First off what is important about finding an n that satisfies b < na
Nothing is important about finding the $n$. What's important is to understand that is you have real positive number $a$, no matter how small, and another positive real number $b$, no matter how big, you will be able to add up some number of the teeny-tiny $a$ and end up something bigger than the huge ginormous gargantuan $b$.
This is important. It means: The integers are not bounded. It means no number so large that in can not be surpassed by adding a smaller value enough times. It means by corrolary that no positive number is so small we can't find a smaller positive number. And it means, also as a corollary that two numbers, no matter how close they are together will always have a number between them.
Now these observations may seem obvious but that are not givens. They have to be proven.
What I don't understand is why is k0 introduced?
We need to rule out that we can't just keep adding $a$s together and never get above $b$. If the set of all integers $k$ where $ak \le b$ is bounded above then it must have a greatest element, $k_0$. If it has a greatest element then we are done; we just take the next element $k_0 +1$ it is not in the set of all such integers. So $a(k_0 + 1) \not \le b$ so $a(k_0 + 1) > b$.
So we would be done with our proof... IF we know the set was bounded above.
Why does E being non-empty depend on whether b
If $b < a$ then $b < 1*a$ and for any $n \in \mathbb N$ we have $n \ge 1$ so $an > b$ and there are no natural numbers where $an \le b$. So $E$ is empty.
If $b\ge a$ then $a*1 \le b$ and $1 \in E$ and $E$ is not empty.
How does it follow that n = k0+1 must satisfy b
$k_0$ is the largest natural number in $E$. $n= k_0 + 1 > k_0$. So $n$ is larger than then largest number in $E$. So $n$ is NOT in $E$. So it is not true that $an \le b$. So $an > b$.
Whats the significance of if b < a then set n=1?
There are two different proofs. There is one proof where $a > b$.
There is another if $a\le b$.
The proof if $a > b$ doesn't deal with $E$ at all. (For one thing, if $a > b$ then $E$ is empty.) For another if $a > b$ the proof is trivial.
I thought we were dealing with the set E. How is E considered non-empty when 1∈E
If $a > b$ then $1\not \in E$. That's the problem. If $a > b$ then we can't do the proof as it is presented If $a > b$ then $E$ is empty. So we have to do another proof. The other proof is one line long.
If $a > b$ then $1*a > b$ and therefore there an $n$ where $na > b$. Because $1$ can be that $n$.
But if $a \le b$ then we have to do the "real" proof.
Let $E$ but the set of all natural numbers, $k$ so that $ak \le b$.
If $a\le b$ then $1*a \le b$ so $1 \in E$. (IF $a \le b$). So $E$ is non-empty.
Let $k \in E$. That means $ka \le b$. So $k \le \frac ba$. So $E$ is bounded above by $\frac ba$. (IF $a \le b$... if $a > b$ then $E$ is empty and none of this matters...)
Since $E$ is bounded above it has a largest element. Call the largest number ..... you know what, DON'T call it $k_0$. That is tripping you up somehow. Call $LOOKATMEIMTHEBIGGESTNUMBERINALLOFE$. $LOOKATMEIMTHEBIGGESTNUMBERINALLOFE$ is the biggest number in all of $E$.
Now let $HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE = LOOKATMEIMTHEBIGGESTNUMBERINALLOFE + 1$. Now $HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE$ is even bigger than $LOOKATMEIMTHEBIGGESTNUMBERINALLOFE$ which is the biggest number in $E$. So $HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE$ is even bigger than the biggest number in $E$ so it is too big to be in $E$.
So $HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE \not \in E$ which means $HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE\cdot a \not \le b$ and $HEYIMEVENBIGGERTHANLOOKATMEIMTHEBIGGESTNUMBERINALLOFE\cdot a> b$
And that's the proof.