Union of Closed Intervals of Positive Reals is Set of Positive Reals

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

For all $x \in \R_{> 0}$, let $B_x$ be the closed real interval $\closedint 0 x$.

Then:
 * $\ds \bigcup_{x \mathop \in \R_{>0} } B_x = \R_{\ge 0}$

Proof
Let $\ds B = \bigcap_{x \mathop \in \R_{>0} } B_x$.

Let $y \in B$.

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

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

or
 * $y \in \R_{>0}$

In either case, $y \in \R_{\ge 0}$

So:
 * $\ds \bigcap_{x \mathop \in \R_{>0} } B_x \subseteq \R_{\ge 0}$

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

If $y = 0$ then $y \in \R_{\ge 0}$ by definition.

Otherwise, by the Archimedean Property:
 * $\exists z \in \N: z > y$

and so:
 * $y \in B_z$

That is by definition of union of family:
 * $\ds y \in \bigcap_{x \mathop \in \R_{\ge 0} } B_x$

So by definition of subset:
 * $\ds \R_{\ge 0} \subseteq \bigcap_{x \mathop \in \R_{>0} } B_x$

By definition of set equality:
 * $\ds \bigcup_{x \mathop \in \R_{>0} } B_x = \R_{\ge 0}$