Counter-intuitive intersection property Consider the following sequence of sets $A_k $ ,Where $A_k  = \{ n \in \mathbb Z \mid n \geqslant k \}$. Now let us consider $\bigcap_{k=1}^{\infty} A_k$. In a book that I was going through it says that this expression is equal to $\varnothing $.
I tried to prove it on my own and I could find an easy argument which I suspect has flaws and I am interested in finding it out. My reason for suspicion is a piece of information, that I already had that, this is in a way related to something called as Caratheodory's lemma which has a different proof.
Anyways the little, making -it-look -like-obvious type of proof goes as follow:
Let us assume that the expression  $\bigcap_{k=1}^{\infty} A_k$ equals to some set  $\ B $ .As we can see every set in the sequence contains its successor . So the expression $\bigcap_{k=1}^{\infty} A_k$ must be a set of the format $A_m $  for some $m \in \mathbb Z$, otherwise if it is a set which is a collection of integers then for every integer we can find a set which cannot have it . Now at the same time if is equal to $\ A_m$ then the set $ A_t $, where $ t $, for any $t >\ m $ cannot contain it. As such the only thing all the sets can have in common is $\varnothing $. 
What are the flaws in this argument, what is a standard proof for it? And if there aren't any flaws in this argument then It gives me a feeling that this points to some connections to the axioms of of set theory, like in defining the null set to be present  in all sets,which can be powerful enough to lead to other such implications. It maybe just a feeling but I would also like to know if this is also true. 
 A: A standard straightforward proof would be:
We prove $\bigcap_{k=1}^\infty A_k = \varnothing$ by contradiction. Namely, assume that the set has an element $x$ and derive a contradiction.
Suppose, therefore, that $x\in\bigcap_{k=1}^\infty A_k$. Then, by definition of intersection, $x\in A_1$ in particular, so $x$ is a positive integer. But then $x+1$ is also a positive integer, and $x\notin A_{x+1}$ which means that $x$ cannot be in $\bigcap_{k=1}^\infty A_k$ after all, a contradiction.
We have proved that it leads to a contradiction to assume that $\bigcap_{k=1}^\infty A_k$ has an element, so it must be the empty set.

None of the many things named after Carathéodory seems to be particularly related to this.
A: $\newcommand{\Int}{\mathbf{Z}}\newcommand{\Empty}{\varnothing}$You've asked several questions, of which I'll address only two: Is the empty set contained in every set (yes, modulo fine print), and what is the standard proof of the fact in your question (supplementary comments to Henning Makholm's (+1) answer).
If $U$ is a set (the universe of whatever context in which you're working) and if $X$ and $Y$ are subsets of $U$, the definition of "$X \subset Y$" is:

For every $x$ in $U$, $x \in X$ implies $x \in Y$.

Loosely, every element of $X$ is an element of $Y$.
Consequently, $\Empty \subset Y$ for every subset $Y \subset U$: For every $x$ in $U$, the statement "$x \in \Empty$" is false (by definition of the empty set), so "$x \in \Empty$ implies $x \in Y$" is vacuously true.
If $I$ is a set and $\{A_{k}\}_{k \in I}$ is some family of subsets of $U$, then by definition
$$
\bigcap_{k \in I} A_{k} = \{x \in U : \text{$x \in A_{k}$ for all $k$ in $I$.}\}
$$
In your example, $U = \Int$, $I$ is the set of positive integers, and $A_{k} = \{n \in \Int : n \geq k\}$ is the set of integers no smaller than $k$. An integer $n$ is in the intersection $\bigcap_{k=1}^{\infty} A_{k}$ if and only if $n \in A_{k}$ for all $k \geq 1$.
Before we pursue this further, I'd like to invite you to play an adversarial game, "Pick the larger integer." Because I'm generous, I'll let you go first.... :)
Now, there's a relationship between this droll game and the intersection of the $A_{k}$: Whatever integer $n$ you pick, we'll test for membership in $\bigcap_{k=1}^{\infty} A_{k}$ by checking whether $n \in A_{k}$ for every positive integer $k$. By definition, $n \in A_{k}$ if and only if $n \geq k$.
So, you pick an $n$ and tell me what it is. I pick a larger positive integer, such as $m = |n| + 1$. Thanks to the definition of the sets $A_{k}$ and my choice of $m$, we have $n \not\in A_{m}$. (You lose!) This means $n \not\in \bigcap_{k=1}^{\infty} A_{k}$. Since $n$ was an arbitrary integer (I have a winning strategy no matter what you do), we have $\bigcap_{k=1}^{\infty} A_{k} = \Empty$.
