Walter Rudin Theorem 1.20 (a) Proof

Question

Theorem 1.20

(a) If $x\in\mathbb{R}$, $y\in\mathbb{R}$ and $x > 0$, then there is a positive integer $n$ such that $nx > y$.

In the proof, set $A$ is assumed to have a least upper bound in $\mathbb{R}$ that implies set $A$ is bounded above.

But I don't known how set $A$ is bounded above, given the constraint, $x$ is a positive real number and $n$ is a natural number.

More clarification:

$$A = \{nx\in\mathbb{R}_{>0} \mid (n\in\mathbb{N})\wedge(x\in\mathbb{R}_{>0})\}$$

My question is: how this set $A$ have a upper bound?

And the whole theorem and proof didn't really make sense to me.

I mean, if this theorem were reversed, i.e. $nx < y$, I wouldn't be able to make difference between these two scenario.

Potential duplicate post: is referring to possibility of set A being empty. In this post, I’m asking(implicitly) about contradiction statement of theorem and bdd property of set. Non empty set is just one of the condition must satisfy to talk about bddness of sets.

Answer 1

It doesn't. So you know $A$ must contain an element larger than $y$.

The proof, presumably, is trying to prove that it doesn't have an upper bound, by contradiction.

Answer 2

I might be wrong, but I think I have spotted a potential area of confusion.

I think throughout the proof, it may help to imagine $x$ as a fixed positive real number, so let $x\in\mathbb{R}$.

With this in mind, let's amend your definition of the set $A$ (because the one you gave lets $x$ vary as well- which is confusing). Let

$$A = \{nx\in\mathbb{R}_{>0} \mid (n\in\mathbb{N})\}.$$

The statement,

(a) is false

means that,

If $y\in\mathbb{R}$ , then there is no positive integer $n$ such that $nx > y$,

i.e.

If $y\in\mathbb{R}$ then for all positive integers $n,$ we have $nx \leq y$,

which is the same as saying,

The set $A$ is bounded above by $y$.

Now go back and see if the above logic works whenever $x$ is replaced by a positive real number. I think it does.

If you agree, then surely you agree that "(a) is false" implies that "the set $A$ is bounded above by $y$".

Note that this only gets us to sentence 2 of the proof, although hopefully it helped clear up some things. If not, then oh well, I'll let the next guy try...

But if you do agree with this, then you should continue on with the proof after sentence 2.

$$$$

Edit (copying from my comments below):

It is easy to assert that "the set $A$ is ("obviously") not bounded above" without proof. However, actually proving it is not so easy, and this is because writing proofs isn't easy. Saying "Set $A$ is not bounded from above because $n$ is a natural number" isn't a convincing argument, because this more-or-less assumes that Theorem 1.20(a) is clearly true here: but you're not allowed to use the theorem you're trying to prove when trying to prove that theorem.

If you think that Set $A$ is not bounded above, you need to provide a convincing argument. And this is exactly what Rudin does with his proof: he provides a convincing logical argument that Set $A$ is not bounded above. He does this by assuming (a) is false, and then $y$ would be an upper bound of $A$ by definition. He then shows that an absurdity arises, and therefore, (a) is true (argument by contradiction).

Note that it is clear from the outset that "the set $A$ is bounded above" is equivalent to saying" (a) is false". And saying "the set $A$ is not bounded above" is equivalent to saying "(a) is true".

However, it isn't immediately obvious whether or not $A$ is actually bounded above or not (this is more or less what we are trying to prove - and Rudin proves that, in fact, $A$ is not bounded above). It might seem obvious to you that Theorem 1.20 is true, but in order to show that it is true you need to, well, prove that it is true.