Intersection of Closed Intervals of Positive Reals is Zero

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 $\left[{0 \,.\,.\, x}\right]$.

Then:
 * $\displaystyle \bigcap_{x \mathop \in \R_{> 0} } B_x = \left\{{0}\right\}$

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

We have that:
 * $\forall x \in \R_{> 0}: 0 \in \left[{0 \,.\,.\, x}\right]$

So by definition of intersection:
 * $0 \in B$

and so by Singleton of Element is Subset:
 * $\displaystyle \left\{{0}\right\} \subseteq \bigcap_{x \mathop \in \R_{> 0} } B_x$

Suppose that $\exists y \in \R_{> 0}: y \in B$.

By definition of $B_x$:
 * $y \notin \left[{0 \,.\,.\, \dfrac y 2}\right] = B_{y/2}$

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

From this contradiction it follows that there can be no elements in $B$ apart from $0$.

That is:
 * $\displaystyle \bigcap_{x \mathop \in \R_{> 0} } A_x \subseteq \left\{{0}\right\}$

By definition of set equality:
 * $\displaystyle \bigcap_{x \mathop \in \R_{> 0} } B_x = \left\{{0}\right\}$