# Real Analysis Proof Help Needed

• Aug 28th 2008, 12:37 PM
Ultros88
Real Analysis Proof Help Needed
Let $\displaystyle A$ be bounded below, and define $\displaystyle B = \{b\in\Re : b$ is a lower bound for $\displaystyle A\}$. Show that $\displaystyle \sup B = \inf A$.

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

Now I need to show that for any other upper bound $\displaystyle s$ for $\displaystyle B$, $\displaystyle r \leq s$. 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 $\displaystyle A$ is bounded below, $\displaystyle B$ is not empty. Every $\displaystyle x \in A$ is an upper bound of $\displaystyle B$. So $\displaystyle B$ is bounded above which implies that $\displaystyle \sup B = \alpha$ exists in $\displaystyle S$ where $\displaystyle A \subset S$ (e.g. assume $\displaystyle S$ has lub property). If $\displaystyle \gamma < \alpha$ then $\displaystyle \gamma$ is not an upper bound of $\displaystyle B$. And so $\displaystyle \gamma \not \in A$. So $\displaystyle \alpha \in B$.

If $\displaystyle \alpha < \beta$ then $\displaystyle \beta \not \in B$. So $\displaystyle \alpha = \inf A = \sup B$.
• Aug 28th 2008, 01:45 PM
Plato
We will stick with the above notation: $\displaystyle \alpha = \inf (A)\,\& \,\beta = \sup (B)$.
Suppose that $\displaystyle \alpha < \beta$ then $\displaystyle \left( {\exists k \in B} \right)\left[ {\alpha < k \le \beta } \right] \Rightarrow \quad \left( {\exists j \in A} \right)\left[ {\alpha \le j < k \le \beta } \right]$.
But that is a contradiction because every $\displaystyle k \in B$ is a lower bound for $\displaystyle A$.
Therefore, we have $\displaystyle \beta \le \alpha$.
Now suppose that $\displaystyle \beta < \alpha$. Define $\displaystyle c = \frac{{\alpha + \beta }}{2} \Rightarrow \quad \beta < c < \alpha$.
This means that $\displaystyle c$ is a lower bound of $\displaystyle A$ which is a contradiction.