Math 2.0

statement theorem

Existence of the infimum

Statement

Let SS be a nonempty subset of the real numbers that is bounded below. Then SS has an infimum in R\mathbb{R}.

Remarks

Proof: the set −S={−s:s∈S}-S = \{-s : s \in S\} is nonempty and bounded above. By the completeness axiom, it has a supremum bb. Then −b-b is the infimum of SS. Another proof: the set of lower bounds of SS is nonempty and bounded above by each element of SS, and its supremum is inf⁡S\inf S.

Explanations

L1 human

If a set of real numbers has a floor, then it has a highest floor. This follows from the completeness axiom by a reflection of the number line through 00, which turns floors into ceilings.

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  1. Real numbers object
  2. Set bounded above definition
  3. Infimum definition
  4. Supremum definition
  5. Completeness axiom statement
  6. Existence of the infimum statement

Sources

  1. Web https://en.wikipedia.org/wiki/Least-upper-bound_property, Statement for real numbers