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

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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:

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


Proof

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

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:

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

$\Box$


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

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

Otherwise, by the Archimedean Property:

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

and so:

$y \in B_z$

That is by definition of union of family:

$\displaystyle y \in \bigcap_{x \mathop \in \R_{\ge 0} } B_x$

So by definition of subset:

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

$\Box$


By definition of set equality:

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

$\blacksquare$


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