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

From ProofWiki
Jump to navigation Jump to search

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:

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


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:

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

$\Box$


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:

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

$\Box$


By definition of set equality:

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

$\blacksquare$


Sources