Nested Interval Property and Infinity I'm having difficulty understanding some of the properties of real numbers which invoke infinity. For example, the fact that $$\bigcap_{n \in \mathbb{N}}\left(0,\frac{1}{n}\right) = \emptyset$$
What confuses me about this is that it seems that it can be proven to be false for arbitrarly large natural numbers. I considered something like this:
Consider the set $\bigcap_{n=1}^{m}\left(0,\frac{1}{n}\right)$ with $m \in \mathbb{N}$. Due to the desnity of the rationals, for any interval there exists a value c such that $0 < \frac{1}{c} < \frac{1}{n}$. So every interval $(0,\frac{1}{n})$ is non-empty. Now, choose c such that $0<\frac{1}{c}<\frac{1}{m}$. Since $\frac{1}{m} < \frac{1}{m-1} < ... < \frac{1}{1}$, then, since $\frac{1}{c}$ is in every interval, then $\frac{1}{c} \in \bigcap_{n=1}^{m}\left(0,\frac{1}{n}\right)$. Since m was an arbitrarily chosen natural number, this should hold for an arbitrarily large natural number.
But, I know that the intersection over all natural number is the empty set, and that means that something must be different between an arbitrarily large natural number and iterating over the entire set of natural numbers.
This breaks my intuition, and I realize I don't have an accurate and rigorous conception of infinity. For example, here are some central questions that I'm unsure how to answer:

*

*What is wrong with viewing infinity as some arbitrarily large natural number?

*How would I prove that while I can always create a smaller fraction than $\frac{1}{n}$ by constructing $\frac{1}{n+1}$, there isn't a fraction between $(0,\frac{1}{n})$ 'at infinity'?

I would appreciate advice on how I should intuitively view infinity, especially regarding the difference between something holding true for an arbitrarily large number but failing 'at infinity.' Thank you for your time!
 A: Since you've shown some confusion about it in the comments, let's start with the meaning of $$\bigcap_{n=0}^\infty A_n = \bigcap_{n\in \Bbb N} A_n$$
The definition of the first expression is the second (not vice versa). And the definition of the second is
$$\bigcap_{n\in \Bbb N} A_n = \{a\mid \forall n \in \Bbb N, a \in A_n\}$$
There is no taking of limits here, and no approximations. It is simple logic: To be in the interesection, it has to be in every set.
The real numbers $\Bbb R$ is the unique (up to isomorphism) complete ordered field. "Ordered field" just means a set with addition and multiplication behaving normally, and an order operation $\le$ that respects the addition and multiplication. By "complete" I mean that the order has the supremum property: If non-empty $A \subset \Bbb R$ is bounded above, then it has a supremum, or least upper bound. Lets play with that:
First, if $\Bbb N$ is bounded above, then it has a supremum $\omega \in \Bbb R$. But then $\omega -1 < \omega$ is not an upper bound, so there must be some $n \in \Bbb N$ with $ n > \omega -1$, and therefore $n + 1 > \omega$. Since $n+1 \in \Bbb N$, this contradicts that $\omega$ is an upper bound. Therefore $\Bbb N$ cannot be bounded. And so for every $x \in \Bbb R$, there is an $n \in \Bbb N$ with $n > x$, the "Archimedean principle".
But then if $\epsilon > 0$, it is invertible, and there is some $n > \frac 1\epsilon$. And that means $\frac 1n < \epsilon$. So for any $\epsilon > 0$, there is an $n \in \Bbb N$ with $\frac 1n < \epsilon$.
If $x \in \bigcap_{n\in \Bbb N} \left(0,\frac 1n\right)$, then $x \in (0,1)$. So $x > 0$. And for all $n, x \in \left(0,\frac 1n\right)$, so for all $n, x < \frac 1n$, in contradiction to the previous result. Hence there cannot be such an $x$, so $$\bigcap_{n\in \Bbb N} \left(0,\frac 1n\right) =\varnothing$$
