This is an exercise from Velleman's "How To Prove It". I would just like feedback on this proof to make sure it is correct.
Prove that there is a unique positive real number $m$ that has the following two properties:
- For every positive real number $x$, $\frac{x}{x+1} < m$
- If $y$ is any positive real number with the property that for every positive real number $x$, $\frac{x}{x+1} < y$, then $m \leq y$
$\exists ! m \in \mathbb{R}^+ \left [ \forall x \in \mathbb{R}^+(\frac{x}{x+1} < m) \wedge \forall y \in \mathbb{R}^+ \left [ \left (\forall x \in \mathbb{R}^+(\frac{x}{x+1}< y) \right ) \rightarrow m \leq y \right ] \right ]$
Proof:
Existence: Let $m = 1 \in \mathbb{R}^+$. Now let $x \in \mathbb{R}^+$ be arbitrary. Suppose $\frac{x}{x+1} \geq m$. Then $x \geq mx + m = x +1$, and $0 \geq 1$, which is a contradiction. Thus, $\frac{x}{x+1} < m$. Now let $y \in \mathbb{R}^+$ be arbitrary such that $\forall x \in \mathbb{R}^+(\frac{x}{x+1}< y)$. Suppose that $m = 1 > y$. Now consider $x = \frac{y}{1-y} \in \mathbb{R}^+$, which is defined since $y < 1$. $\frac{x}{x+1} = \frac{y}{1-y} \big / \left ( \frac{y}{1-y}+1 \right ) = \frac{y}{1-y} \big /\frac{1}{y-1} = y \geq y $. Thus, we have found an $x \in \mathbb{R}^+$ such that $\frac{x}{x+1} \geq y$, which is a contradiction. It follows that $m \leq y$.
Uniqueness: Let $n \in \mathbb{R}^+$ be arbitrary such that $\forall x \in \mathbb{R}^+(\frac{x}{x+1}< n)$ and $\forall y \in \mathbb{R}^+ \left [ \left (\forall x \in \mathbb{R}^+(\frac{x}{x+1}< y) \right ) \rightarrow n \leq y \right ]$. Since $m \in \mathbb{R}^+$ and $\forall x \in \mathbb{R}^+(\frac{x}{x+1}<m)$, we have $n \leq m$. Similar steps show that $m \leq n$. Therefore, $m = n$. $\square$