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:

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

$\Box$

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

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

Otherwise, by the Axiom of Archimedes:

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

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:

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

$\Box$

By definition of set equality:

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

$\blacksquare$

Sources

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