Union of Open Intervals of Positive Reals is Set of Strictly Positive Reals

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:
 * $\displaystyle \bigcup_{x \mathop \in \R_{> 0} } A_x = \R_{> 0}$

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

Let $y \in A$.

Then by definition of union of family:
 * $\exists x \in \R_{> 0}: y \in A_x$

As $A_x \subseteq \R_{> 0}$ it follows by definition of subset that:
 * $y \in \R_{> 0}$

So:
 * $\displaystyle \bigcap_{x \mathop \in \R_{> 0} } A_x \subseteq \R_{> 0}$

Let $y \in \R_{> 0}$.

By the Archimedean Property:
 * $\exists z \in \N: z > y$

and so:
 * $y \in A_z$

That is by definition of union of family:
 * $y \in \displaystyle \bigcap_{x \mathop \in \R_{> 0} } A_x$

So by definition of subset:
 * $\displaystyle \R_{> 0} \subseteq \bigcap_{x \mathop \in \R_{> 0} } A_x$

By definition of set equality:
 * $\displaystyle \bigcup_{x \mathop \in \R_{> 0} } A_x = \R_{> 0}$