Intersection of Open Intervals of Positive Reals is Empty

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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:

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


Proof

Let $\displaystyle A = \bigcap_{x \mathop \in \R_{> 0} } A_x$.

Aiming for a contradiction, suppose $A \ne \O$.

Then:

$\exists y \in \R_{> 0}: y \in A$

By definition of open interval:

$y \notin \openint 0 y = A_y$

and so by definition of intersection of family:

$y \notin A$

From this contradiction it follows that $A$ has no elements.

That is:

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

$\blacksquare$


Sources