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 ?

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- Sep 22nd 2013, 07:42 AMmrmaaza123Arbitrary intersection of open sets
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 ? - Sep 22nd 2013, 09:51 AMPlatoRe: Arbitrary intersection of open sets
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 $\displaystyle \exists n\in\mathbb{Z}^+\left[n^{-1}<r\right]$ - Sep 22nd 2013, 01:41 PMHallsofIvyRe: Arbitrary intersection of open sets
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.)

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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.)

Quote:

Could someone please explain, why is it like this ?

- Sep 22nd 2013, 06:57 PMmrmaaza123Re: Arbitrary intersection of open sets
Could you please explain this in terms of neighborhoods of x?