upper bound

**particlejohn** · July 4th 2008, 07:17 AM

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?

**Plato** · July 4th 2008, 08:17 AM

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 $\alpha = \beta$ .

**particlejohn** · July 4th 2008, 09:35 AM

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 $B$ ?

**particlejohn** · July 4th 2008, 09:42 AM

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

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