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Answer by rschwieb for How is a complete lattice defined solely by a least-upper bound?

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The problem is here:

fine by their definition to define that $\{0\}$ is the lub of $\{\}$.

I think you meant "that $0$ is the greatest lower bound of $\{\}$." But $1$ is a lower bound of $\{\}$ as well: for all $x\in \{\}, x\geq 1$. So $0$ couldn't be the greatest lower bound.

The upshot is that for $\Bbb N$ you did not actually find a suitable candidate for the glb of $\{\}$, nor will you be able to. If you think about the last paragraph long enough, you'll see that that any element is a lower bound of the emptyset, so if we truly did have greatest lower bounds, the glb would have to be the greatest element of the entire poset.

So while nonempty subsets of $\Bbb N$ are cooperative with greatest lower bounds, the emptyset is not so cooperative :)

how to prove the existence of the glb given just that lub

The next paragraph in the book you're citing tells you everything you need to know. (I think you've reversed glb and lub again here too.) Just verify that the glb of all upper bounds to a set is actually the lub of the set. It exists because glb's exist for all subsets.


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