Proof regarding indexed families of sets and intervals

Question

I have the following problem:

Let $I$ be the set of real numbers that are greater than $0$. For each $x \in I$, let $A_x$ be the open interval $(0,x)$. Prove that $\cap_{x \in I} A_x = \emptyset$.

My reasoning is that this is true because $0$ is not contained and for every interval $(0,x)$ you can always find a smaller interval, let's say $(0,\frac{x}{2})$ so that the intersection of all such intervals is the empty set. I'm having trouble putting such reasoning in a formal proof.

Any ideas?

Thank you in advance.

Answer 1

Suppose the intersection is not empty. Then it contains some element (greater than zero), let's say $x$. What is the formal definition of intersection of sets? Try to derive a contradiction from here.

Answer 2

So, I should assume $\cap_{x \in I} A_x \neq \emptyset$, so:

(1) There exists some element $p \in A_x$ for every index $x$ in $I$.

(2) But $p \notin A_x$ if index $x < p$.

(3) So $p \notin A_x$ for at least one index $x \in I$.

(4) $p \notin \cap_{x \in I} A_x$, which contradicts (1).

Something like that? Also, is it legal to compare $x$ and $p$ since they are elements of different sets? ($p \in A_x$ and $x \in I$).

Thank you.