Proving on the completeness theorem of real number

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In summary, the question asks whether the supremum of a set D, defined as the set of all elements that are twice the elements in a non-empty subset A of the real numbers, is equal to 2 times the supremum of A. Using the completeness axiom, we can prove that 2 times the supremum of A is an upper bound for D and that there is no upper bound less than 2 times the supremum of A. This means that 2 times the supremum of A is indeed the supremum of D.
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Lily@pie
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Homework Statement


Let A be a non empty subset of R that is bounded above

Set D:={2a|a (belongs to) A}
Is it necessarily true that the sup D = 2 sup A? Either prove or provide a counterexample.


Homework Equations


The completeness axiom


The Attempt at a Solution


I am seriously clueless on how to approach... but I still tried something
Let sup D = y and sup A = x
d=2a; d<=y; a<=x

or can I say choose a as the largest value in A, so 2a=d is the largest value in D. Since both are the upper bound for each set and for all upper bound and real number, they are the smallest. so, d is sup D and a is sup A.
Therefore, d=2a => sup D = 2 sup A


But these method seems a bit weird...
 
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  • #2
No, you can't say "choose a as the largest value in A". The whole point of "sup" is that the set may not have a largest value. And you can't say "Let sup D = y and sup A = x d=2a; d<=y; a<=x" because you didn't say what "a" was.

"sup" has two properties- it is an upper bound (so there are no member of the set larger than the sup) and there is no smaller upper bound (so there are members of the set within any "[itex]epsilon[/itex]" distance of the sup.

Let S be the sup of set A. We can prove that 2S is an upper bound for 2A by contradiction:
suppose there exist b in 2A such that b> 2S. But b= 2a for some a in A. That says 2a> 2S so that a> S which contradicts the fact that S is an upper bound for A.

Similarly, you can prove by contradiction that there is no upper bound less than 2S.
 
  • #3
But how do we relate this to the prove of sup D = 2 sup A?

Does 2S means D?

Sorry, I'm a bit confused now
 

FAQ: Proving on the completeness theorem of real number

What is the completeness theorem of real number?

The completeness theorem of real numbers states that every non-empty set of real numbers that is bounded above has a least upper bound. This means that there is a maximum value in the set, regardless of whether or not it is a part of the set itself.

Why is the completeness theorem important?

The completeness theorem is important because it allows for the construction of the real numbers, which are essential in many areas of mathematics. It also provides a rigorous foundation for calculus and other mathematical concepts.

How is the completeness theorem proven?

The completeness theorem is proven using a mathematical proof, which involves using logical reasoning and previously established theorems to show that the statement is true. This proof can be quite complex and requires a solid understanding of real analysis and mathematical logic.

What are some applications of the completeness theorem?

The completeness theorem has many applications in mathematics, including in the construction of the real numbers, the formulation of the Heine-Borel theorem, and the development of calculus. It also has applications in other fields such as economics and physics.

Are there any limitations to the completeness theorem?

Although the completeness theorem is incredibly powerful and useful, it does have some limitations. It only applies to the real numbers and cannot be extended to other number systems, such as complex numbers. Additionally, there are some sets of real numbers that do not have a least upper bound, making the completeness theorem inapplicable in those cases.

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