Families of sets How can I show that 
$$\bigcap_{n\in\Bbb N}\left(0,\frac{1}{n}\right] = \emptyset?$$ 
I understand that I have to make sure the set on the right is a subset of the set on the left, and visa verse. But I don't understand how to do this with the empty set. 
 A: You show that each $x\in\mathbb R$ is not an element of your set.
So, you take an arbitrary $x\in\mathbb R$, and you show that $x\notin \bigcap_{n\in\mathbb N}(0,\frac1n]$.

Hint:
$$x\notin \bigcap_{i\in I} A_i\iff \exists i\in I: x\notin A_i$$
This means that in your case, you have to prove the statement

For all $x\in \mathbb R$, there exists some $n$ such that $x\notin (0,\frac1n]$.

This claim is obviously true for all negative numbers, all for numbers greater than $1$ and for $0$, and it shouldn't be hard to prove the claim for positive numbers between $0$ and $1$, either.
A: Let us prove your claim by using proof by contradiction.
Suppose that $\bigcap_{n\in\Bbb N}\left(0,\frac{1}{n}\right]\neq \emptyset$. Let $x\in\bigcap_{n\in\Bbb N}\left(0,\frac{1}{n}\right]$. Then using the definition of intersection, we get $x\in\left(0,\frac{1}{n}\right]$ for all $n\in\Bbb N$. This means that $x>0$ and $x\leq \frac{1}{n}$ for all $n\in\Bbb N$. But by using the Archimedean Property (applied to $x>0$), we can find $N\in\Bbb N$ such that $\frac{1}{N}<x$. We obtain a contradiction. This proves your claim.
A: You can prove that $$\bigcup_{n\in\mathbb{N}}(1/n,+\infty)=\mathbb{R}_{>0}\setminus\left( \bigcap_{n\in\mathbb{N}}(0,1/n]\right)=\mathbb{R}_{>0}$$
So you need to take $x \in\mathbb{R}_{>0}$ and find $n\in\mathbb{N}$ such that $1/n<x$. But this $n$ is given by Archimedean property so you win.
A: Suppose that $\bigcap_{n\in\Bbb N}\left(0,\frac{1}{n}\right]\neq \emptyset$. Let $x\in\bigcap_{n\in\Bbb N}\left(0,\frac{1}{n}\right]$. So  $x\in (0,1/n] $ for all $n\in \Bbb N$. But we have $x/2 <x$ and $x\notin (x/2, 1/n)$ that is contradiction. Then $\bigcap_{n\in\Bbb N}\left(0,\frac{1}{n}\right]= \emptyset$.
