Prove that $\phi\leq\Phi$ where $\phi$ is the least upper bound of all lower sums and $\Phi$ is the greatest lower bound of all upper sums.

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Lesson Parent: 
The Logical Framework For Definite Integrals
Problem: 

Prove that $\phi\leq\Phi$ where $\phi$ is the least upper bound of all lower sums and $\Phi$ is the greatest lower bound of all upper sums.

Answer: 

It is true that $\phi\leq\Phi$.

SAMPLE SOLUTION

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Solution: 

We define $\phi$ to be the least upper bound of all lower sums and $\Phi$ to be the greatest lower bound of all upper sums.

Let's prove this by contradiction. Let's assume that <Sign in to see all the formulas> and show that it contradicts something we already know is true.

Since $\Phi$ is a greatest lower bound and it is less than $\phi$ we know that there is an upper sum that is less than $\phi$. How do we know this? Because if this were not true then $\Phi$ would only be a lower bound and not the greatest lower bound. More specifically we know that there is an upper sum in the interval <Sign in to see all the formulas> where <Sign in to see all the formulas>. Again, if this were not the case then $\Phi$ would only be a lower bound and not the greatest lower bound.

Similarly, there is an lower sum in the interval <Sign in to see all the formulas>. There fore, there exists two partitions where <Sign in to see all the formulas>.

But this violates the fact that for any two arbitrary partitions $P$ and $Q$, <Sign in to see all the formulas>. Therefore, we must have <Sign in to see all the formulas>.