Clarification regarding the Archimedean Property This is kind of a general question concerning the Archimedean property in the context of real analysis. I understand that for every real number, there exists a natural number that is greater than or equal to it.
Specifically in this example, I am trying to understand what it means when the book states that $\frac{1}{K} \lt \epsilon$. Then it says $n \ge K $. $1/n \le 1/K \lt \epsilon$.

*

*Is this saying that you can always find a number $n$ greater than or
equal to any natural number $K$?


*How is $1/n \le 1/K \lt \epsilon$ to be interpreted?


*What number system is $n$ from? Is it the naturals?


*What does the book mean when it states that $1/K \lt \epsilon$? What is epsilon representing in this case?
I am using this to do some proofs on sequences of functions and uniform continuity and this seems to be a big part of it. I need some help interpreting what is actually going in this statement. Any help is appreciated.
 A: What this is trying to prove is that $\frac 1n \rightarrow 0$ as $n\rightarrow \infty $.  What that means is intuitively easy to comprehend but hard to put into words.  When considering all the $1/n$ as $n $ gets larger and larger, there comes a point where all the $1/n $ are really close to $0$.
Obviously that language is not good math.  What does "really close to zero" mean.  It means for any really small number the absolute value of all these $1/n $ at this point are even smaller.  We use $\epsilon $ to indicate any arbitrarily small value.  $\epsilon$'s only purpose in life is to be as small as we can imagine, simply so that we can show all $1/n $ at some point are even smaller.
What do I mean "at some point"?  That means there is some natural number $K $ (which might need to be very large) that for all the $n \ge K $, then all the $1/n $ are smaller than the really small $\epsilon$ we picked.
So in techical math language: $\frac 1n\rightarrow 0$ means for any $\epsilon >0$ (no matter how small) there is some $K $ ( maybe very large) so that for all $n$ that are larger than $K $, we have $|\frac 1n|<\epsilon $.
So that is what the proof is trying to show. 
So to your questions.
1) is this saying for any natural $K $ we can find an $n $ larger?  Not really.  Since we can find $n =K+1$ that would be trivial.  
What it is saying is that for any $\epsilon $ (no matter how small) we can find a $K $ so that $0 <1/K<\epsilon$.
The archemidean principal says: for any real number $x >0$ (no matter how small) and any real value $M $ (no matter how big) there is an integer $n $ so that $nx > M $.
So for any $\epsilon >0$ (no matter how small) we can find a natural number $K$ so that $K\epsilon >1$.
Which means $0  < 1/K < \epsilon $.
But that's only one natural number where $1/K <\epsilon $.
We need to prove that from now on out (i.e. for all $n \ge K $) we will also have $|1/n|<\epsilon $.
2) how is $\frac 1n \le \frac 1K  < \epsilon $ to be interpreted?  Well,... $1/n $ is a number.  It happens to be smaller or equal to  $1/K $ which is another number.  $1/K $ is smaller than a third number $\epsilon $.  So $1/n < \epsilon $.
(The proof doesn't state it, but since $n>0$, $1/n >0$ and $0 < 1/n \le 1/K <\epsilon $.)
3) what is $n $? $n $ is an arbitrary natural number that is equal or larger than $K $.
4) What does $\epsilon $ represent?  It is a very small number, as small as we like, but no matter how small we choose it to be, there'll come a point that all the $1/n $ are smaller.
Okay, let's do an example.  I want to show if I take a really small number, say the small number $\epsilon = 17/5,698,394$, there comes a point where all the $1/n $ are less than that.
If $n=2$, $1/2$ isn't smaller than $17/5,698,394$ so we haven't gotten that point yet.  If $n=5,678$, $1/5,678$ isn't either so we still haven't gotten to the point yet.  But if we get to the point $K = 4,000,000$ then for all $n\ge K $ we have $1/n < 17/5,698,394$ and we have gotten to the point.
So for that really small value there is a number $K $ where for all the $n \ge K $ we know $1/n $ is smaller.
If we took an even smaller number, say $\epsilon = 8.9 \times 10^{-567} $ we could still find a point, $K= 10^{568} $ where for all $n \ge K =10^{568} $ we have $1/n < \epsilon = 8.9\times 10^{-567} $.
A: It might help if I write out the definition of convergence for a sequence in $\mathbb{R}$ using logical symbols. 
DEFINITION. Let $L \in \mathbb{R}$. A sequence $(s_n)$ in $\mathbb{R}$ converges to $L$ if: $$\forall \varepsilon > 0 \, \exists K \in \mathbb{N} \, \forall n \in \mathbb{N} \, (n \geq K \longrightarrow |s_n - L| < \varepsilon).$$ Notice the "$\forall n \in \mathbb{N}$"! This universal quantifier is known to the authors of textbooks on real analysis, but the authors almost never mention this universal quantifier.
Now to your questions.


*

*It is true that if you have a natural number $K$, you can always find a natural number $n$ such that $n \geq K$. (Obviously, there are infinitely many such $n$.) The part of the definition of convergence mentioning $n \geq K$ simply means that if you take any natural $n$ that satisfies $n \geq K$, then the statement $|s_n - L| < \varepsilon$ must be true.

*For your example, $(s_n)$ is given by $s_n = 1/n$. Now, once you have said, "Let $n \in \mathbb{N}$, and suppose $n \geq K$," then you try to show that $|s_n - L| < \varepsilon$ must be true, i.e., $|1/n - 0| < \varepsilon$ must be true. That's exactly what is being shown by writing $1/n \leq 1/K < \varepsilon$.

*Again, the "$\forall n \in \mathbb{N}$" is rarely (if at all) emphasized in real analysis textbooks, but it should be (omitting it causes students/readers to raise your very question). So $n$ is taken from $\mathbb{N}$.

*The statement $1/K < \varepsilon$ was used to help show that $1/n < \varepsilon$ is true.


One more comment. Look at the definition of convergence as I've written it above. To prove the convergence of a sequence with your bare hands using only this definition, you must look at the quantifiers, the logical connectives, and their ordering that is given in the definition. Therefore, the structure of your proof should be like so:

"Let $\varepsilon > 0$."
        [A choice for $K$ goes here.]
              "Let $n \in \mathbb{N}$."
                    "Suppose $n \geq K$."
                    [Proof that $|s_n - L| < \varepsilon$ goes here.]
  "Thus, $(s_n)$ converges to $L$."

The proof for your example:
"Let $\varepsilon > 0$. Choose a $K \in \mathbb{N}$ such that $K > 1/\varepsilon$. (Such a $K$ exists by the Archimedean property.) Let $n \in \mathbb{N}$, and suppose $n \geq K$. Then $1/n \leq 1/K$, so $$|s_n - 0| = \biggl| \dfrac{1}{n} \biggr| = \dfrac{1}{n} \leq \dfrac{1}{K} < \varepsilon.$$ Therefore, $\lim 1/n = 0$."
A: The Archimedean proprty is that for every positive $x, y$ there exists $n\in \mathbb N$ such that $nx>y$. In particular if $x=1$ and $y=1/\epsilon>0$ there exists $n\in \mathbb N$ such that $n=nx>y=1/\epsilon,$ that is $1/n<\epsilon.$ And hence,also,  $1/n'<\epsilon$ whenever  $n<n'\in \mathbb N.$
A: 1: By the Archimedean property, there is some natural number $K$ such that $K>{1\over \epsilon}$. The rest of this part is just algebra: $\epsilon>0$ by assumption, so ${1\over\epsilon}>0$ as well (exercise), and so $K>0$ as well since $K>{1\over \epsilon}$. But this means that when we take the reciprocal of each side of the inequality "$K>{1\over\epsilon}$," we get "${1\over K}<\epsilon$."
2: Exactly as written: ${1\over n}$ is less than or equal to ${1\over K}$, which in turn is strictly less than $\epsilon$. I'm not sure what your confusion is here?
3: Yes, $n$ is a natural number here. Remember you're computing the limit of the sequence $({1\over n})_{n\in\mathbb{N}}$.
4: Look at the beginning where it says "If $\epsilon>0$ is given ..." So $\epsilon$ is an arbitrary positive real number. Remember that you're trying to prove a statement of the form "For every $\epsilon>0$, ..." - the way you do this is show how, for any such $\epsilon$, you can prove the desired result.
A: *

*It is true that for any natural number $K$ you can find a larger natural number, but for that you don't need the Archimedean property; it's part of the very definition of the natural numbers.
Indeed for the formulation of the Archimedean property you used, it could not be derived from it, as it only states that there exists a natural number that is greater or equal to a given number; since $K$ already is a natural number, the existence of a natural number $K'\ge K$ is already satisfied by $K'=K$ as $K\ge K$ and $K$ is by definition natural.
The reason why you consider numbers $n>K$ is because of the definition of limits: A sequence $a_n$ converges to the limit $a$ if for every $\epsilon>0$ there exists a natural number $K$ so that for each natural number $n>K$, $\left|a_n-a\right|<\epsilon$. The Archimedean property is used to argue that $K$ exists.

*$1/n < 1/K \le \epsilon$ means $1/n < 1/K \land 1/K \le \epsilon$. The first inequality is equivalent to $n>K$ which you need from the definition of limits. The second inequality is equivalent to $K>1/\epsilon$ which is what you get by using the Archimedean property.

*Yes, $n\in\mathbb N$. It is the index to the sequence $(1/n)_{n\in\mathbb N}$.

*$\epsilon$ is an arbitrary positive real number, as required by the definition of the limit of a sequence.
