Infinite intersection with real numbers I cannot wrap my head around something I did read in a math book. Maybe someone can tell me at which point I went wrong.
Background
The author defines the infinite intersection as follows:

$$\bigcap_{n \in M} A_{n} := \{ x\,|\,x \in A_{n}\:\text{ for all } n \in M \}$$

Later on the author brings up an example:

$$
M = \mathbb{N}, A_{n} := \Big\{ x \in \mathbb{R}\, \Big| \,0 < x < \frac{1}{n} \Big\}$$

And for this example the author states, that

$$
\bigcap_{n \in M} A_{n} = \emptyset
$$

Problem
I do not understand why the infinite intersection results in the empty set.
My thoughts

*

*$n$ ranges from $1$ to infinity (never reaching infinity) $\implies$ No $A_{n}$ should be empty, as it should contain at least $\frac{1}{n+1}$


*$A_n$ is the set of real numbers between $0$ and $\frac{1}{n}$ (both exclusive)


*$A_{n} \subsetneq A_{n-1}$ should be true


*So for any $j$, $k\in \mathbb{N}$ with $k \leq j$: $A_k$ contains at least $\frac{1}{j+1}$. Therefore the intersection of $A_n$ should contain at least this lowest element.
Example:
$A_1$ from $0$ to $1$
$A_2$ from $0$ to $\frac{1}{2}$
$A_3$ from $0$ to $\frac{1}{3}$
All of them contain the element $\frac{1}{4}$.
So basically my problem is, that I do not understand where I did make the mistake. Since my thoughts conflict with the statement, that the infinite intersection equals the empty set. Because in my imagination it contains at least one element which goes towards 0 but never reaches 0.
Thanks in advance for your time and effort.
 A: Yes, but $A_5$ doesn't contain $\frac14$.  For any $x>0$ there is an integer $N$ such that $x>\frac1N$.  Then for $n>N$, $x\notin A_n$.  Thus, there is no number in all of the $A_n$.
A: Let's suppose there was at least one number $x$ in the infinite intersection
$$
x \in \bigcap_{n \in M}A_n
$$
by the Archimedean property of real numbers, there is some natural number $N$ such that
$$
\frac{1}{N} < x
$$
and so ${x \notin A_N}$. Meaning we are forced to conclude
$$
x \notin \bigcap_{n \in M}A_n
$$
$x$ was arbitrary. This argument shows that the infinite intersection could not possibly have any elements inside, and so
$$
\bigcap_{n \in M}A_n = \emptyset
$$
TL;DR: for every number you believe is in the intersection, you can find some $A_N$ such that this number is not in $A_N$. Hence the intersection must be empty.

EDIT: maybe another way to think about it is this. In the above answer, I gave you a proof that the infinite intersection must be empty, and in the comments I tried to sort of "play to your intuition". Instead, let's look at how we might go about defining some of these things. Expressions such as
$$
\bigcap_{n=1}^{\infty}\left(0,\frac{1}{n}\right)
$$
(which is the same as in your question) don't have any "self-evident" answer. We are required to expand upon what we mean by this sort of expression in order to try and conclude a "reasonable" answer, just like with infinite sums, where we are required to define limits and things to try to make sense of it. If we take everything we know about real numbers as "given" (including, as my proof above uses, the Archimedean property) - and expand the infinite intersection as meaning "any element in the above intersection must belong to each set in the intersection", then the proof I did is correct, and the intersection is empty. Since as we said, for any number $x$, we can always find a set ${\left(0,\frac{1}{N}\right)}$ in the intersection such that $x$ is not in this set, and so couldn't be in the intersection.
If the answer was non-empty, it would be a contradiction to my assumptions. Meaning ${(1)}$ what we know about the behaviour of real numbers is broken, or ${(2)}$ I expanded what we meant by the infinite intersection incorrectly. And I don't think there is anything wrong with either assumption. Of course, you could disagree with me, and use alternative definitions - and then you could arrive at different answers. But if you follow the assumptions I made, the only logically sound result is ${\emptyset}$. You could override this if you like, and say "the result is non-empty, there is some number in there" - but then things become inconsistent, since we would start contradicting ourselves.
I think should also help explain your confusion in the comments, as you say

I can find ${A_N}$ so that ${x_1 \notin A_N}$, but I can also find ${x_2 \in A_N}$. Why is one stronger than the other?

and the answer is because the latter is not how we define an infinite intersection. Not only that, but I don't think this would really be helpful at all in trying to define what we are trying to represent when we say ${\bigcap_{n=1}^{\infty}\left(0,\frac{1}{n}\right)}$. Sure, ${x_2}$ is in ${A_N}$ - but intuitively we can then find a set for a bigger ${N'>N}$ so that ${x_2}$ is once again not there. So that would not be a helpful definition.
Also, I want to further clarify my example in the comments of
$$
1 + 2 + 3 + \dots
$$
I am definitely not saying infinity is a number when I say "we would call the result of this ${\infty}$" <- infinity is not a number, and in the comment I do further state "this is not a positive integer". The ONLY point I was trying to make with this example is that if we follow our definition of the infinite sum, and try ${1 + 2 + 3 + \dots}$ - we end up with something that couldn't possibly be a number. In the same way - we followed our definition of an infinite intersection of a bunch of sets that were subsets of eachother, and ended up with something that was not one of the sets (none of the ${A_N}$ are ${\emptyset}$). Perhaps this example was not helpful and could add more confusion.
