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:
- $\ds \bigcup_{x \mathop \in \R_{>0} } B_x = \R_{\ge 0}$
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 $5$