Understanding Upper Bound & Sup in Theorem Proving

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SUMMARY

The discussion focuses on the mathematical concepts of "upper bound" and "supremum" (sup) as they relate to theorem proving. An upper bound is defined as any value that is greater than or equal to all elements in a set, while the supremum is the least upper bound of that set. The example provided illustrates that the supremum of the range of the sigmoid function \(\frac{1}{1+e^{-x}}\) is 1, which is not an actual value of the function but serves as an upper bound. Understanding these terms is crucial for grasping the proofs in calculus and related fields.

PREREQUISITES
  • Basic understanding of set theory and functions
  • Familiarity with calculus concepts, particularly limits
  • Knowledge of the sigmoid function and its properties
  • Experience with theorem proving techniques
NEXT STEPS
  • Study the concept of limits in calculus
  • Learn about the properties of the sigmoid function and its applications
  • Explore the differences between upper bounds and lower bounds
  • Research theorem proving methods in mathematical analysis
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Students of mathematics, particularly those studying calculus and analysis, as well as educators and anyone involved in theorem proving who seeks to deepen their understanding of upper bounds and supremum.

Miike012
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This calc book that I am reading uses words like "upper bound" and "sup" a lot when proving theorems. I have never heared these terms before so it makes it hard for me to understand the proofs.

I think it has to deal with max's values of a graph: For example given a set S of all elements c in a ≤ c ≤ b would the the upper bound in the following graph in [a,b] be c? and also c = Sup S...
is this correct?
 

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Upper bound is any value that is greater than or equal to all the values in a set.

sup is short for supremum, the least upper bound of a set of numbers. The range of a function might be such a set. The supremum is not necessarily a member of the set: for example, the supremum of the range of the sigmoid function \frac{1}{1+e^{-x}} is 1, but that is not a value of the function. 1 is an upper bound to this function, but so are 2, 100 and 100!2
 

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