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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.

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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.

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  • $\begingroup$ Looking at the question, it seems the difficulty lies in the methodology. So I upvoted this answer as it emphasises the “how to prove an intersection is empty” part. $\endgroup$ – Michaël Le Barbier Mar 9 at 11:43
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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 $\displaystyle \bigcap_{n \in N} \left(0, 1 + \frac{1}{n}\right) = \emptyset$.

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  • $\begingroup$ Thanks for the good feedback. Yes, the inequality was a typo. I see what you're saying about (0,1+1/n) but the presence of the 1 means that each set contains the interval (0,1] so yes for that interval the intersection would be nonempty. But it would seem to me that each interval (0,1/n) tends to zero as n increases and therefore there couldn't be a common element to all sets in the family. $\endgroup$ – jinks908 Mar 9 at 3:49
  • $\begingroup$ Yes, $\lim_{n \to \infty} 1/n = 0$ implies that for every $x > 0$ there is some $n$ such that $\dfrac{1}{n} \le x$, and that ensures that the intersection is empty. But without writing down something like that, your proof is incomplete. $\endgroup$ – Robert Israel Mar 9 at 4:19
  • $\begingroup$ Thanks for the feedback! $\endgroup$ – jinks908 Mar 9 at 4:23
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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

$$\bigcap\mathscr{C}=\bigcap_{n\in\Bbb Z^+}\left(0,\frac1{10}+\frac1n\right)=\left(0,\frac1{10}\right]\ne\varnothing\,.$$

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  • $\begingroup$ Okay, this is similar to the two arguments above and now I think I understand. I was thinking too specifically about this particular case. But for the reason to be valid, is it correct to say that it would need to apply to any case where the inequality a < b holds, no matter what a and b are? $\endgroup$ – jinks908 Mar 9 at 3:54
  • $\begingroup$ @jinks908: I’m not sure what you mean. In the original problem one proves that the intersection is empty by showing that if $x$ is any real number, there is an $n\in\Bbb Z^+$ such that $x\notin\left(0,\frac1n\right)$, and that basically uses the Archimedean property of $\Bbb R$. $\endgroup$ – Brian M. Scott Mar 9 at 4:05

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