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To quote from my lecture notes:

When every subset of A has a lub and glb, we say that the order is a complete lattice, but this takes us beyond the syllabus. It is notable that $\mathbb{Q}$, ordered by $\leq$, is not a complete lattice but $\mathbb{R}$, ordered by $\leq$, is a complete lattice. This is the fundamental difference between $\mathbb{Q}$ and $\mathbb{R}$.

Please can someone explain why this is true? I can't see what the lub of $\mathbb{R}$ would be, in the same way I can't see a lub for $\mathbb{Q}$.

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You are right. In the usual sense of lattice completeness, the reals under the ordinary order do not form a complete lattice. – André Nicolas Dec 18 '12 at 20:25

The book is mistaken; for instance, $A=\mathbb{R}\subseteq \mathbb{R}$ has no g.l.b. or l.u.b. in $\mathbb{R}$, so $\mathbb{R}$ is not a complete lattice. However, $\widehat{\mathbb{R}} = \mathbb{R} \cup \{ - \infty, \infty \}$ is a complete lattice, where we declare $-\infty < r < \infty$ for all $r \in \mathbb{R}$.

But even this modification doesn't help in the case of $\widehat{\mathbb{Q}} = \mathbb{Q} \cup \{ -\infty, \infty \}$.

Consider the set $$A = \{ x \in \mathbb{Q}\, :\, x^2 > 2 \} \subseteq \widehat{\mathbb{Q}}$$ Does it have a g.l.b. in $\widehat{\mathbb{Q}}$?


No it doesn't. Considered as a subset of $\mathbb{R}$, $\bigwedge A = \sqrt{2}$, which is not a rational number; and if $q<\sqrt{2}$ is rational (or $-\infty$) then there exists a rational $q'$ with $q<q'<\sqrt{2}$, so no $q<\sqrt{2}$ can be a g.l.b. Likewise, no $q>\sqrt{2}$ can be a g.l.b. since then $q \in A$ and there would exist $q' \in A$ with $q'<q$.

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You are right. Under the usual definition of lattice completeness, the reals under the ordinary order do not form a complete lattice, since there are unbounded sets.

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Your lecture notes are confusingly written: You want to distinguish between "complete lattice" and a "Dedekind complete" partial order.

Both $\mathbb Q$ and $\mathbb R$ are incomplete lattices, meaning there is no supremum (least upper bound) and no infimum (greatest lower bound) unless you add them by adding $\{-\infty, \infty\}$.

But one main difference between the two is that one of them is Dedekind complete while the other isn't: $\mathbb R$ is Dedekind complete while $\mathbb Q$ is not. To see this, verify that the pair $\langle \{q \in \mathbb Q : q < \sqrt{2}\}, \{q \in \mathbb Q : q > \sqrt{2}\}\rangle$ is a gap in $\mathbb Q$. See for example Just / Weese, page 54.

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It's easier to imagine when we consider specific subsets.

Let's consider the subset $A=${$1, 1.4, 1.41, 1.414, 1.4142, ...$}, where the decimals approach $\sqrt{2}$.

The least upper bound of this subset, in $\mathbb{R}$, is $\sqrt{2}$. Every decimal in $A$ is less than $\sqrt{2}$, so $\sqrt{2}$ is an upper bound. And since there are decimals in $A$ that get arbitrarily close to $\sqrt{2}$, no number less than $\sqrt{2}$ can be an upper bound for $A$. Therefore, $\sqrt{2}$ is the least upper bound of $A$.

On the other hand, in $\mathbb{Q}$, there is no least upper bound. Why? Suppose there were a least upper bound $L$. Clearly $L>\sqrt{2}$. But then there's another rational $L'$ between $L$ and $\sqrt{2}$, and $L'$ is also an upper bound for $A$. So $L$ wasn't the least such upper bound after all, contradiction.

It was easy to show there that $\mathbb{Q}$ doesn't always have least upper bounds, but showing that any bounded subset in $\mathbb{R}$ does have a least upper bound is much trickier. See, e.g., a bottom-up construction of the reals by Dedekind cuts if you want a formal proof of that.

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