Def. 1.10 Baby Rudin: why doesn't $\mathbb{Q}$ fulfill the definition of having a least upper bound?

Question

I know similar questions have been asked on this site, but I have gone through all that I could find, but I'm still confused. In Baby Rudin, defintion 1.10 states that

"An ordered set $S$ is said to have the least-upper-bound property if the following is true: $E\subset S$ is not empty, $E$ is bounded above, then $\sup(E)$ exists in $S$"

I'm confused about this definition, because according to me, it seems like $Q$ would possess that property since e.g. set $A=\{x \in \mathbb{Q} : x^2 < 2\}$ is a non-empty subset of an ordered field $\mathbb{Q}$ and it is bounded above (by e.g. 5). So what in my reasoning is flaud since $\mathbb{Q}$ does not possess the least upper bound property?

Answer 1

Since $A\subset\Bbb Q$ and it has an upper bound, if $\Bbb Q$ had the least-upper-bound property, then $\sup A$ would exist in $\Bbb Q$. But it doesn't: $\sup A$ would have to be a non-negative square root of $2$, and no element of $\Bbb Q$ exists with that property.

Answer 2

It does have an upper bound, you are correct about that.

However, it does not have a least upper bound (lub). For example, you state that $5$ is an upper bound, but so is $4 \in \Bbb Q$. So $5$ is not an lub.

Is $4$? Well, no, since $2 \in \Bbb Q$ is even smaller.

And the fact is: Given any upper bound $s \in \Bbb Q$, there exists an upper bound $t \in \Bbb Q$ such that $t < s$. Rudin explicitly shows how one can construct $t$ from $s$.

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Due to this, $\Bbb Q$ does not have the lub property since we have a set that is bounded above but has no least upper bount.

Answer 3

An ordered set $S$ is said to have the least upper bound property if, for every non-empty subset $E$ which is bounded above in $S$, the supremum of $E$ exists in $S$.

So let's look at your question and the example in the description of your question. The set you have taken is $\mathbb{Q}$ which is indeed ordered. The subset of $\mathbb{Q}$ you have taken is $A=\{x \in \mathbb{Q} : x^2<2\}$, which is indeed non-empty. You are also right in inferring that $A$ is bounded above.

So we are left with questioning whether $A$ has a supremum in $\mathbb{Q}$. Note that this is usually not enough to prove a statement, but since we just want a counter-example here, this will suffice.

The supremum of $A$ if considered as a subset of $\mathbb{R}$ is $\sqrt{2}$. But $\sqrt{2}$ is not a rational number. This should be enough.

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If you want to prove that $A$ does not have a supremum in $\mathbb{Q}$, you can do so using the method of contradiction. Assume there exists a rational number $r$ which is the supremum of $A$.

Case 1: $r$ is less than $\sqrt{2}$
Since between any two distinct real numbers there is a rational number, there exists a rational number greater than $r$ which belongs to $A$. This contradicts our assumption that $r$ is the supremum of $A$.

Case 2: $r$ is greater than $\sqrt{2}$
Since between any two distinct real numbers there is a rational number, there exists a rational number lesser than $r$ which is an upper bound of $A$. This contradicts our assumption that $r$ is the supremum of $A$.