i.e. b in U). Then sup E ≤ b by the definition of supremum. Therefore, sup E is a lower bound of U. Let ß > sup E. We need an upper bound of E (say b*) where ß > b* ≥ sup E. Take b* = sup E. Then by the definition of infimum, sup E = inf U.

Is that even close to a decent proof? This class is really killing me. :|
Any help at all is greatly appreciated.

Thank you for your time.
 
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Yes, your attempt at a proof is very close to being a decent proof. It could use some more details, though. Here is a suggested proof:Proof: Let E be a subset of R, E is not empty, and E is bounded above. U is the set of all upper bounds of E. Since E is bounded above, U is not empty. For any b in U, sup E ≤ b by the definition of supremum. Therefore, sup E is a lower bound of U. To show that sup E is also the infimum of U, let ß > sup E. We need to find an upper bound of E (say b*) where ß > b* ≥ sup E. Since sup E is the least upper bound of E, we can take b* = sup E. Since ß > b* ≥ sup E, it follows from the definition of infimum that sup E is the infimum of U. Hence, sup E = inf U.
 

What does "sup E = inf U" mean?

Sup E = inf U is a mathematical notation that represents the supremum (or least upper bound) of a set E, which is equal to the infimum (or greatest lower bound) of a set U. It is a common notation used in analysis and calculus to denote the maximum and minimum values of a function or set.

Why is scratchwork necessary for proving "sup E = inf U"?

Scratchwork is necessary for proving "sup E = inf U" because it allows you to work out the steps and manipulations needed to arrive at the final proof. It helps you organize your thoughts and ensures that your proof is logically sound and accurate.

How do I finish the proof of "sup E = inf U"?

To finish the proof of "sup E = inf U", you need to show that the supremum of E is equal to the infimum of U. This can be done by using the definitions of supremum and infimum, as well as the properties of real numbers such as transitivity, monotonicity, and completeness.

What are some common mistakes to avoid when proving "sup E = inf U"?

Some common mistakes to avoid when proving "sup E = inf U" include assuming that the supremum and infimum exist, using incorrect definitions or properties, and incorrect manipulations or simplifications. It is important to carefully follow the steps and check for errors along the way.

Can you provide an example of a proof for "sup E = inf U"?

Yes, here is an example proof for "sup E = inf U":
Let E and U be non-empty sets of real numbers.
By definition, sup E is the least upper bound of E, and inf U is the greatest lower bound of U.
Suppose sup E = a and inf U = b.
Since a is the supremum of E, then a ≥ x for all x ∈ E. Similarly, b ≤ x for all x ∈ U.
Therefore, a ≥ b.
Now, suppose there exists c such that c < b. Then, c is not a lower bound of U, which means there exists x ∈ U such that x < c.
But this contradicts the fact that b is the infimum of U.
So, b is the greatest lower bound of U, and a = sup E = inf U = b.
Hence, sup E = inf U.

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