I want to know if my proofs about the supremum are correct.

Question

Let $x \in \Bbb R$. Prove that $x = \sup\{q \in \Bbb Q: q \lt x\}$.

Proof 1:

Let $x \in \Bbb R$ and let $S=\{q \in \Bbb Q: q \lt x\}$.

Since $\Bbb Q$ is dense in $\Bbb R$, there exists an $r \in \Bbb Q$ such that: $$q \lt r \lt x$$

Since $x$ is an upper bound and $r \gt q$ with $r \in S$, $x$ must be the least upper bound.

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Proof 2:

Let $x \in \Bbb R$ and let $S=\{q \in \Bbb Q: q \lt x\}$.

Suppose $y = \sup S$. We want to show that $y=x$.

Suppose $ y \gt x$. By the density of $\Bbb Q$ in $\Bbb R$, there exists an $r \in \Bbb Q$ such that:$$x \lt r \lt y$$ We said $y$ is the least upper bound but $r \lt y$, which is a contradiction.

Suppose $y \lt x$. By the density of $\Bbb Q$ in $\Bbb R$, there exists an $r \in \Bbb Q$ such that:$$y \lt r \lt x.$$

We said $y$ is the least upper bound, but $r \in S$, which is a contradiction to $y$ being the least upper bound.

$\therefore$ We must have that $y=x=\sup S$.

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Are both these proofs correct? Which one is better/ more precise?

Answer 1

Proof 1:

Let x∈R and let S={q∈Q:q

.

Since Q is dense in R, there exists an r∈Q such that: q

What is $q$? You need to state "Let $q\in S$".

Since x is an upper bound and r>q with r∈S, x

must be the least upper bound.

Why? This shows that $q$ is not an upper bound. You must show that no number (whether or not is $S$) less than $x$ can be an upper bound. You must show if $y < x$ then there is a rational $q$ so that $y < q < x$ and so $q \in S$ so $y$ can not be an upper bound.

But you have to show $x$ is an upper bound.

Actually you never stated that $x$ is an upper bound. But it is. As for every $q \in S$, $q< x$.

Fix those two things and this is the better proof. It's a straight definition.

Proof 2:

Suppose y=supS

How do you know $\sup S$ exists. Is $S$ non-empty? Is $S$ bounded above. Both those must be shown.

Suppose y>x . By the density of Q in R, there exists an r∈Q such that: x

, which is a contradiction.

A contradiction of what? Real numbers less then $y$ exist.

Are you claiming that $r$ is an upper bound of $S$? That would be a contradiction. But why are you concluding $r$ is an upper bound of $S$?

(You can claim it as $r> x$ would imply for all $q \in S$ that $q < x < r$.)

Suppose y

We said y is the least upper bound, but r∈S, which is a contradiction to y

being the least upper bound.

That works but... Why introduce $y$ and then fit it to $x$. Just work with finding $\sup S$ directly.

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FWIW, my solution:

$\mathbb Q$ is unbounded so for any $x\in \mathbb R$ there are $q < x$ so $S$ is not empty. For any $q\in S$, $q < x$ so is an upper bound of $S$. If $r< x$ then because $\mathbb Q$ is dense in $\mathbb R$ there exists a rational $q$ so that $r < q < x$. So $q\in S$ and $r$ is not an upper bound so $x =\sup S$.

Answer 2

First one is both shorter and more clear -- an obvious winner to me.

Answer 3

The first proof doesn't make sense. You claim that $q<r<x$ without saying what $q$ is.

The second one is just fine.