Greatest Lower Bound Property

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S \subset \R$ be a non-empty subset of the set of real numbers such that $S$ is bounded below.

Then $S$ admits an infimum in $\R$.

This is known as the greatest lower bound property of the real numbers.


Proof 1

Let $T = \set {x \in \R: -x \in S}$.

By Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above:

$T$ is bounded above.

Thus by the Least Upper Bound Property, $T$ admits a supremum in $\R$.

From Negative of Supremum is Infimum of Negatives:

$\ds -\sup_{x \mathop \in T} x = \map {\inf_{x \mathop \in T} } {-x}$

That is, by definition of $T$:

$\ds -\sup_{x \mathop \in T} x = \inf_{x \mathop \in S} x$

and so $S$ admits an infimum in $\R$.

$\blacksquare$


Proof 2

Let $T$ be the set of lower bounds of $S$:

$T = \set {x \in \R: x \le \forall y \in S}$


Since $S$ is bounded below, $T$ is non-empty.

Now, every $x \in T$ and $y \in S$ satisfy $x \leq y$.

That is, $T$ is bounded above by every element of $S$.

By the Least Upper Bound Property, $T$ admits a supremum in $\R$.

Let $B = \sup T$.


Since every element of $S$ is an upper bound of $T$ it follows from the definition of supremum that:

$\forall y \in S : y \ge B$

Thus $B$ is a lower bound of $S$.


Because $B$ is an upper bound of $T$:

$\forall x \in T : x \le B$

and so $B$ is the infimum of $S$.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Greatest_Lower_Bound_Property&oldid=640009"