Is my reasoning sound for proving that the intersection over a family of sets is empty? So as a homework problem I was asked to find the intersection of the following family of sets. I was told that my reasoning was invalid as I did not correctly prove that the intersection is the empty set. Is the fact that $\displaystyle \frac{1}{n} \lt \frac{1}{n+1}$ not sufficient to show that there are no common elements over the entire family of sets and thus the intersection contains no elements?
1. Find the union and intersection of each of the following families.
(g) $\displaystyle \mathscr{A} =\{A_{n} :n\in \mathbb{N}\}$, where $\displaystyle A_{n} =\left( 0,\frac{1}{n}\right)$ for each natural number $\displaystyle n$.
Solution.
Since $\displaystyle n\in \mathbb{N}$, and $\displaystyle 0$ is not in any set because it is an open interval, it follows that the quantity $\displaystyle \frac{1}{n} \leq 1$. Therefore, the union of the family of sets $\displaystyle \mathscr{A}$ is $\displaystyle \bigcup _{n\in \mathbb{N}} A_{n} =( 0,1)$.
Since for each set $\displaystyle A_{n}$ the quantity $\displaystyle \frac{1}{n} \lt \frac{1}{n+1}$, and $\displaystyle 0$ is not in any set because it is an open interval, it follows that the intersection of the family of sets is $\displaystyle \bigcap _{n\in \mathbb{N}} A_{n} =\emptyset $, since no two sets have a common element.
 A: Your argument is not valid because it could just as easily be applied to the family of sets $A_n = (0, 1 + 1/n)$.
Here is a correct argument:
$\bigcap\limits_{n \in \mathbb{N}} A_n$ is empty for the following reason: suppose we have some $x \in \bigcap\limits_{n \in \mathbb{N}} A_n$. Then $x \in A_1 = (0, 1/1)$; then $x > 0$. Therefore, $1/x$ is defined. By the Archimedean property of real numbers, there exists a natural number $n > 1/x$. In this case, we see that $1/n < x$, since the reciprocal function is decreasing on $(0, \infty)$. Because $x \in \bigcap\limits_{n \in \mathbb{N}} A_n $, it follows that $x \in A_n = (0, 1/n)$ and therefore $x < 1/n < x$. This is a contradiction.
A: The point you're missing is that for any $x > 0$ there is some $n$ such that $\dfrac{1}{n} \le x$.  The mere fact that $\dfrac{1}{n} > \dfrac{1}{n+1}$ (btw, you wrote $<$, but I guess that was just a typo) is not sufficient.  For example,
you could also say $1 + \dfrac{1}{n} > 1 + \dfrac{1}{n+1}$, but it's not true
that $\displaystyle \bigcap_{n \in N} \left(0, 1 + \frac{1}{n}\right) = \emptyset$.
A: For starters, $\frac1n$ is larger than $\frac1{n+1}$: you have the inequality backwards. But the argument is inadequate even after you correct the inequality. Consider instead the family of intervals $\mathscr{C}=\{C_n:n\in\Bbb Z^+\}$, where
$$C_n=\left(0,\frac1{10}+\frac1n\right)\,.$$
If your argument were valid, it would apply equally well to $\mathscr{C}$, since $$\frac1{10}+\frac1{n+1}<\frac1{10}+\frac1n$$ for each $n\in\Bbb Z^+$, yet
$$\bigcap\mathscr{C}=\bigcap_{n\in\Bbb Z^+}\left(0,\frac1{10}+\frac1n\right)=\left(0,\frac1{10}\right]\ne\varnothing\,.$$
