Intro to Real Analysis: Supremum

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SUMMARY

The supremum of the open interval E = (0, 1) is definitively 1. This conclusion is derived from the definition of an upper bound, where for any upper bound M, it must hold that 1 ≤ M. The least upper bound property confirms that for every ε > 0, there exists an element a in (0, 1) such that s - ε < a, reinforcing that the supremum is not contained within the interval itself.

PREREQUISITES
  • Understanding of open intervals in real analysis
  • Familiarity with the concept of upper bounds
  • Knowledge of the least upper bound property
  • Basic proficiency in mathematical proofs and ε-δ definitions
NEXT STEPS
  • Study the properties of open intervals in real analysis
  • Learn about the completeness property of the real numbers
  • Explore examples of supremum and infimum in various sets
  • Review the formal definitions of upper bounds and least upper bounds
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Students of real analysis, mathematicians, and anyone seeking to understand the concepts of supremum and upper bounds in mathematical sets.

doubleaxel195
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Homework Statement


Find the supremum of E=(0,1)


Homework Equations





The Attempt at a Solution


By definition of open interval, x<1 for all x in E. So 1 is an upper bound. Let M be any upper bound. We must show 1&lt;=M. Can I just say that any upper bound of M must be greater than or equal to one based on the definition of open interval again?

I'm just not sure if that last line is completely correct. Thanks.
 
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Use your theorems.

s is the least upperbound if for all \epsilon &gt; 0, there exists an a \in (0, 1) satisfying s - \epsilon &lt; a.
 
Since you are only required to find the sup(E) you can just state it.

Hint: sup(E) ∉ E.
 

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