Math 2.0

Real functions Theorem

Existence of the infimum

Builds on Completeness axiomInfimum

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.

Reading path

To understand this page from the ground up, read these 6 pages in order. The foundations come first.

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Real numbers, object, no evidence levelReal numbersobjectSet bounded above, definition, no evidence levelSet bounded abovedefinitionInfimum, definition, no evidence levelInfimumdefinitionSupremum, definition, no evidence levelSupremumdefinitionCompleteness axiom, statement, no evidence levelCompleteness axiomstatementExistence of the infimum, statement, no evidence levelExistence of the infi…statement
Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
  • no evidence level
  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

Further reading

  1. Web https://en.wikipedia.org/wiki/Least-upper-bound_property, Statement for real numbers
About this page
Tag0EDH403M21
URIm2://0EDH403M21
Kindstatement, theorem
WritingW0 stub
MSC26A03
StatusThis copy of the site has no build data. Run m2 status 0EDH403M21.
UsesCompleteness axiom, Infimum

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