Arbitrary intersection of the open set (-1/n,1/n) where n is a natural number is zero.
Also the arbitrary intersection of the set (-r,r) where r is a positive real numbers is zero.
Could someone please explain, why is it like this ?
Arbitrary intersection of the open set (-1/n,1/n) where n is a natural number is zero.
Also the arbitrary intersection of the set (-r,r) where r is a positive real numbers is zero.
Could someone please explain, why is it like this ?
This says $\displaystyle \bigcap\limits_{n \in {\mathbb{Z}^+ }} {\left( { - {n^{ - 1}},{n^{ - 1}}} \right)} = \left\{ 0 \right\}$.
To see that note that $\displaystyle \forall{n \in {\mathbb{Z}^ + }}\left[ {0\in\left( { - {n^{ - 1}},{n^{ - 1}}} \right) \right]$
To see the next note that if $\displaystyle r>0$ then $\displaystyle \exists n\in\mathbb{Z}^+\left[n^{-1}<r\right]$
Suppose x is a positive number. Then 1/x is positive number and so there exist a positive integer, N, such that N> 1/x. But then x> 1/N and so x is NOT in (-1/N, 1/N) for that N so is not in the intersection.
Suppose x is a negative number. Then -1/x is a positive number and the above applies. But if there is N such that N> -1/x, then x< -1/N and so x is NOT in (-1/N, 1/N).
(The statement, as given, is not true- the intersection is NOT "zero", it is the set whose only member is 0.)
Much the same. Suppose x is a positive number. Then there exist a positive real number r< x. x is not in (-r, r) for that r and so not in the intersection. Suppose x is a negative number. Then -x is a postivie number, so there exist a positive real number r< -x. Then x< -r so is not in (-r, r) and so is not in the intersection.Also the arbitrary intersection of the set (-r,r) where r is a positive real numbers is zero.
(Again, the intersection is NOT "zero"- it is the set whose only member is 0.)
Could someone please explain, why is it like this ?