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Consider the set

$$ S = { \frac{1}{x} : x \in \mathbb{N} } $$

This set has the Least upper bound property since every non empty subset has a least upper bound in S. How is the greatest lower bound property implied in this case? Consider the subset $E = S$, and since $0 \notin S $, it doesn't have a lower bound in $S$, so by definition, glb property does not hold true.

Rudin says on page 4 that, every ordered set with the least-upper bound property also has the greatest-lower bound property.

urmish
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  • Does Rudin give a proof? Also a quote from Rudin about when a set is said to have the least upper bound property would be good. [Not a paraphrase but the actual wording, along with any related definitions and examples given before his claim appears.] – coffeemath Sep 21 '22 at 04:53
  • Does the set S not have a lub property? Consider set of all x in S such that x < 1/pi. The lub of such a set is 1/pi and that doesn't exist in S. Is constructing such a set for a counter example allowed? – urmish Sep 21 '22 at 05:18
  • Please look more carefully at Rudin. It's likely what he says, when spelled out, is that a set S has the least upper bound property provided every nonempty subset T of S which has an upper bound in S has a least upper bound in S. And similarly S has the greatest lower bound property if every nonempty subset U of S which has a lower bound in S has a greatest lower bound in S. In your example, the subset E=S of S does not have a lower bound in S, so there is no contradiction to the greatest lower bound property here. [Yu cannot use $0$ as the ower bound for E sinc $0$ is not in S. – coffeemath Sep 21 '22 at 13:05

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