Difference between revisions of "Least upper bound"

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Given a [[subset]] <math>S</math> in some larger [[partially ordered set]] <math>R</math>, a '''least upper bound''' or '''supremum''', for <math>S</math> is an [[element]] <math>M \in R</math> such that <math>s \leq M</math> for every <math>s \in S</math> and there is no <math>m < M</math> with this same property.
 
Given a [[subset]] <math>S</math> in some larger [[partially ordered set]] <math>R</math>, a '''least upper bound''' or '''supremum''', for <math>S</math> is an [[element]] <math>M \in R</math> such that <math>s \leq M</math> for every <math>s \in S</math> and there is no <math>m < M</math> with this same property.
  
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The fact that <math>\mathbb{R}</math> is complete is something intuitively clear but impossible to prove using only the field and order properties of <math>\mathbb{R}</math>
 
The fact that <math>\mathbb{R}</math> is complete is something intuitively clear but impossible to prove using only the field and order properties of <math>\mathbb{R}</math>
  
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==See also==
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*[[Greatest lower bound]]
  
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[[Category:Definition]]
 
[[Category:Definition]]

Latest revision as of 13:08, 5 March 2022

Given a subset $S$ in some larger partially ordered set $R$, a least upper bound or supremum, for $S$ is an element $M \in R$ such that $s \leq M$ for every $s \in S$ and there is no $m < M$ with this same property.

If the least upper bound $M$ of $S$ is an element of $S$, it is also the maximum of $S$. If $M \not\in S$, then $S$ has no maximum.


Completeness: This is one of the fundamental axioms of real analysis.

A set $S$ is said to be complete if any nonempty subset of $S$ that is bounded above has a supremum.

The fact that $\mathbb{R}$ is complete is something intuitively clear but impossible to prove using only the field and order properties of $\mathbb{R}$

See also

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