Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb{Q}$.

share|improve this question
    
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

4 Answers 4

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 $\widehat{\mathbb{Q}}$?

Answer:

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 $q'<q$.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.