Math 2.0

statement proposition

Approximation property of the supremum

Statement

Let SS be a nonempty subset of the real numbers that is bounded above, and let b=sup⁡Sb = \sup S be its supremum. For each real number c<bc < b, there is an element ss of SS with c<s≤bc < s \le b.

Remarks

Proof: if no such ss exists, then cc is an upper bound of SS that is less than bb. This contradicts the definition of the supremum. An equivalent form: for each ε>0\varepsilon > 0, there is ss in SS with b−ε<sb - \varepsilon < s. For an infimum a=inf⁡Sa = \inf S and each c>ac > a, there is ss in SS with a≤s<ca \le s < c.

Explanations

L1 human

You can find elements of a set as close to its supremum as you want. If you lower the ceiling even a little, then some element of the set sticks out above it.

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Real numbers, object, no evidence levelReal numbersobjectSet bounded above, definition, no evidence levelSet bounded abovedefinitionSupremum, definition, no evidence levelSupremumdefinitionCompleteness axiom, statement, no evidence levelCompleteness axiomstatementApproximation property of the supremum, statement, no evidence levelApproximation propert…statement
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  1. Real numbers object
  2. Set bounded above definition
  3. Supremum definition
  4. Completeness axiom statement
  5. Approximation property of the supremum statement

Sources

  1. Web https://en.wikipedia.org/wiki/Infimum_and_supremum, Identifying infima and suprema

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