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 $\openint 0 x$.

Then:

$\ds \bigcup_{x \mathop \in \R_{>0} } A_x = \R_{>0}$

Proof

Let $\ds A = \bigcup_{x \mathop \in \R_{>0} } A_x$.

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:

$\ds \bigcup_{x \mathop \in \R_{>0} } A_x \subseteq \R_{>0}$

$\Box$

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

By the Axiom of Archimedes:

$\exists z \in \N: z > y$

and so:

$y \in A_z$

That is by definition of union of family:

$y \in \ds \bigcup_{x \mathop \in \R_{>0} } A_x$

So by definition of subset:

$\ds \R_{>0} \subseteq \bigcup_{x \mathop \in \R_{>0} } A_x$

$\Box$

By definition of set equality:

$\ds \bigcup_{x \mathop \in \R_{>0} } A_x = \R_{>0}$

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $5$