# Real Analysis Proof Help Needed

• Aug 28th 2008, 12:37 PM
Ultros88
Real Analysis Proof Help Needed
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
• Aug 28th 2008, 12:54 PM
particlejohn
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 $\alpha = \inf A = \sup B$.
• Aug 28th 2008, 01:45 PM
Plato
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.