Real Analysis Proof Help Needed

**Ultros88** · August 28th 2008, 12:37 PM

Let $A$ be bounded below, and define $B = \{b\in\Re : b$ is a lower bound for $A\}$ . Show that $\sup B = \inf A$ .

This is what I have:
Let $r=\inf A$ . Let $b\in B$ be any lower bound for $A$ . It follows that $r\geq b$ . Since $r\geq$ for all $b \in B$ , $r$ is an upper bound for $B$ .

Now I need to show that for any other upper bound $s$ for $B$ , $r \leq s$ . I think this has to do with showing that any upper bound for B is an element of A, but I can't figure out how to show this.

Thanks for the help,
Ultros

**particlejohn** · August 28th 2008, 12:54 PM

Since $A$ is bounded below, $B$ is not empty. Every $x \in A$ is an upper bound of $B$ . So $B$ is bounded above which implies that $\sup B = \alpha$ exists in $S$ where $A \subset S$ (e.g. assume $S$ has lub property). If $\gamma < \alpha$ then $\gamma$ is not an upper bound of $B$ . And so $\gamma \not \in A$ . So $\alpha \in B$ .

If $\alpha < \beta$ then $\beta \not \in B$ . So $\alpha = \inf A = \sup B$ .

**Plato** · August 28th 2008, 01:45 PM

We will stick with the above notation: $\alpha = \inf (A)\,\& \,\beta = \sup (B)$ .
Suppose that $\alpha < \beta$ then $\left( {\exists k \in B} \right)\left[ {\alpha < k \le \beta } \right] \Rightarrow \quad \left( {\exists j \in A} \right)\left[ {\alpha \le j < k \le \beta } \right]$ .
But that is a contradiction because every $k \in B$ is a lower bound for $A$ .
Therefore, we have $\beta \le \alpha$ .
Now suppose that $\beta < \alpha$ . Define $c = \frac{{\alpha + \beta }}{2} \Rightarrow \quad \beta < c < \alpha$ .
This means that $c$ is a lower bound of $A$ which is a contradiction.

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