Prove no member of $S = \left\{x \mid x \in \mathbb{Q^+}, x^2 < 2 \right\}$ can be an upper bound for $S$

Question

Prove no member of $S = \left\{x \mid x \in \mathbb{Q^+}, x^2 < 2 \right\}$ can be an upper bound for $S$.

So at first, I tried to prove that this statement was false; that a member of $S = \left\{x \mid x \in \mathbb{Q}, x^2 < 2 \right\}$ CAN be an upper bound for $S$.

Say $r \in S$, $r \in \mathbb{Q^+}$, and $r^2 < 2$. I can show there is a $r' \in \mathbb{Q^+}$ with $r' \in S$ and $r'^2 < 2$ such that $r' > r$.

Let $r' = \frac{4r}{r^2 + 2}$.

First, I prove that $r' > r$. $r' \in \mathbb{Q^+}$, so $r' > 0$. We know $r^2 < 2$, so $r^3 < 2r \rightarrow r^3 + 2r < 2r + 2r \rightarrow r^3 + 2r < 4r \rightarrow r(r^2 + 2) < 4r \rightarrow r < \frac{4r}{r^2+2} = r'$. So $r < r'$.

Next, I show that $r' \in S$. $r^2 < 2 \rightarrow (r^2 - 2)^2 > 0 \rightarrow r^4 - 4r^2 + 4 > 0 \rightarrow r^4 + 4r^2 + 4 > 8r^2 \rightarrow 2(r^4 + 4r^2 + 4) > 16r^2 \rightarrow 2 > \frac{16r^2}{r^4 + 4r + 4} = r'^2$, so $2 > r'^2$.

So since $r' \in S$, $r' \in \mathbb{Q^+}$, $r'^2 < 2$, and $r' > r \in S$, $r'$ is an upper bound for $S$.

But the statement (which is true) says that $r'$ cannot be an upper bound for $S$, although it feels like I have just shown it is. Where is my fallacy?

Answer 1

You have proven that given $r\in S$ you can find $r'\in S$, $r'>r$. This means that $S$ has not an upper bound.

To prove that $r'$ is an upper bound you should instead prove that all $r \in S$ satisfy $r<r'$.

Answer 2

You have proven that we can always obtain an upper bound greater than a given upper bound.

I think that the idea to obtain a proof is assume that an upper bound exists in $S$, then obtain a contradiction from that upper bound.

Proof. Suppose for sake a contradiction that $M$ is an upper bound for $S$ and $M\in S$. So $M^2<2$. Let $0<\epsilon<1$ be a small number; then we have $$(M+\epsilon)^2=M^2+2\epsilon M+\epsilon^2< M^2+4\epsilon+\epsilon=M^2+5\epsilon$$ since $M<2$ (because if $M\ge2$, then $M^2\ge4\ge2$ and hence $M\notin S$) and $\epsilon^2<\epsilon$. Since $M^2<2$, we can choose an $0<\epsilon<1$ such that $M^2+5\epsilon<2$, thus $(M+\epsilon)^2<2$. This means that $M+\epsilon\in S$; but this contradicts the fact that $M$ is an upper bound of $S$.