Least Upper Bound property - A set $X$ is said to have this property if every non empty subset $A$ of $X$ which is bounded above has the least upper bound. (This does not imply the least upper bound must belong to $A$, which is my inference from the definition of least upper bound)

The definition for the Greatest Lower Bound property is analogous.

Let $X=(0,1]$ and hence $( X, \leq)$ is a partially ordered set where $\leq$ has the usual meaning.

Now clearly $X$ has the least upper bound property as its closed above.

But if we take $A=X\subseteq X$ then what is the greatest lower bound of $A$? How come then $X$ has the greatest lower bound property?

  • $\begingroup$ Do you mean $A = X = (0, 1]$? $\endgroup$ – Taroccoesbrocco Aug 17 at 19:04
  • $\begingroup$ @Taroccoesbrocco yes $\endgroup$ – Abhay Aug 17 at 19:17

An ordered set $(X, \leq)$ has the greatest lower bound property if every non-empty subset of $X$ with a lower bound (in $X$) has a greatest lower bound (infimum) in $X$.

If we take $A = X = (0,1]$ then $A \subseteq X$ but there is no lower bound of $A$ in $X$ (note that $0 \notin X$), so $A$ is not a counter-example to the greatest lower bound property for $X = (0, 1]$.

  • $\begingroup$ Thank you very much $\endgroup$ – Abhay Aug 17 at 19:18
  • $\begingroup$ @Abhay - You are welcome! $\endgroup$ – Taroccoesbrocco Aug 17 at 19:21

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