1. ## upper bound

Let $A \subset \mathbb{R}$ be bounded above and $B = \{x: x \ \text{is an upper bound of} \ A \}$. Prove $\sup A = \inf B$.

Since $B$ is bounded below, it has a greatest lower bound $\inf B$. To show that $\sup A = \inf B$, we need to show that $\forall x \in A, x \leq \sup A \Longleftrightarrow \forall x \in B, x \geq \inf B$? It this all there is to it?

2. Suppose that $\alpha = \sup (A)\,\& \,\beta = \inf (B)$.
If $\beta < \alpha$ then by definition of supremum and infimum $\beta$ is not an upper bound of A, so $\left( {\exists c \in A} \right)\left[ {\beta < c \le \alpha } \right]$ but c is not a lower bound of B. That means $\left( {\exists d \in B} \right)\left[ {\beta \le d < c} \right]$, but that means d is a upper bound of A which is impossible because ${d < c \le \alpha }$. That means that $\alpha \le \beta$.
This time suppose that $\alpha < \beta$ which implies $\alpha < \frac{{\alpha + \beta }}{2} < \beta$.
But this means that $\frac{{\alpha + \beta }}{2}$ is an upper bound for A but it cannot belong to B. That is a contradiction. Thus $\alpha = \beta$.

3. Originally Posted by Plato
Suppose that $\alpha = \sup (A)\,\& \,\beta = \inf (B)$.
If $\beta < \alpha$ then by definition of supremum and infimum $\beta$ is not an upper bound of A, so $\left( {\exists c \in A} \right)\left[ {\beta < c \le \alpha } \right]$ but c is not a lower bound of B. That means $\left( {\exists d \in B} \right)\left[ {\beta \le d < c} \right]$, but that means d is a upper bound of A which is impossible because ${d < c \le \alpha }$. That means that $\alpha \le \beta$.
This time suppose that $\alpha < \beta$ which implies $\alpha < \frac{{\alpha + \beta }}{2} < \beta$.
But this means that $\frac{{\alpha + \beta }}{2}$ is an upper bound for A but it cannot belong to B. That is a contradiction. Thus $\alpha = \beta$.
Why can't $\frac{\alpha + \beta}{2}$ belong to $B$?

4. oh ok because $\frac{\alpha + \beta}{2} < \beta$.