Doubt regarding greatest lower bound property implies least upper bound property

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?

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