Intersection of Open Intervals of Positive Reals is Empty

Theorem
Let $\R_{> 0}$ be the set of strictly positive real numbers.

For all $x \in \R_{> 0}$, let $A_x$ be the open real interval $\left({0 \,.\,.\, x}\right)$.

Then:
 * $\displaystyle \bigcap_{x \mathop \in \R_{> 0} } A_x = \varnothing$

Proof
Let $\displaystyle A = \bigcap_{x \mathop \in \R_{> 0} } A_x$.

Suppose that $A \ne \varnothing$.

Then:
 * $\exists y \in \R_{> 0}: y \in A$

By definition of open interval:
 * $y \notin \left({0 \,.\,.\, y}\right) = A_y$

and so by definition of intersection of family:
 * $y \notin A$

From this contradiction it follows that there can be no elements in $A$.

That is:
 * $\displaystyle \bigcap_{x \mathop \in \R_{> 0} } A_x = \varnothing$